As noted in the figure caption, the chapter-opening image is of the Whirlpool Galaxy, which we examine in the first section of this chapter. Galaxies are as immense as atoms are small, yet the same laws of physics describe both, along with all the rest of nature—an indication of the underlying unity in the universe. The laws of physics are surprisingly few, implying an underlying simplicity to nature’s apparent complexity. In this text, you learn about the laws of physics. Galaxies and atoms may seem far removed from your daily life, but as you begin to explore this broad-ranging subject, you may soon come to realize that physics plays a much larger role in your life than you first thought, no matter your life goals or career choice.
]]>Physics is devoted to the understanding of all natural phenomena. In physics, we try to understand physical phenomena at all scales—from the world of subatomic particles to the entire universe. Despite the breadth of the subject, the various subfields of physics share a common core. The same basic training in physics will prepare you to work in any area of physics and the related areas of science and engineering. In this section, we investigate the scope of physics; the scales of length, mass, and time over which the laws of physics have been shown to be applicable; and the process by which science in general, and physics in particular, operates.
Take another look at the chapter-opening image. The Whirlpool Galaxy contains billions of individual stars as well as huge clouds of gas and dust. Its companion galaxy is also visible to the right. This pair of galaxies lies a staggering billion trillion miles $latex (1.4\times {10}^{21}\text{mi}) $ from our own galaxy (which is called the Milky Way). The stars and planets that make up the Whirlpool Galaxy might seem to be the furthest thing from most people’s everyday lives, but the Whirlpool is a great starting point to think about the forces that hold the universe together. The forces that cause the Whirlpool Galaxy to act as it does are thought to be the same forces we contend with here on Earth, whether we are planning to send a rocket into space or simply planning to raise the walls for a new home. The gravity that causes the stars of the Whirlpool Galaxy to rotate and revolve is thought to be the same as what causes water to flow over hydroelectric dams here on Earth. When you look up at the stars, realize the forces out there are the same as the ones here on Earth. Through a study of physics, you may gain a greater understanding of the interconnectedness of everything we can see and know in this universe.
Think, now, about all the technological devices you use on a regular basis. Computers, smartphones, global positioning systems (GPSs), MP3 players, and satellite radio might come to mind. Then, think about the most exciting modern technologies you have heard about in the news, such as trains that levitate above tracks, “invisibility cloaks” that bend light around them, and microscopic robots that fight cancer cells in our bodies. All these groundbreaking advances, commonplace or unbelievable, rely on the principles of physics. Aside from playing a significant role in technology, professionals such as engineers, pilots, physicians, physical therapists, electricians, and computer programmers apply physics concepts in their daily work. For example, a pilot must understand how wind forces affect a flight path; a physical therapist must understand how the muscles in the body experience forces as they move and bend. As you will learn in this text, the principles of physics are propelling new, exciting technologies, and these principles are applied in a wide range of careers.
The underlying order of nature makes science in general, and physics in particular, interesting and enjoyable to study. For example, what do a bag of chips and a car battery have in common? Both contain energy that can be converted to other forms. The law of conservation of energy (which says that energy can change form but is never lost) ties together such topics as food calories, batteries, heat, light, and watch springs. Understanding this law makes it easier to learn about the various forms energy takes and how they relate to one another. Apparently unrelated topics are connected through broadly applicable physical laws, permitting an understanding beyond just the memorization of lists of facts.Science consists of theories and laws that are the general truths of nature, as well as the body of knowledge they encompass. Scientists are continuously trying to expand this body of knowledge and to perfect the expression of the laws that describe it. Physics, which comes from the Greek phúsis, meaning “nature,” is concerned with describing the interactions of energy, matter, space, and time to uncover the fundamental mechanisms that underlie every phenomenon. This concern for describing the basic phenomena in nature essentially defines the scope of physics.
Physics aims to understand the world around us at the most basic level. It emphasizes the use of a small number of quantitative laws to do this, which can be useful to other fields pushing the performance boundaries of existing technologies. Consider a smartphone (Figure). Physics describes how electricity interacts with the various circuits inside the device. This knowledge helps engineers select the appropriate materials and circuit layout when building a smartphone. Knowledge of the physics underlying these devices is required to shrink their size or increase their processing speed. Or, think about a GPS. Physics describes the relationship between the speed of an object, the distance over which it travels, and the time it takes to travel that distance. When you use a GPS in a vehicle, it relies on physics equations to determine the travel time from one location to another.
Knowledge of physics is useful in everyday situations as well as in nonscientific professions. It can help you understand how microwave ovens work, why metals should not be put into them, and why they might affect pacemakers. Physics allows you to understand the hazards of radiation and to evaluate these hazards rationally and more easily. Physics also explains the reason why a black car radiator helps remove heat in a car engine, and it explains why a white roof helps keep the inside of a house cool. Similarly, the operation of a car’s ignition system as well as the transmission of electrical signals throughout our body’s nervous system are much easier to understand when you think about them in terms of basic physics.
Physics is a key element of many important disciplines and contributes directly to others. Chemistry, for example—since it deals with the interactions of atoms and molecules—has close ties to atomic and molecular physics. Most branches of engineering are concerned with designing new technologies, processes, or structures within the constraints set by the laws of physics. In architecture, physics is at the heart of structural stability and is involved in the acoustics, heating, lighting, and cooling of buildings. Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, and heat transfer within Earth. Some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines.
Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties of cells and their environments. On the macroscopic level, it explains the heat, work, and power associated with the human body and its various organ systems. Physics is involved in medical diagnostics, such as radiographs, magnetic resonance imaging, and ultrasonic blood flow measurements. Medical therapy sometimes involves physics directly; for example, cancer radiotherapy uses ionizing radiation. Physics also explains sensory phenomena, such as how musical instruments make sound, how the eye detects color, and how lasers transmit information.
It is not necessary to study all applications of physics formally. What is most useful is knowing the basic laws of physics and developing skills in the analytical methods for applying them. The study of physics also can improve your problem-solving skills. Furthermore, physics retains the most basic aspects of science, so it is used by all the sciences, and the study of physics makes other sciences easier to understand.
From the discussion so far, it should be clear that to accomplish your goals in any of the various fields within the natural sciences and engineering, a thorough grounding in the laws of physics is necessary. The reason for this is simply that the laws of physics govern everything in the observable universe at all measurable scales of length, mass, and time. Now, that is easy enough to say, but to come to grips with what it really means, we need to get a little bit quantitative. So, before surveying the various scales that physics allows us to explore, let’s first look at the concept of “order of magnitude,” which we use to come to terms with the vast ranges of length, mass, and time that we consider in this text (Figure).
The order of magnitude of a number is the power of 10 that most closely approximates it. Thus, the order of magnitude refers to the scale (or size) of a value. Each power of 10 represents a different order of magnitude. For example, $latex {10}^{1},{10}^{2},{10}^{3}, $ and so forth, are all different orders of magnitude, as are $latex {10}^{0}=1,{10}^{-1},{10}^{-2}, $ and $latex {10}^{-3}. $ To find the order of magnitude of a number, take the base-10 logarithm of the number and round it to the nearest integer, then the order of magnitude of the number is simply the resulting power of 10. For example, the order of magnitude of 800 is 10^{3} because $latex {\text{log}}_{10}800\approx 2.903, $ which rounds to 3. Similarly, the order of magnitude of 450 is 10^{3} because $latex {\text{log}}_{10}450\approx 2.653, $ which rounds to 3 as well. Thus, we say the numbers 800 and 450 are of the same order of magnitude: 10^{3}. However, the order of magnitude of 250 is 10^{2} because $latex {\text{log}}_{10}250\approx 2.397, $ which rounds to 2.
An equivalent but quicker way to find the order of magnitude of a number is first to write it in scientific notation and then check to see whether the first factor is greater than or less than $latex \sqrt{10}={10}^{0.5}\approx 3. $ The idea is that $latex \sqrt{10}={10}^{0.5} $ is halfway between $latex 1={10}^{0} $ and $latex 10={10}^{1} $ on a log base-10 scale. Thus, if the first factor is less than $latex \sqrt{10}, $ then we round it down to 1 and the order of magnitude is simply whatever power of 10 is required to write the number in scientific notation. On the other hand, if the first factor is greater than $latex \sqrt{10}, $ then we round it up to 10 and the order of magnitude is one power of 10 higher than the power needed to write the number in scientific notation. For example, the number 800 can be written in scientific notation as $latex 8\times {10}^{2}. $ Because 8 is bigger than $latex \sqrt{10}\approx 3, $ we say the order of magnitude of 800 is $latex {10}^{2+1}={10}^{3}. $ The number 450 can be written as $latex 4.5\times {10}^{2}, $ so its order of magnitude is also 10^{3} because 4.5 is greater than 3. However, 250 written in scientific notation is $latex 2.5\times {10}^{2} $ and 2.5 is less than 3, so its order of magnitude is $latex {10}^{2}. $
The order of magnitude of a number is designed to be a ballpark estimate for the scale (or size) of its value. It is simply a way of rounding numbers consistently to the nearest power of 10. This makes doing rough mental math with very big and very small numbers easier. For example, the diameter of a hydrogen atom is on the order of 10^{−10} m, whereas the diameter of the Sun is on the order of 10^{9} m, so it would take roughly $latex {10}^{9}\text{/}{10}^{-10}={10}^{19} $ hydrogen atoms to stretch across the diameter of the Sun. This is much easier to do in your head than using the more precise values of $latex 1.06\times {10}^{-10}\text{m} $ for a hydrogen atom diameter and $latex 1.39\times {10}^{9}\text{m} $ for the Sun’s diameter, to find that it would take $latex 1.31\times {10}^{19} $ hydrogen atoms to stretch across the Sun’s diameter. In addition to being easier, the rough estimate is also nearly as informative as the precise calculation.
The vastness of the universe and the breadth over which physics applies are illustrated by the wide range of examples of known lengths, masses, and times (given as orders of magnitude) in Figure. Examining this table will give you a feeling for the range of possible topics in physics and numerical values. A good way to appreciate the vastness of the ranges of values in Figure is to try to answer some simple comparative questions, such as the following:
In studying Figure, take some time to come up with similar questions that interest you and then try answering them. Doing this can breathe some life into almost any table of numbers.
Visit this site to explore interactively the vast range of length scales in our universe. Scroll down and up the scale to view hundreds of organisms and objects, and click on the individual objects to learn more about each one.
How did we come to know the laws governing natural phenomena? What we refer to as the laws of nature are concise descriptions of the universe around us. They are human statements of the underlying laws or rules that all natural processes follow. Such laws are intrinsic to the universe; humans did not create them and cannot change them. We can only discover and understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle, triumph, and disappointment inherent in any creative effort (Figure). The cornerstone of discovering natural laws is observation; scientists must describe the universe as it is, not as we imagine it to be.
A model is a representation of something that is often too difficult (or impossible) to display directly. Although a model is justified by experimental tests, it is only accurate in describing certain aspects of a physical system. An example is the Bohr model of single-electron atoms, in which the electron is pictured as orbiting the nucleus, analogous to the way planets orbit the Sun (Figure). We cannot observe electron orbits directly, but the mental image helps explain some of the observations we can make, such as the emission of light from hot gases (atomic spectra). However, other observations show that the picture in the Bohr model is not really what atoms look like. The model is “wrong,” but is still useful for some purposes. Physicists use models for a variety of purposes. For example, models can help physicists analyze a scenario and perform a calculation or models can be used to represent a situation in the form of a computer simulation. Ultimately, however, the results of these calculations and simulations need to be double-checked by other means—namely, observation and experimentation.
The word theory means something different to scientists than what is often meant when the word is used in everyday conversation. In particular, to a scientist a theory is not the same as a “guess” or an “idea” or even a “hypothesis.” The phrase “it’s just a theory” seems meaningless and silly to scientists because science is founded on the notion of theories. To a scientist, a theory is a testable explanation for patterns in nature supported by scientific evidence and verified multiple times by various groups of researchers. Some theories include models to help visualize phenomena whereas others do not. Newton’s theory of gravity, for example, does not require a model or mental image, because we can observe the objects directly with our own senses. The kinetic theory of gases, on the other hand, is a model in which a gas is viewed as being composed of atoms and molecules. Atoms and molecules are too small to be observed directly with our senses—thus, we picture them mentally to understand what the instruments tell us about the behavior of gases. Although models are meant only to describe certain aspects of a physical system accurately, a theory should describe all aspects of any system that falls within its domain of applicability. In particular, any experimentally testable implication of a theory should be verified. If an experiment ever shows an implication of a theory to be false, then the theory is either thrown out or modified suitably (for example, by limiting its domain of applicability).
A law uses concise language to describe a generalized pattern in nature supported by scientific evidence and repeated experiments. Often, a law can be expressed in the form of a single mathematical equation. Laws and theories are similar in that they are both scientific statements that result from a tested hypothesis and are supported by scientific evidence. However, the designation law is usually reserved for a concise and very general statement that describes phenomena in nature, such as the law that energy is conserved during any process, or Newton’s second law of motion, which relates force (F), mass (m), and acceleration (a) by the simple equation $latex F=ma. $ A theory, in contrast, is a less concise statement of observed behavior. For example, the theory of evolution and the theory of relativity cannot be expressed concisely enough to be considered laws. The biggest difference between a law and a theory is that a theory is much more complex and dynamic. A law describes a single action whereas a theory explains an entire group of related phenomena. Less broadly applicable statements are usually called principles (such as Pascal’s principle, which is applicable only in fluids), but the distinction between laws and principles often is not made carefully.
The models, theories, and laws we devise sometimes imply the existence of objects or phenomena that are as yet unobserved. These predictions are remarkable triumphs and tributes to the power of science. It is the underlying order in the universe that enables scientists to make such spectacular predictions. However, if experimentation does not verify our predictions, then the theory or law is wrong, no matter how elegant or convenient it is. Laws can never be known with absolute certainty because it is impossible to perform every imaginable experiment to confirm a law for every possible scenario. Physicists operate under the assumption that all scientific laws and theories are valid until a counterexample is observed. If a good-quality, verifiable experiment contradicts a well-established law or theory, then the law or theory must be modified or overthrown completely.
The study of science in general, and physics in particular, is an adventure much like the exploration of an uncharted ocean. Discoveries are made; models, theories, and laws are formulated; and the beauty of the physical universe is made more sublime for the insights gained.
Physics is the science concerned with describing the interactions of energy, matter, space, and time to uncover the fundamental mechanisms that underlie every phenomenon.
Some have described physics as a “search for simplicity.” Explain why this might be an appropriate description.
If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assuming both use accepted rules of logic)?
No, neither of these two theories is more valid than the other. Experimentation is the ultimate decider. If experimental evidence does not suggest one theory over the other, then both are equally valid. A given physicist might prefer one theory over another on the grounds that one seems more simple, more natural, or more beautiful than the other, but that physicist would quickly acknowledge that he or she cannot say the other theory is invalid. Rather, he or she would be honest about the fact that more experimental evidence is needed to determine which theory is a better description of nature.
What determines the validity of a theory?
Certain criteria must be satisfied if a measurement or observation is to be believed. Will the criteria necessarily be as strict for an expected result as for an unexpected result?
Probably not. As the saying goes, “Extraordinary claims require extraordinary evidence.”
Can the validity of a model be limited or must it be universally valid? How does this compare with the required validity of a theory or a law?
Find the order of magnitude of the following physical quantities. (a) The mass of Earth’s atmosphere: $latex 5.1\times {10}^{18}\text{kg;} $ (b) The mass of the Moon’s atmosphere: 25,000 kg; (c) The mass of Earth’s hydrosphere: $latex 1.4\times {10}^{21}\text{kg;} $ (d) The mass of Earth: $latex 5.97\times {10}^{24}\text{kg;} $ (e) The mass of the Moon: $latex 7.34\times {10}^{22}\text{kg;} $ (f) The Earth–Moon distance (semimajor axis): $latex 3.84\times {10}^{8}\text{m;} $ (g) The mean Earth–Sun distance: $latex 1.5\times {10}^{11}\text{m;} $ (h) The equatorial radius of Earth: $latex 6.38\times {10}^{6}\text{m;} $ (i) The mass of an electron: $latex 9.11\times {10}^{-31}\text{kg;} $ (j) The mass of a proton: $latex 1.67\times {10}^{-27}\text{kg;} $ (k) The mass of the Sun: $latex 1.99\times {10}^{30}\text{kg.} $
Use the orders of magnitude you found in the previous problem to answer the following questions to within an order of magnitude. (a) How many electrons would it take to equal the mass of a proton? (b) How many Earths would it take to equal the mass of the Sun? (c) How many Earth–Moon distances would it take to cover the distance from Earth to the Sun? (d) How many Moon atmospheres would it take to equal the mass of Earth’s atmosphere? (e) How many moons would it take to equal the mass of Earth? (f) How many protons would it take to equal the mass of the Sun?
a. 10^{3}; b. 10^{5}; c. 10^{2}; d. 10^{15}; e. 10^{2}; f. 10^{57}
For the remaining questions, you need to use Figure to obtain the necessary orders of magnitude of lengths, masses, and times.
Roughly how many heartbeats are there in a lifetime?
A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?
10^{2} generations
Roughly how many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human?
Calculate the approximate number of atoms in a bacterium. Assume the average mass of an atom in the bacterium is 10 times the mass of a proton.
10^{11} atoms
(a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is 10 times the mass of a bacterium. (b) Making the same assumption, how many cells are there in a human?
Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second?
10^{3} nerve impulses/s
About how many floating-point operations can a supercomputer perform each year?
Roughly how many floating-point operations can a supercomputer perform in a human lifetime?
10^{26} floating-point operations per human lifetime
As we saw previously, the range of objects and phenomena studied in physics is immense. From the incredibly short lifetime of a nucleus to the age of Earth, from the tiny sizes of subnuclear particles to the vast distance to the edges of the known universe, from the force exerted by a jumping flea to the force between Earth and the Sun, there are enough factors of 10 to challenge the imagination of even the most experienced scientist. Giving numerical values for physical quantities and equations for physical principles allows us to understand nature much more deeply than qualitative descriptions alone. To comprehend these vast ranges, we must also have accepted units in which to express them. We shall find that even in the potentially mundane discussion of meters, kilograms, and seconds, a profound simplicity of nature appears: all physical quantities can be expressed as combinations of only seven base physical quantities.
We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other measurements. For example, we might define distance and time by specifying methods for measuring them, such as using a meter stick and a stopwatch. Then, we could define average speed by stating that it is calculated as the total distance traveled divided by time of travel.
Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length of a race, which is a physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners). Without standardized units, it would be extremely difficult for scientists to express and compare measured values in a meaningful way (Figure).
Two major systems of units are used in the world: SI units (for the French Système International d’Unités), also known as the metric system, and English units (also known as the customary or imperial system). English units were historically used in nations once ruled by the British Empire and are still widely used in the United States. English units may also be referred to as the foot–pound–second (fps) system, as opposed to the centimeter–gram–second (cgs) system. You may also encounter the term SAE units, named after the Society of Automotive Engineers. Products such as fasteners and automotive tools (for example, wrenches) that are measured in inches rather than metric units are referred to as SAE fasteners or SAE wrenches.
Virtually every other country in the world (except the United States) now uses SI units as the standard. The metric system is also the standard system agreed on by scientists and mathematicians.
In any system of units, the units for some physical quantities must be defined through a measurement process. These are called the base quantities for that system and their units are the system’s base units. All other physical quantities can then be expressed as algebraic combinations of the base quantities. Each of these physical quantities is then known as a derived quantity and each unit is called a derived unit. The choice of base quantities is somewhat arbitrary, as long as they are independent of each other and all other quantities can be derived from them. Typically, the goal is to choose physical quantities that can be measured accurately to a high precision as the base quantities. The reason for this is simple. Since the derived units can be expressed as algebraic combinations of the base units, they can only be as accurate and precise as the base units from which they are derived.
Based on such considerations, the International Standards Organization recommends using seven base quantities, which form the International System of Quantities (ISQ). These are the base quantities used to define the SI base units. Figure lists these seven ISQ base quantities and the corresponding SI base units.
ISQ Base Quantity | SI Base Unit |
---|---|
Length | meter (m) |
Mass | kilogram (kg) |
Time | second (s) |
Electrical current | ampere (A) |
Thermodynamic temperature | kelvin (K) |
Amount of substance | mole (mol) |
Luminous intensity | candela (cd) |
You are probably already familiar with some derived quantities that can be formed from the base quantities in Figure. For example, the geometric concept of area is always calculated as the product of two lengths. Thus, area is a derived quantity that can be expressed in terms of SI base units using square meters $latex (\text{m}\times \text{m}={\text{m}}^{2}). $ Similarly, volume is a derived quantity that can be expressed in cubic meters $latex ({\text{m}}^{3}). $ Speed is length per time; so in terms of SI base units, we could measure it in meters per second (m/s). Volume mass density (or just density) is mass per volume, which is expressed in terms of SI base units such as kilograms per cubic meter (kg/m^{3}). Angles can also be thought of as derived quantities because they can be defined as the ratio of the arc length subtended by two radii of a circle to the radius of the circle. This is how the radian is defined. Depending on your background and interests, you may be able to come up with other derived quantities, such as the mass flow rate (kg/s) or volume flow rate (m^{3}/s) of a fluid, electric charge $latex (\text{A}\cdot \text{s}), $ mass flux density $latex \text{[kg/}({\text{m}}^{2}\cdot \text{s)],} $ and so on. We will see many more examples throughout this text. For now, the point is that every physical quantity can be derived from the seven base quantities in Figure, and the units of every physical quantity can be derived from the seven SI base units.
For the most part, we use SI units in this text. Non-SI units are used in a few applications in which they are in very common use, such as the measurement of temperature in degrees Celsius $latex (^\circ\text{C}), $ the measurement of fluid volume in liters (L), and the measurement of energies of elementary particles in electron-volts (eV). Whenever non-SI units are discussed, they are tied to SI units through conversions. For example, 1 L is $latex {10}^{-3}{\,\text{m}}^{3}. $
Check out a comprehensive source of information on SI units at the National Institute of Standards and Technology (NIST) Reference on Constants, Units, and Uncertainty.
The initial chapters in this textbook are concerned with mechanics, fluids, and waves. In these subjects all pertinent physical quantities can be expressed in terms of the base units of length, mass, and time. Therefore, we now turn to a discussion of these three base units, leaving discussion of the others until they are needed later.
The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more precise. The meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement was improved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum–iridium bar now kept near Paris. By 1960, it had become possible to define the meter even more accurately in terms of the wavelength of light, so it was again redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meter was given its current definition (in part for greater accuracy) as the distance light travels in a vacuum in 1/299,792,458 of a second (Figure). This change came after knowing the speed of light to be exactly 299,792,458 m/s. The length of the meter will change if the speed of light is someday measured with greater accuracy.
The SI unit for mass is the kilogram (abbreviated kg); it is defined to be the mass of a platinum–iridium cylinder kept with the old meter standard at the International Bureau of Weights and Measures near Paris. Exact replicas of the standard kilogram are also kept at the U.S. National Institute of Standards and Technology (NIST), located in Gaithersburg, Maryland, outside of Washington, DC, and at other locations around the world. Scientists at NIST are currently investigating two complementary methods of redefining the kilogram (see Figure). The determination of all other masses can be traced ultimately to a comparison with the standard mass.
SI units are part of the metric system, which is convenient for scientific and engineering calculations because the units are categorized by factors of 10. Figure lists the metric prefixes and symbols used to denote various factors of 10 in SI units. For example, a centimeter is one-hundredth of a meter (in symbols, 1 cm = 10^{–2} m) and a kilometer is a thousand meters (1 km = 10^{3} m). Similarly, a megagram is a million grams (1 Mg = 10^{6} g), a nanosecond is a billionth of a second (1 ns = 10^{–9} s), and a terameter is a trillion meters (1 Tm = 10^{12} m).
Prefix | Symbol | Meaning | Prefix | Symbol | Meaning |
---|---|---|---|---|---|
yotta- | Y | 10^{24} | yocto- | y | 10^{–24} |
zetta- | Z | 10^{21} | zepto- | z | 10^{–21} |
exa- | E | 10^{18} | atto- | a | 10^{–18} |
peta- | P | 10^{15} | femto- | f | 10^{–15} |
tera- | T | 10^{12} | pico- | p | 10^{–12} |
giga- | G | 10^{9} | nano- | n | 10^{–9} |
mega- | M | 10^{6} | micro- | $latex \mu $ | 10^{–6} |
kilo- | k | 10^{3} | milli- | m | 10^{–3} |
hecto- | h | 10^{2} | centi- | c | 10^{–2} |
deka- | da | 10^{1} | deci- | d | 10^{–1} |
The only rule when using metric prefixes is that you cannot “double them up.” For example, if you have measurements in petameters (1 Pm = 10^{15} m), it is not proper to talk about megagigameters, although $latex {10}^{6}\times {10}^{9}={10}^{15}. $ In practice, the only time this becomes a bit confusing is when discussing masses. As we have seen, the base SI unit of mass is the kilogram (kg), but metric prefixes need to be applied to the gram (g), because we are not allowed to “double-up” prefixes. Thus, a thousand kilograms (10^{3} kg) is written as a megagram (1 Mg) since
As we see in the next section, metric systems have the advantage that conversions of units involve only powers of 10. There are 100 cm in 1 m, 1000 m in 1 km, and so on. In nonmetric systems, such as the English system of units, the relationships are not as simple—there are 12 in. in 1 ft, 5280 ft in 1 mi, and so on.
Another advantage of metric systems is that the same unit can be used over extremely large ranges of values simply by scaling it with an appropriate metric prefix. The prefix is chosen by the order of magnitude of physical quantities commonly found in the task at hand. For example, distances in meters are suitable in construction, whereas distances in kilometers are appropriate for air travel, and nanometers are convenient in optical design. With the metric system there is no need to invent new units for particular applications. Instead, we rescale the units with which we are already familiar.
Replacing the k in kilogram with a factor of 10^{3}, we find that
From Figure, we see that 10^{16} is between “peta-” (10^{15}) and “exa-” (10^{18}). If we use the “peta-” prefix, then we find that $latex 1.93\times {10}^{16}\text{g}=1.93\times {10}^{1}\text{Pg}, $ since $latex 16=1+15. $ Alternatively, if we use the “exa-” prefix we find that $latex 1.93\times {10}^{16}\text{g}=1.93\times {10}^{-2}\text{Eg}, $ since $latex 16=-2+18. $ Because the problem asks for the numerical value between one and 1000, we use the “peta-” prefix and the answer is 19.3 Pg.
If this mass arose from a calculation, we would also want to check to determine whether a mass this large makes any sense in the context of the problem. For this, Figure might be helpful.
Restate $latex 4.79\times {10}^{5}\text{kg} $ using a metric prefix such that the resulting number is bigger than one but less than 1000.
$latex 4.79\times {10}^{2} $ Mg or 479 Mg
Identify some advantages of metric units.
Conversions between units require factors of 10 only, which simplifies calculations. Also, the same basic units can be scaled up or down using metric prefixes to sizes appropriate for the problem at hand.
What are the SI base units of length, mass, and time?
What is the difference between a base unit and a derived unit? (b) What is the difference between a base quantity and a derived quantity? (c) What is the difference between a base quantity and a base unit?
a. Base units are defined by a particular process of measuring a base quantity whereas derived units are defined as algebraic combinations of base units. b. A base quantity is chosen by convention and practical considerations. Derived quantities are expressed as algebraic combinations of base quantities. c. A base unit is a standard for expressing the measurement of a base quantity within a particular system of units. So, a measurement of a base quantity could be expressed in terms of a base unit in any system of units using the same base quantities. For example, length is a base quantity in both SI and the English system, but the meter is a base unit in the SI system only.
For each of the following scenarios, refer to Figure and Figure to determine which metric prefix on the meter is most appropriate for each of the following scenarios. (a) You want to tabulate the mean distance from the Sun for each planet in the solar system. (b) You want to compare the sizes of some common viruses to design a mechanical filter capable of blocking the pathogenic ones. (c) You want to list the diameters of all the elements on the periodic table. (d) You want to list the distances to all the stars that have now received any radio broadcasts sent from Earth 10 years ago.
The following times are given using metric prefixes on the base SI unit of time: the second. Rewrite them in scientific notation without the prefix. For example, 47 Ts would be rewritten as $latex 4.7\times {10}^{13}\text{s.} $ (a) 980 Ps; (b) 980 fs; (c) 17 ns; (d) $latex 577\,\mu \text{s}. $
The following times are given in seconds. Use metric prefixes to rewrite them so the numerical value is greater than one but less than 1000. For example, $latex 7.9\times {10}^{-2}\text{s} $ could be written as either 7.9 cs or 79 ms. (a) $latex 9.57\times {10}^{5}\text{s;} $ (b) 0.045 s; (c) $latex 5.5\times {10}^{-7}\text{s;} $ (d) $latex 3.16\times {10}^{7}\text{s.} $
a. 957 ks; b. 4.5 cs or 45 ms; c. 550 ns; d. 31.6 Ms
The following lengths are given using metric prefixes on the base SI unit of length: the meter. Rewrite them in scientific notation without the prefix. For example, 4.2 Pm would be rewritten as $latex 4.2\times {10}^{15}\text{m.} $ (a) 89 Tm; (b) 89 pm; (c) 711 mm; (d) $latex 0.45\,\mu \text{m}\text{.} $
The following lengths are given in meters. Use metric prefixes to rewrite them so the numerical value is bigger than one but less than 1000. For example, $latex 7.9\times {10}^{-2}\text{m} $ could be written either as 7.9 cm or 79 mm. (a) $latex 7.59\times {10}^{7}\text{m;} $ (b) 0.0074 m; (c) $latex 8.8\times {10}^{-11}\text{m;} $ (d) $latex 1.63\times {10}^{13}\text{m.} $
a. 75.9 Mm; b. 7.4 mm; c. 88 pm; d. 16.3 Tm
The following masses are written using metric prefixes on the gram. Rewrite them in scientific notation in terms of the SI base unit of mass: the kilogram. For example, 40 Mg would be written as $latex 4\times {10}^{4}\text{kg.} $ (a) 23 mg; (b) 320 Tg; (c) 42 ng; (d) 7 g; (e) 9 Pg.
The following masses are given in kilograms. Use metric prefixes on the gram to rewrite them so the numerical value is bigger than one but less than 1000. For example, $latex 7\times {10}^{-4}\text{kg} $ could be written as 70 cg or 700 mg. (a) $latex 3.8\times {10}^{-5}\text{kg;} $ (b) $latex 2.3\times {10}^{17}\text{kg;} $ (c) $latex 2.4\times {10}^{-11}\text{kg;} $ (d) $latex 8\times {10}^{15}\text{kg;} $ (e) $latex 4.2\times {10}^{-3}\text{kg.} $
a. 3.8 cg or 38 mg; b. 230 Eg; c. 24 ng; d. 8 Eg e. 4.2 g
It is often necessary to convert from one unit to another. For example, if you are reading a European cookbook, some quantities may be expressed in units of liters and you need to convert them to cups. Or perhaps you are reading walking directions from one location to another and you are interested in how many miles you will be walking. In this case, you may need to convert units of feet or meters to miles.
Let’s consider a simple example of how to convert units. Suppose we want to convert 80 m to kilometers. The first thing to do is to list the units you have and the units to which you want to convert. In this case, we have units in meters and we want to convert to kilometers. Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio that expresses how many of one unit are equal to another unit. For example, there are 12 in. in 1 ft, 1609 m in 1 mi, 100 cm in 1 m, 60 s in 1 min, and so on. Refer to Appendix B for a more complete list of conversion factors. In this case, we know that there are 1000 m in 1 km. Now we can set up our unit conversion. We write the units we have and then multiply them by the conversion factor so the units cancel out, as shown:
Note that the unwanted meter unit cancels, leaving only the desired kilometer unit. You can use this method to convert between any type of unit. Now, the conversion of 80 m to kilometers is simply the use of a metric prefix, as we saw in the preceding section, so we can get the same answer just as easily by noting that
since “kilo-” means 10^{3} (see Figure) and $latex 1=-2+3. $ However, using conversion factors is handy when converting between units that are not metric or when converting between derived units, as the following examples illustrate.
Light travels about 9 Pm in a year. Given that a year is about $latex 3\times {10}^{7}\text{s}, $ what is the speed of light in meters per second?
$latex 3\times {10}^{8}\text{m/s} $
We know from Figure that the diameter of Earth is on the order of 10^{7} m, so the order of magnitude of its surface area is 10^{14} m^{2}. What is that in square kilometers (that is, km^{2})? (Try doing this both by converting 10^{7} m to km and then squaring it and then by converting 10^{14} m^{2} directly to square kilometers. You should get the same answer both ways.)
$latex 1{0}^{8}{\,\text{km}}^{2} $
Unit conversions may not seem very interesting, but not doing them can be costly. One famous example of this situation was seen with the Mars Climate Orbiter. This probe was launched by NASA on December 11, 1998. On September 23, 1999, while attempting to guide the probe into its planned orbit around Mars, NASA lost contact with it. Subsequent investigations showed a piece of software called SM_FORCES (or “small forces”) was recording thruster performance data in the English units of pound-seconds (lb-s). However, other pieces of software that used these values for course corrections expected them to be recorded in the SI units of newton-seconds (N-s), as dictated in the software interface protocols. This error caused the probe to follow a very different trajectory from what NASA thought it was following, which most likely caused the probe either to burn up in the Martian atmosphere or to shoot out into space. This failure to pay attention to unit conversions cost hundreds of millions of dollars, not to mention all the time invested by the scientists and engineers who worked on the project.
Given that 1 lb (pound) is 4.45 N, were the numbers being output by SM_FORCES too big or too small?
The numbers were too small, by a factor of 4.45.
The volume of Earth is on the order of 10^{21} m^{3}. (a) What is this in cubic kilometers (km^{3})? (b) What is it in cubic miles (mi^{3})? (c) What is it in cubic centimeters (cm^{3})?
The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this?
a. 27.8 m/s; b. 62 mi/h
A car is traveling at a speed of 33 m/s. (a) What is its speed in kilometers per hour? (b) Is it exceeding the 90 km/h speed limit?
In SI units, speeds are measured in meters per second (m/s). But, depending on where you live, you’re probably more comfortable of thinking of speeds in terms of either kilometers per hour (km/h) or miles per hour (mi/h). In this problem, you will see that 1 m/s is roughly 4 km/h or 2 mi/h, which is handy to use when developing your physical intuition. More precisely, show that (a) $latex 1.0\,\text{m/s}=3.6\,\text{km/h} $ and (b) $latex 1.0\,\text{m/s}=2.2\,\text{mi/h}. $
a. 3.6 km/h; b. 2.2 mi/h
American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 m = 3.281 ft.)
Soccer fields vary in size. A large soccer field is 115 m long and 85.0 m wide. What is its area in square feet? (Assume that 1 m = 3.281 ft.)
$latex 1.05\times {10}^{5}{\,\text{ft}}^{2} $
What is the height in meters of a person who is 6 ft 1.0 in. tall?
Mount Everest, at 29,028 ft, is the tallest mountain on Earth. What is its height in kilometers? (Assume that 1 m = 3.281 ft.)
8.847 km
The speed of sound is measured to be 342 m/s on a certain day. What is this measurement in kilometers per hour?
Tectonic plates are large segments of Earth’s crust that move slowly. Suppose one such plate has an average speed of 4.0 cm/yr. (a) What distance does it move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?
a. $latex 1.3\times {10}^{-9}\text{m}; $ b. 40 km/My
The average distance between Earth and the Sun is $latex 1.5\times {10}^{11}\text{m.} $ (a) Calculate the average speed of Earth in its orbit (assumed to be circular) in meters per second. (b) What is this speed in miles per hour?
The density of nuclear matter is about 10^{18} kg/m^{3}. Given that 1 mL is equal in volume to cm^{3}, what is the density of nuclear matter in megagrams per microliter (that is, $latex \text{Mg/}\mu \text{L} $)?
$latex {10}^{6}\text{Mg/}\mu \text{L} $
The density of aluminum is 2.7 g/cm^{3}. What is the density in kilograms per cubic meter?
A commonly used unit of mass in the English system is the pound-mass, abbreviated lbm, where 1 lbm = 0.454 kg. What is the density of water in pound-mass per cubic foot?
62.4 lbm/ft^{3}
A furlong is 220 yd. A fortnight is 2 weeks. Convert a speed of one furlong per fortnight to millimeters per second.
It takes $latex 2\pi $ radians (rad) to get around a circle, which is the same as 360°. How many radians are in 1°?
0.017 rad
Light travels a distance of about $latex 3\times {10}^{8}\text{m/s.} $ A light-minute is the distance light travels in 1 min. If the Sun is $latex 1.5\times {10}^{11}\text{m} $ from Earth, how far away is it in light-minutes?
A light-nanosecond is the distance light travels in 1 ns. Convert 1 ft to light-nanoseconds.
1 light-nanosecond
An electron has a mass of $latex 9.11\times {10}^{-31}\text{kg.} $ A proton has a mass of $latex 1.67\times {10}^{-27}\text{kg}\text{.} $ What is the mass of a proton in electron-masses?
A fluid ounce is about 30 mL. What is the volume of a 12 fl-oz can of soda pop in cubic meters?
$latex 3.6\times {10}^{-4}{\text{m}}^{3} $
The dimension of any physical quantity expresses its dependence on the base quantities as a product of symbols (or powers of symbols) representing the base quantities. Figure lists the base quantities and the symbols used for their dimension. For example, a measurement of length is said to have dimension L or L^{1}, a measurement of mass has dimension M or M^{1}, and a measurement of time has dimension T or T^{1}. Like units, dimensions obey the rules of algebra. Thus, area is the product of two lengths and so has dimension L^{2}, or length squared. Similarly, volume is the product of three lengths and has dimension L^{3}, or length cubed. Speed has dimension length over time, L/T or LT^{–1}. Volumetric mass density has dimension M/L^{3} or ML^{–3}, or mass over length cubed. In general, the dimension of any physical quantity can be written as $latex {\text{L}}^{a}{\text{M}}^{b}{\text{T}}^{c}{\text{I}}^{d}{\text{Θ}}^{e}{\text{N}}^{f}{\text{J}}^{g} $ for some powers $latex a,b,c,d,e,f, $ and g. We can write the dimensions of a length in this form with $latex a=1 $ and the remaining six powers all set equal to zero: $latex {\text{L}}^{1}={\text{L}}^{1}{\text{M}}^{0}{\text{T}}^{0}{\text{I}}^{0}{\text{Θ}}^{0}{\text{N}}^{0}{\text{J}}^{0}. $ Any quantity with a dimension that can be written so that all seven powers are zero (that is, its dimension is $latex {\text{L}}^{0}{\text{M}}^{0}{\text{T}}^{0}{\text{I}}^{0}{\text{Θ}}^{0}{\text{N}}^{0}{\text{J}}^{0} $) is called dimensionless (or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists often call dimensionless quantities pure numbers.
Base Quantity | Symbol for Dimension |
---|---|
Length | L |
Mass | M |
Time | T |
Current | I |
Thermodynamic temperature | Θ |
Amount of substance | N |
Luminous intensity | J |
Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity. For example, if $latex r $ is the radius of a cylinder and $latex h $ is its height, then we write $latex [r]=\text{L} $ and $latex [h]=\text{L} $ to indicate the dimensions of the radius and height are both those of length, or L. Similarly, if we use the symbol $latex A $ for the surface area of a cylinder and $latex V $ for its volume, then [A] = L^{2} and [V] = L^{3}. If we use the symbol $latex m $ for the mass of the cylinder and $latex \rho $ for the density of the material from which the cylinder is made, then $latex [m]=\text{M} $ and $latex [\rho ]={\text{ML}}^{-3}. $
The importance of the concept of dimension arises from the fact that any mathematical equation relating physical quantities must be dimensionally consistent, which means the equation must obey the following rules:
If either of these rules is violated, an equation is not dimensionally consistent and cannot possibly be a correct statement of physical law. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of physics, and even to suggest the form that new laws of physics might take. This last use of dimensions is beyond the scope of this text, but is something you will undoubtedly learn later in your academic career.
We know the dimension of area is L^{2}. Now, the dimension of the expression $latex \pi {r}^{2} $ is
since the constant $latex \pi $ is a pure number and the radius $latex r $ is a length. Therefore, $latex \pi {r}^{2} $ has the dimension of area. Similarly, the dimension of the expression $latex 2\pi r $ is
since the constants $latex 2 $ and $latex \pi $ are both dimensionless and the radius $latex r $ is a length. We see that $latex 2\pi r $ has the dimension of length, which means it cannot possibly be an area.
We rule out $latex 2\pi r $ because it is not dimensionally consistent with being an area. We see that $latex \pi {r}^{2} $ is dimensionally consistent with being an area, so if we have to choose between these two expressions, $latex \pi {r}^{2} $ is the one to choose.
Suppose we want the formula for the volume of a sphere. The two expressions commonly mentioned in elementary discussions of spheres are $latex 4\pi {r}^{2} $ and $latex 4\pi {r}^{3}\text{/}3. $ One is the volume of a sphere of radius r and the other is its surface area. Which one is the volume?
$latex 4\pi {r}^{3}\text{/}3 $
The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation is another example of “nonsense.”
Is the equation v = at dimensionally consistent?
yes
One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t, we have that the dimension of the derivative of v with respect to t is just the ratio of the dimension of v over that of t:
Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t:
By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.
A student is trying to remember some formulas from geometry. In what follows, assume $latex A $ is area, $latex V $ is volume, and all other variables are lengths. Determine which formulas are dimensionally consistent. (a) $latex V=\pi {r}^{2}h; $ (b) $latex A=2\pi {r}^{2}+2\pi rh; $ (c) $latex V=0.5bh; $ (d) $latex V=\pi {d}^{2}; $ (e) $latex V=\pi {d}^{3}\text{/}6. $
Consider the physical quantities s, v, a, and t with dimensions $latex [s]=\text{L}, $ $latex [v]={\text{LT}}^{-1}, $ $latex [a]={\text{LT}}^{-2}, $ and $latex [t]=\text{T}. $ Determine whether each of the following equations is dimensionally consistent. (a) $latex {v}^{2}=2as; $ (b) $latex s=v{t}^{2}+0.5a{t}^{2}; $ (c) $latex v=s\text{/}t; $ (d) $latex a=v\text{/}t. $
a. Yes, both terms have dimension L^{2}T^{-2} b. No. c. Yes, both terms have dimension LT^{-1} d. Yes, both terms have dimension LT^{-2}
Consider the physical quantities $latex m, $ $latex s, $ $latex v, $ $latex a, $ and $latex t $ with dimensions [m] = M, [s] = L, [v] = LT^{–1}, [a] = LT^{–2}, and [t] = T. Assuming each of the following equations is dimensionally consistent, find the dimension of the quantity on the left-hand side of the equation: (a) F = ma; (b) K = 0.5mv^{2}; (c) p = mv; (d) W = mas; (e) L = mvr.
Suppose quantity $latex s $ is a length and quantity $latex t $ is a time. Suppose the quantities $latex v $ and $latex a $ are defined by v = ds/dt and a = dv/dt. (a) What is the dimension of v? (b) What is the dimension of the quantity a? What are the dimensions of (c) $latex \int vdt, $ (d) $latex \int adt, $ and (e) da/dt?
a. [v] = LT^{–1}; b. [a] = LT^{–2}; c. $latex [\int vdt]=\text{L;} $ d. $latex [\int adt]={\text{LT}}^{–1}\text{;} $ e. $latex [\frac{da}{dt}]={\text{LT}}^{–3} $
Suppose [V] = L^{3}, $latex [\rho ]={\text{ML}}^{–3}, $ and [t] = T. (a) What is the dimension of $latex \int \rho dV? $ (b) What is the dimension of dV/dt? (c) What is the dimension of $latex \rho (dV\text{/}dt)? $
The arc length formula says the length $latex s $ of arc subtended by angle $latex Ɵ $ in a circle of radius $latex r $ is given by the equation $latex s=rƟ. $ What are the dimensions of (a) s, (b) r, and (c) $latex \text{Ɵ?} $
a. L; b. L; c. L^{0} = 1 (that is, it is dimensionless)
On many occasions, physicists, other scientists, and engineers need to make estimates for a particular quantity. Other terms sometimes used are guesstimates, order-of-magnitude approximations, back-of-the-envelope calculations, or Fermi calculations. (The physicist Enrico Fermi mentioned earlier was famous for his ability to estimate various kinds of data with surprising precision.) Will that piece of equipment fit in the back of the car or do we need to rent a truck? How long will this download take? About how large a current will there be in this circuit when it is turned on? How many houses could a proposed power plant actually power if it is built? Note that estimating does not mean guessing a number or a formula at random. Rather, estimation means using prior experience and sound physical reasoning to arrive at a rough idea of a quantity’s value. Because the process of determining a reliable approximation usually involves the identification of correct physical principles and a good guess about the relevant variables, estimating is very useful in developing physical intuition. Estimates also allow us perform “sanity checks” on calculations or policy proposals by helping us rule out certain scenarios or unrealistic numbers. They allow us to challenge others (as well as ourselves) in our efforts to learn truths about the world.
Many estimates are based on formulas in which the input quantities are known only to a limited precision. As you develop physics problem-solving skills (which are applicable to a wide variety of fields), you also will develop skills at estimating. You develop these skills by thinking more quantitatively and by being willing to take risks. As with any skill, experience helps. Familiarity with dimensions (see Figure) and units (see Figure and Figure), and the scales of base quantities (see Figure) also helps.
To make some progress in estimating, you need to have some definite ideas about how variables may be related. The following strategies may help you in practicing the art of estimation:
We estimate the surface area of Earth (and hence the surface area of Earth’s oceans) to be roughly
Next, using our average depth estimate of $latex D=3\times {10}^{3}\text{m,} $ which was obtained by bounding, we estimate the volume of Earth’s oceans to be
Last, we estimate the mass of the world’s oceans to be
Thus, we estimate that the order of magnitude of the mass of the planet’s oceans is 10^{21} kg.
Figure says the mass of the atmosphere is 10^{19} kg. Assuming the density of the atmosphere is 1 kg/m^{3}, estimate the height of Earth’s atmosphere. Do you think your answer is an underestimate or an overestimate? Explain why.
$latex 3\times {10}^{4}\text{m} $ or 30 km. It is probably an underestimate because the density of the atmosphere decreases with altitude. (In fact, 30 km does not even get us out of the stratosphere.)
For practice estimating relative lengths, areas, and volumes, check out this PhET simulation, titled “Estimation.”
Assuming the human body is made primarily of water, estimate the volume of a person.
Assuming the human body is primarily made of water, estimate the number of molecules in it. (Note that water has a molecular mass of 18 g/mol and there are roughly 10^{24} atoms in a mole.)
Estimate the mass of air in a classroom.
Estimate the number of molecules that make up Earth, assuming an average molecular mass of 30 g/mol. (Note there are on the order of 10^{24} objects per mole.)
10^{51} molecules
Estimate the surface area of a person.
Roughly how many solar systems would it take to tile the disk of the Milky Way?
10^{16} solar systems
(a) Estimate the density of the Moon. (b) Estimate the diameter of the Moon. (c) Given that the Moon subtends at an angle of about half a degree in the sky, estimate its distance from Earth.
The average density of the Sun is on the order 10^{3} kg/m^{3}. (a) Estimate the diameter of the Sun. (b) Given that the Sun subtends at an angle of about half a degree in the sky, estimate its distance from Earth.
Estimate the mass of a virus.
A floating-point operation is a single arithmetic operation such as addition, subtraction, multiplication, or division. (a) Estimate the maximum number of floating-point operations a human being could possibly perform in a lifetime. (b) How long would it take a supercomputer to perform that many floating-point operations?
a. A reasonable estimate might be one operation per second for a total of 10^{9} in a lifetime.; b. about (10^{9})(10^{–17} s) = 10^{–8} s, or about 10 ns
Figure shows two instruments used to measure the mass of an object. The digital scale has mostly replaced the double-pan balance in physics labs because it gives more accurate and precise measurements. But what exactly do we mean by accurate and precise? Aren’t they the same thing? In this section we examine in detail the process of making and reporting a measurement.
Science is based on observation and experiment—that is, on measurements. Accuracy is how close a measurement is to the accepted reference value for that measurement. For example, let’s say we want to measure the length of standard printer paper. The packaging in which we purchased the paper states that it is 11.0 in. long. We then measure the length of the paper three times and obtain the following measurements: 11.1 in., 11.2 in., and 10.9 in. These measurements are quite accurate because they are very close to the reference value of 11.0 in. In contrast, if we had obtained a measurement of 12 in., our measurement would not be very accurate. Notice that the concept of accuracy requires that an accepted reference value be given.
The precision of measurements refers to how close the agreement is between repeated independent measurements (which are repeated under the same conditions). Consider the example of the paper measurements. The precision of the measurements refers to the spread of the measured values. One way to analyze the precision of the measurements is to determine the range, or difference, between the lowest and the highest measured values. In this case, the lowest value was 10.9 in. and the highest value was 11.2 in. Thus, the measured values deviated from each other by, at most, 0.3 in. These measurements were relatively precise because they did not vary too much in value. However, if the measured values had been 10.9 in., 11.1 in., and 11.9 in., then the measurements would not be very precise because there would be significant variation from one measurement to another. Notice that the concept of precision depends only on the actual measurements acquired and does not depend on an accepted reference value.
The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let’s consider an example of a GPS attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull’s-eye target and think of each GPS attempt to locate the restaurant as a black dot. In Figure(a), we see the GPS measurements are spread out far apart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target. This indicates a low-precision, high-accuracy measuring system. However, in Figure(b), the GPS measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high-precision, low-accuracy measuring system.
The precision of a measuring system is related to the uncertainty in the measurements whereas the accuracy is related to the discrepancy from the accepted reference value. Uncertainty is a quantitative measure of how much your measured values deviate from one another. There are many different methods of calculating uncertainty, each of which is appropriate to different situations. Some examples include taking the range (that is, the biggest less the smallest) or finding the standard deviation of the measurements. Discrepancy (or “measurement error”) is the difference between the measured value and a given standard or expected value. If the measurements are not very precise, then the uncertainty of the values is high. If the measurements are not very accurate, then the discrepancy of the values is high.
Recall our example of measuring paper length; we obtained measurements of 11.1 in., 11.2 in., and 10.9 in., and the accepted value was 11.0 in. We might average the three measurements to say our best guess is 11.1 in.; in this case, our discrepancy is 11.1 – 11.0 = 0.1 in., which provides a quantitative measure of accuracy. We might calculate the uncertainty in our best guess by using the range of our measured values: 0.3 in. Then we would say the length of the paper is 11.1 in. plus or minus 0.3 in. The uncertainty in a measurement, A, is often denoted as δA (read “delta A”), so the measurement result would be recorded as A ± δA. Returning to our paper example, the measured length of the paper could be expressed as 11.1 ± 0.3 in. Since the discrepancy of 0.1 in. is less than the uncertainty of 0.3 in., we might say the measured value agrees with the accepted reference value to within experimental uncertainty.
Some factors that contribute to uncertainty in a measurement include the following:
In our example, such factors contributing to the uncertainty could be the smallest division on the ruler is 1/16 in., the person using the ruler has bad eyesight, the ruler is worn down on one end, or one side of the paper is slightly longer than the other. At any rate, the uncertainty in a measurement must be calculated to quantify its precision. If a reference value is known, it makes sense to calculate the discrepancy as well to quantify its accuracy.
Another method of expressing uncertainty is as a percent of the measured value. If a measurement A is expressed with uncertainty δA, the percent uncertainty is defined as
We then determine the average weight of the 5-lb bag of apples is 5.1 ± 0.2 lb. What is the percent uncertainty of the bag’s weight?
Substitute the values into the equation:
SignificanceWe can conclude the average weight of a bag of apples from this store is 5.1 lb ± 4%. Notice the percent uncertainty is dimensionless because the units of weight in $latex \delta A=0.2 $ lb canceled those in A = 5.1 lb when we took the ratio.
A high school track coach has just purchased a new stopwatch. The stopwatch manual states the stopwatch has an uncertainty of ±0.05 s. Runners on the track coach’s team regularly clock 100-m sprints of 11.49 s to 15.01 s. At the school’s last track meet, the first-place sprinter came in at 12.04 s and the second-place sprinter came in at 12.07 s. Will the coach’s new stopwatch be helpful in timing the sprint team? Why or why not?
No, the coach’s new stopwatch will not be helpful. The uncertainty in the stopwatch is too great to differentiate between the sprint times effectively.
Uncertainty exists in anything calculated from measured quantities. For example, the area of a floor calculated from measurements of its length and width has an uncertainty because the length and width have uncertainties. How big is the uncertainty in something you calculate by multiplication or division? If the measurements going into the calculation have small uncertainties (a few percent or less), then the method of adding percents can be used for multiplication or division. This method states the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to make the calculation. For example, if a floor has a length of 4.00 m and a width of 3.00 m, with uncertainties of 2% and 1%, respectively, then the area of the floor is 12.0 m^{2} and has an uncertainty of 3%. (Expressed as an area, this is 0.36 m^{2} [$latex 12.0{\,\text{m}}^{2}\times 0.03 $], which we round to 0.4 m^{2} since the area of the floor is given to a tenth of a square meter.)
An important factor in the precision of measurements involves the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, a standard ruler can measure length to the nearest millimeter whereas a caliper can measure length to the nearest 0.01 mm. The caliper is a more precise measuring tool because it can measure extremely small differences in length. The more precise the measuring tool, the more precise the measurements.
When we express measured values, we can only list as many digits as we measured initially with our measuring tool. For example, if we use a standard ruler to measure the length of a stick, we may measure it to be 36.7 cm. We can’t express this value as 36.71 cm because our measuring tool is not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement. For example, the person measuring the length of a stick with a ruler notices the stick length seems to be somewhere in between 36.6 cm and 36.7 cm, and he or she must estimate the value of the last digit. Using the method of significant figures, the rule is that the last digit written down in a measurement is the first digit with some uncertainty. To determine the number of significant digits in a value, start with the first measured value at the left and count the number of digits through the last digit written on the right. For example, the measured value 36.7 cm has three digits, or three significant figures. Significant figures indicate the precision of the measuring tool used to measure a value.
Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant because they are placeholders that locate the decimal point. There are two significant figures in 0.053. The zeros in 10.053 are not placeholders; they are significant. This number has five significant figures. The zeros in 1300 may or may not be significant, depending on the style of writing numbers. They could mean the number is known to the last digit or they could be placeholders. So 1300 could have two, three, or four significant figures. To avoid this ambiguity, we should write 1300 in scientific notation as $latex 1.3\times {10}^{3}, $ $latex 1.30\times {10}^{3}, $ or $latex 1.300\times {10}^{3}, $ depending on whether it has two, three, or four significant figures. Zeros are significant except when they serve only as placeholders.
When combining measurements with different degrees of precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least-precise measured value. There are two different rules, one for multiplication and division and the other for addition and subtraction.
In this text, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significant figures are used in all worked examples. An answer given to three digits is based on input good to at least three digits, for example. If the input has fewer significant figures, the answer will also have fewer significant figures. Care is also taken that the number of significant figures is reasonable for the situation posed. In some topics, particularly in optics, more accurate numbers are needed and we use more than three significant figures. Finally, if a number is exact, such as the two in the formula for the circumference of a circle, C = 2πr, it does not affect the number of significant figures in a calculation. Likewise, conversion factors such as 100 cm/1 m are considered exact and do not affect the number of significant figures in a calculation.
Percent uncertainty | $latex \text{Percent uncertainty}=\frac{\delta A}{A}\times 100% $ |
(a) What is the relationship between the precision and the uncertainty of a measurement? (b) What is the relationship between the accuracy and the discrepancy of a measurement?
a. Uncertainty is a quantitative measure of precision. b. Discrepancy is a quantitative measure of accuracy.
Consider the equation 4000/400 = 10.0. Assuming the number of significant figures in the answer is correct, what can you say about the number of significant figures in 4000 and 400?
2 kg
A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?
An infant’s pulse rate is measured to be 130 ± 5 beats/min. What is the percent uncertainty in this measurement?
4%
(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 years? (b) In 2.00 years? (c) In 2.000 years?
A can contains 375 mL of soda. How much is left after 308 mL is removed?
67 mL
State how many significant figures are proper in the results of the following calculations: (a) $latex (106.7)(98.2)/(46.210)(1.01); $ (b) $latex {(18.7)}^{2}; $ (c) $latex (1.60\times {10}^{-19})(3712) $
(a) How many significant figures are in the numbers 99 and 100.? (b) If the uncertainty in each number is 1, what is the percent uncertainty in each? (c) Which is a more meaningful way to express the accuracy of these two numbers: significant figures or percent uncertainties?
a. The number 99 has 2 significant figures; 100. has 3 significant figures. b. 1.00%; c. percent uncertainties
(a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h, what is the percent uncertainty? (b) If it has the same percent uncertainty when it reads 60 km/h, what is the range of speeds you could be going?
(a) A person’s blood pressure is measured to be $latex 120±2\,\text{mm Hg}. $ What is its percent uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mm Hg?
A person measures his or her heart rate by counting the number of beats in 30 s. If 40 ± 1 beats are counted in 30.0 ± 0.5 s, what is the heart rate and its uncertainty in beats per minute?
What is the area of a circle 3.102 cm in diameter?
7.557 cm^{2}
Determine the number of significant figures in the following measurements: (a) 0.0009, (b) 15,450.0, (c) 6×10^{3}, (d) 87.990, and (e) 30.42.
Perform the following calculations and express your answer using the correct number of significant digits. (a) A woman has two bags weighing 13.5 lb and one bag with a weight of 10.2 lb. What is the total weight of the bags? (b) The force F on an object is equal to its mass m multiplied by its acceleration a. If a wagon with mass 55 kg accelerates at a rate of 0.0255 m/s^{2}, what is the force on the wagon? (The unit of force is called the newton and it is expressed with the symbol N.)
a. 37.2 lb; because the number of bags is an exact value, it is not considered in the significant figures; b. 1.4 N; because the value 55 kg has only two significant figures, the final value must also contain two significant figures
Problem-solving skills are clearly essential to success in a quantitative course in physics. More important, the ability to apply broad physical principles—usually represented by equations—to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday life.
As you are probably well aware, a certain amount of creativity and insight is required to solve problems. No rigid procedure works every time. Creativity and insight grow with experience. With practice, the basics of problem solving become almost automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and then progressing to the more difficult. After you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.
Although there is no simple step-by-step method that works for every problem, the following three-stage process facilitates problem solving and makes it more meaningful. The three stages are strategy, solution, and significance. This process is used in examples throughout the book. Here, we look at each stage of the process in turn.
Strategy is the beginning stage of solving a problem. The idea is to figure out exactly what the problem is and then develop a strategy for solving it. Some general advice for this stage is as follows:
The solution stage is when you do the math. Substitute the knowns (along with their units) into the appropriate equation and obtain numerical solutions complete with units. That is, do the algebra, calculus, geometry, or arithmetic necessary to find the unknown from the knowns, being sure to carry the units through the calculations. This step is clearly important because it produces the numerical answer, along with its units. Notice, however, that this stage is only one-third of the overall problem-solving process.
After having done the math in the solution stage of problem solving, it is tempting to think you are done. But, always remember that physics is not math. Rather, in doing physics, we use mathematics as a tool to help us understand nature. So, after you obtain a numerical answer, you should always assess its significance:
The three stages of the process for solving physics problems used in this book are as follows:
What information do you need to choose which equation or equations to use to solve a problem?
What should you do after obtaining a numerical answer when solving a problem?
Check to make sure it makes sense and assess its significance.
Consider the equation y = mt +b, where the dimension of y is length and the dimension of t is time, and m and b are constants. What are the dimensions and SI units of (a) m and (b) b?
Consider the equation $latex s={s}_{0}+{v}_{0}t+{a}_{0}{t}^{2}\text{/}2+{j}_{0}{t}^{3}\text{/}6+{S}_{0}{t}^{4}\text{/}24+c{t}^{5}\text{/}120, $ where s is a length and t is a time. What are the dimensions and SI units of (a) $latex {s}_{0}, $ (b) $latex {v}_{0}, $ (c) $latex {a}_{0}, $ (d) $latex {j}_{0}, $ (e) $latex {S}_{0}, $ and (f) c?
(a) A car speedometer has a 5% uncertainty. What is the range of possible speeds when it reads 90 km/h? (b) Convert this range to miles per hour. Note 1 km = 0.6214 mi.
A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the percent uncertainty in the elapsed time. (c) What is the average speed in meters per second? (d) What is the uncertainty in the average speed?
a. 0.059%; b. 0.01%; c. 4.681 m/s; d. 0.07%, 0.003 m/s
a. 0.02%; b. 1×10^{4} lbm
The length and width of a rectangular room are measured to be 3.955 ± 0.005 m and 3.050 ± 0.005 m. Calculate the area of the room and its uncertainty in square meters.
A car engine moves a piston with a circular cross-section of 7.500 ± 0.002 cm in diameter a distance of 3.250 ± 0.001 cm to compress the gas in the cylinder. (a) By what amount is the gas decreased in volume in cubic centimeters? (b) Find the uncertainty in this volume.
a. 143.6 cm^{3}; b. 0.2 cm^{3} or 0.14%
The purpose of this problem is to show the entire concept of dimensional consistency can be summarized by the old saying “You can’t add apples and oranges.” If you have studied power series expansions in a calculus course, you know the standard mathematical functions such as trigonometric functions, logarithms, and exponential functions can be expressed as infinite sums of the form $latex \sum _{n=0}^{\infty }{a}_{n}{x}^{n}={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+{a}_{3}{x}^{3}+\cdots , $ where the $latex {a}_{n} $ are dimensionless constants for all $latex n=0,1,2,\cdots $ and x is the argument of the function. (If you have not studied power series in calculus yet, just trust us.) Use this fact to explain why the requirement that all terms in an equation have the same dimensions is sufficient as a definition of dimensional consistency. That is, it actually implies the arguments of standard mathematical functions must be dimensionless, so it is not really necessary to make this latter condition a separate requirement of the definition of dimensional consistency as we have done in this section.
Since each term in the power series involves the argument raised to a different power, the only way that every term in the power series can have the same dimension is if the argument is dimensionless. To see this explicitly, suppose [x] = L^{a}M^{b}T^{c}. Then, [x^{n}] = [x]^{n} = L^{an}M^{bn}T^{cn}. If we want [x] = [x^{n}], then an = a, bn = b, and cn = c for all n. The only way this can happen is if a = b = c = 0.
Percent uncertainty | $latex \text{Percent uncertainty}=\frac{\delta A}{A}\times 100% $ |
The three stages of the process for solving physics problems used in this book are as follows:
Some have described physics as a “search for simplicity.” Explain why this might be an appropriate description.
If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assuming both use accepted rules of logic)?
What determines the validity of a theory?
Certain criteria must be satisfied if a measurement or observation is to be believed. Will the criteria necessarily be as strict for an expected result as for an unexpected result?
Can the validity of a model be limited or must it be universally valid? How does this compare with the required validity of a theory or a law?
What are the SI base units of length, mass, and time?
What is the difference between a base unit and a derived unit? (b) What is the difference between a base quantity and a derived quantity? (c) What is the difference between a base quantity and a base unit?
For each of the following scenarios, refer to Figure 1.4 and Table 1.2 to determine which metric prefix on the meter is most appropriate for each of the following scenarios. (a) You want to tabulate the mean distance from the Sun for each planet in the solar system. (b) You want to compare the sizes of some common viruses to design a mechanical filter capable of blocking the pathogenic ones. (c) You want to list the diameters of all the elements on the periodic table. (d) You want to list the distances to all the stars that have now received any radio broadcasts sent from Earth 10 years ago.
(a) What is the relationship between the precision and the uncertainty of a measurement? (b) What is the relationship between the accuracy and the discrepancy of a measurement?
What information do you need to choose which equation or equations to use to solve a problem?
Find the order of magnitude of the following physical quantities. (a) The mass of Earth’s atmosphere: 5.1×1018kg;5.1×1018kg; (b) The mass of the Moon’s atmosphere: 25,000 kg; (c) The mass of Earth’s hydrosphere: 1.4×1021kg;1.4×1021kg; (d) The mass of Earth: 5.97×1024kg;5.97×1024kg; (e) The mass of the Moon: 7.34×1022kg;7.34×1022kg; (f) The Earth–Moon distance (semimajor axis): 3.84×108m;3.84×108m;(g) The mean Earth–Sun distance: 1.5×1011m;1.5×1011m; (h) The equatorial radius of Earth: 6.38×106m;6.38×106m; (i) The mass of an electron: 9.11×10−31kg;9.11×10−31kg; (j) The mass of a proton: 1.67×10−27kg;1.67×10−27kg; (k) The mass of the Sun: 1.99×1030kg.1.99×1030kg.
Use the orders of magnitude you found in the previous problem to answer the following questions to within an order of magnitude. (a) How many electrons would it take to equal the mass of a proton? (b) How many Earths would it take to equal the mass of the Sun? (c) How many Earth–Moon distances would it take to cover the distance from Earth to the Sun? (d) How many Moon atmospheres would it take to equal the mass of Earth’s atmosphere? (e) How many moons would it take to equal the mass of Earth? (f) How many protons would it take to equal the mass of the Sun?
For the remaining questions, you need to use Figure 1.4 to obtain the necessary orders of magnitude of lengths, masses, and times.
Roughly how many heartbeats are there in a lifetime?
A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?
Roughly how many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human?
Calculate the approximate number of atoms in a bacterium. Assume the average mass of an atom in the bacterium is 10 times the mass of a proton.
(a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is 10 times the mass of a bacterium. (b) Making the same assumption, how many cells are there in a human?
Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second?
About how many floating-point operations can a supercomputer perform each year?
The following times are given using metric prefixes on the base SI unit of time: the second. Rewrite them in scientific notation without the prefix. For example, 47 Ts would be rewritten as 4.7×1013s.4.7×1013s. (a) 980 Ps; (b) 980 fs; (c) 17 ns; (d) 577μs.577μs.
The following times are given in seconds. Use metric prefixes to rewrite them so the numerical value is greater than one but less than 1000. For example, 7.9×10−2s7.9×10−2s could be written as either 7.9 cs or 79 ms. (a) 9.57×105s;9.57×105s; (b) 0.045 s; (c) 5.5×10−7s;5.5×10−7s; (d) 3.16×107s.3.16×107s.
The following lengths are given using metric prefixes on the base SI unit of length: the meter. Rewrite them in scientific notation without the prefix. For example, 4.2 Pm would be rewritten as 4.2×1015m.4.2×1015m. (a) 89 Tm; (b) 89 pm; (c) 711 mm; (d) 0.45μm.0.45μm.
The following lengths are given in meters. Use metric prefixes to rewrite them so the numerical value is bigger than one but less than 1000. For example, 7.9×10−2m7.9×10−2m could be written either as 7.9 cm or 79 mm. (a) 7.59×107m;7.59×107m; (b) 0.0074 m; (c) 8.8×10−11m;8.8×10−11m; (d) 1.63×1013m.1.63×1013m.
The following masses are written using metric prefixes on the gram. Rewrite them in scientific notation in terms of the SI base unit of mass: the kilogram. For example, 40 Mg would be written as 4×104kg.4×104kg. (a) 23 mg; (b) 320 Tg; (c) 42 ng; (d) 7 g; (e) 9 Pg.
The following masses are given in kilograms. Use metric prefixes on the gram to rewrite them so the numerical value is bigger than one but less than 1000. For example, 7×10−4kg7×10−4kg could be written as 70 cg or 700 mg. (a) 3.8×10−5kg;3.8×10−5kg; (b) 2.3×1017kg;2.3×1017kg; (c) 2.4×10−11kg;2.4×10−11kg; (d) 8×1015kg;8×1015kg; (e) 4.2×10−3kg.4.2×10−3kg.
The volume of Earth is on the order of 10^{21} m^{3}. (a) What is this in cubic kilometers (km^{3})? (b) What is it in cubic miles (mi^{3})? (c) What is it in cubic centimeters (cm^{3})?
The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this?
A car is traveling at a speed of 33 m/s. (a) What is its speed in kilometers per hour? (b) Is it exceeding the 90 km/h speed limit?
In SI units, speeds are measured in meters per second (m/s). But, depending on where you live, you’re probably more comfortable of thinking of speeds in terms of either kilometers per hour (km/h) or miles per hour (mi/h). In this problem, you will see that 1 m/s is roughly 4 km/h or 2 mi/h, which is handy to use when developing your physical intuition. More precisely, show that (a) 1.0m/s=3.6km/h1.0m/s=3.6km/h and (b) 1.0m/s=2.2mi/h.1.0m/s=2.2mi/h.
American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 m = 3.281 ft.)
Soccer fields vary in size. A large soccer field is 115 m long and 85.0 m wide. What is its area in square feet? (Assume that 1 m = 3.281 ft.)
What is the height in meters of a person who is 6 ft 1.0 in. tall?
Mount Everest, at 29,028 ft, is the tallest mountain on Earth. What is its height in kilometers? (Assume that 1 m = 3.281 ft.)
The speed of sound is measured to be 342 m/s on a certain day. What is this measurement in kilometers per hour?
Tectonic plates are large segments of Earth’s crust that move slowly. Suppose one such plate has an average speed of 4.0 cm/yr. (a) What distance does it move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?
The average distance between Earth and the Sun is 1.5×1011m.1.5×1011m. (a) Calculate the average speed of Earth in its orbit (assumed to be circular) in meters per second. (b) What is this speed in miles per hour?
The density of nuclear matter is about 10^{18} kg/m^{3}. Given that 1 mL is equal in volume to cm^{3}, what is the density of nuclear matter in megagrams per microliter (that is, Mg/μLMg/μL)?
The density of aluminum is 2.7 g/cm^{3}. What is the density in kilograms per cubic meter?
A commonly used unit of mass in the English system is the pound-mass, abbreviated lbm, where 1 lbm = 0.454 kg. What is the density of water in pound-mass per cubic foot?
A furlong is 220 yd. A fortnight is 2 weeks. Convert a speed of one furlong per fortnight to millimeters per second.
It takes 2π2π radians (rad) to get around a circle, which is the same as 360°. How many radians are in 1°?
Light travels a distance of about 3×108m/s.3×108m/s. A light-minute is the distance light travels in 1 min. If the Sun is 1.5×1011m1.5×1011m from Earth, how far away is it in light-minutes?
An electron has a mass of 9.11×10−31kg.9.11×10−31kg. A proton has a mass of 1.67×10−27kg.1.67×10−27kg. What is the mass of a proton in electron-masses?
A student is trying to remember some formulas from geometry. In what follows, assume AA is area, VV is volume, and all other variables are lengths. Determine which formulas are dimensionally consistent. (a) V=πr2h;V=πr2h; (b) A=2πr2+2πrh;A=2πr2+2πrh; (c) V=0.5bh;V=0.5bh; (d) V=πd2;V=πd2; (e) V=πd3/6.V=πd3/6.
Consider the physical quantities s, v, a, and t with dimensions [s]=L,[s]=L, [v]=LT−1,[v]=LT−1, [a]=LT−2,[a]=LT−2, and [t]=T.[t]=T. Determine whether each of the following equations is dimensionally consistent. (a) v2=2as;v2=2as; (b) s=vt2+0.5at2;s=vt2+0.5at2; (c) v=s/t;v=s/t; (d) a=v/t.a=v/t.
Consider the physical quantities m,m, s,s, v,v, a,a, and tt with dimensions [m] = M, [s] = L, [v] = LT^{–1}, [a] = LT^{–2}, and [t] = T. Assuming each of the following equations is dimensionally consistent, find the dimension of the quantity on the left-hand side of the equation: (a) F = ma; (b) K = 0.5mv^{2}; (c) p = mv; (d) W = mas; (e) L = mvr.
Suppose quantity ss is a length and quantity tt is a time. Suppose the quantities vv and aa are defined by v = ds/dt and a =dv/dt. (a) What is the dimension of v? (b) What is the dimension of the quantity a? What are the dimensions of (c) ∫vdt,∫vdt, (d) ∫adt,∫adt, and (e) da/dt?
Suppose [V] = L^{3}, [ρ]=ML–3,[ρ]=ML–3, and [t] = T. (a) What is the dimension of ∫ρdV?∫ρdV? (b) What is the dimension of dV/dt? (c) What is the dimension of ρ(dV/dt)?ρ(dV/dt)?
The arc length formula says the length ss of arc subtended by angle ƟƟ in a circle of radius rr is given by the equation s=rƟ.s=rƟ. What are the dimensions of (a) s, (b) r, and (c) Ɵ?Ɵ?
Assuming the human body is made primarily of water, estimate the volume of a person.
Assuming the human body is primarily made of water, estimate the number of molecules in it. (Note that water has a molecular mass of 18 g/mol and there are roughly 10^{24} atoms in a mole.)
Estimate the mass of air in a classroom.
Estimate the number of molecules that make up Earth, assuming an average molecular mass of 30 g/mol. (Note there are on the order of 10^{24} objects per mole.)
Estimate the surface area of a person.
(a) Estimate the density of the Moon. (b) Estimate the diameter of the Moon. (c) Given that the Moon subtends at an angle of about half a degree in the sky, estimate its distance from Earth.
The average density of the Sun is on the order 10^{3} kg/m^{3}. (a) Estimate the diameter of the Sun. (b) Given that the Sun subtends at an angle of about half a degree in the sky, estimate its distance from Earth.
Estimate the mass of a virus.
A floating-point operation is a single arithmetic operation such as addition, subtraction, multiplication, or division. (a) Estimate the maximum number of floating-point operations a human being could possibly perform in a lifetime. (b) How long would it take a supercomputer to perform that many floating-point operations?
Consider the equation 4000/400 = 10.0. Assuming the number of significant figures in the answer is correct, what can you say about the number of significant figures in 4000 and 400?
Suppose your bathroom scale reads your mass as 65 kg with a 3% uncertainty. What is the uncertainty in your mass (in kilograms)?
A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?
An infant’s pulse rate is measured to be 130 ± 5 beats/min. What is the percent uncertainty in this measurement?
(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 years? (b) In 2.00 years? (c) In 2.000 years?
State how many significant figures are proper in the results of the following calculations: (a) (106.7)(98.2)/(46.210)(1.01);(106.7)(98.2)/(46.210)(1.01); (b) (18.7)2;(18.7)2; (c) (1.60×10−19)(3712)(1.60×10−19)(3712)
(a) How many significant figures are in the numbers 99 and 100.? (b) If the uncertainty in each number is 1, what is the percent uncertainty in each? (c) Which is a more meaningful way to express the accuracy of these two numbers: significant figures or percent uncertainties?
(a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h, what is the percent uncertainty? (b) If it has the same percent uncertainty when it reads 60 km/h, what is the range of speeds you could be going?
(a) A person’s blood pressure is measured to be 120±2mm Hg.120±2mm Hg. What is its percent uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mm Hg?
A person measures his or her heart rate by counting the number of beats in 30 s. If 40 ± 1 beats are counted in 30.0 ± 0.5 s, what is the heart rate and its uncertainty in beats per minute?
Determine the number of significant figures in the following measurements: (a) 0.0009, (b) 15,450.0, (c) 6×10^{3}, (d) 87.990, and (e) 30.42.
Perform the following calculations and express your answer using the correct number of significant digits. (a) A woman has two bags weighing 13.5 lb and one bag with a weight of 10.2 lb. What is the total weight of the bags? (b) The force F on an object is equal to its mass m multiplied by its acceleration a. If a wagon with mass 55 kg accelerates at a rate of 0.0255 m/s^{2}, what is the force on the wagon? (The unit of force is called the newton and it is expressed with the symbol N.)
Consider the equation y = mt +b, where the dimension of y is length and the dimension of t is time, and m and b are constants. What are the dimensions and SI units of (a) m and (b) b?
Consider the equation s=s0+v0t+a0t2/2+j0t3/6+S0t4/24+ct5/120,s=s0+v0t+a0t2/2+j0t3/6+S0t4/24+ct5/120, where s is a length and t is a time. What are the dimensions and SI units of (a) s0,s0, (b) v0,v0, (c) a0,a0, (d) j0,j0, (e) S0,S0, and (f) c?
(a) A car speedometer has a 5% uncertainty. What is the range of possible speeds when it reads 90 km/h? (b) Convert this range to miles per hour. Note 1 km = 0.6214 mi.
A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the percent uncertainty in the elapsed time. (c) What is the average speed in meters per second? (d) What is the uncertainty in the average speed?
The sides of a small rectangular box are measured to be 1.80 ± 0.1 cm, 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm long. Calculate its volume and uncertainty in cubic centimeters.
When nonmetric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was used, where 1 lbm = 0.4539 kg. (a) If there is an uncertainty of 0.0001 kg in the pound-mass unit, what is its percent uncertainty? (b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?
The length and width of a rectangular room are measured to be 3.955 ± 0.005 m and 3.050 ± 0.005 m. Calculate the area of the room and its uncertainty in square meters.
A car engine moves a piston with a circular cross-section of 7.500 ± 0.002 cm in diameter a distance of 3.250 ± 0.001 cm to compress the gas in the cylinder. (a) By what amount is the gas decreased in volume in cubic centimeters? (b) Find the uncertainty in this volume.
The first atomic bomb was detonated on July 16, 1945, at the Trinity test site about 200 mi south of Los Alamos. In 1947, the U.S. government declassified a film reel of the explosion. From this film reel, British physicist G. I. Taylor was able to determine the rate at which the radius of the fireball from the blast grew. Using dimensional analysis, he was then able to deduce the amount of energy released in the explosion, which was a closely guarded secret at the time. Because of this, Taylor did not publish his results until 1950. This problem challenges you to recreate this famous calculation. (a) Using keen physical insight developed from years of experience, Taylor decided the radius r of the fireball should depend only on time since the explosion, t, the density of the air, ρ,ρ, and the energy of the initial explosion, E. Thus, he made the educated guess that r=kEaρbtcr=kEaρbtc for some dimensionless constant k and some unknown exponents a, b, and c. Given that [E] = ML^{2}T^{–2}, determine the values of the exponents necessary to make this equation dimensionally consistent. (Hint: Notice the equation implies that k=rE−aρ−bt−ck=rE−aρ−bt−c and that [k]=1.[k]=1.) (b) By analyzing data from high-energy conventional explosives, Taylor found the formula he derived seemed to be valid as long as the constant k had the value 1.03. From the film reel, he was able to determine many values of r and the corresponding values of t. For example, he found that after 25.0 ms, the fireball had a radius of 130.0 m. Use these values, along with an average air density of 1.25 kg/m^{3}, to calculate the initial energy release of the Trinity detonation in joules (J). (Hint: To get energy in joules, you need to make sure all the numbers you substitute in are expressed in terms of SI base units.) (c) The energy released in large explosions is often cited in units of “tons of TNT” (abbreviated “t TNT”), where 1 t TNT is about 4.2 GJ. Convert your answer to (b) into kilotons of TNT (that is, kt TNT). Compare your answer with the quick-and-dirty estimate of 10 kt TNT made by physicist Enrico Fermi shortly after witnessing the explosion from what was thought to be a safe distance. (Reportedly, Fermi made his estimate by dropping some shredded bits of paper right before the remnants of the shock wave hit him and looked to see how far they were carried by it.)
The purpose of this problem is to show the entire concept of dimensional consistency can be summarized by the old saying “You can’t add apples and oranges.” If you have studied power series expansions in a calculus course, you know the standard mathematical functions such as trigonometric functions, logarithms, and exponential functions can be expressed as infinite sums of the form ∑n=0∞anxn=a0+a1x+a2x2+a3x3+⋯,∑n=0∞anxn=a0+a1x+a2x2+a3x3+⋯, where the anan are dimensionless constants for all n=0,1,2,⋯n=0,1,2,⋯ and x is the argument of the function. (If you have not studied power series in calculus yet, just trust us.) Use this fact to explain why the requirement that all terms in an equation have the same dimensions is sufficient as a definition of dimensional consistency. That is, it actually implies the arguments of standard mathematical functions must be dimensionless, so it is not really necessary to make this latter condition a separate requirement of the definition of dimensional consistency as we have done in this section.
Vectors are essential to physics and engineering. Many fundamental physical quantities are vectors, including displacement, velocity, force, and electric and magnetic vector fields. Scalar products of vectors define other fundamental scalar physical quantities, such as energy. Vector products of vectors define still other fundamental vector physical quantities, such as torque and angular momentum. In other words, vectors are a component part of physics in much the same way as sentences are a component part of literature.
In introductory physics, vectors are Euclidean quantities that have geometric representations as arrows in one dimension (in a line), in two dimensions (in a plane), or in three dimensions (in space). They can be added, subtracted, or multiplied. In this chapter, we explore elements of vector algebra for applications in mechanics and in electricity and magnetism. Vector operations also have numerous generalizations in other branches of physics.
]]>Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100 m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of “number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.
Scalar quantities that have the same physical units can be added or subtracted according to the usual rules of algebra for numbers. For example, a class ending 10 min earlier than 50 min lasts $latex 50\,\text{min}-10\,\text{min}=40\,\text{min} $. Similarly, a 60-cal serving of corn followed by a 200-cal serving of donuts gives $latex 60\,\text{cal}+200\,\text{cal}=260\,\text{cal} $ of energy. When we multiply a scalar quantity by a number, we obtain the same scalar quantity but with a larger (or smaller) value. For example, if yesterday’s breakfast had 200 cal of energy and today’s breakfast has four times as much energy as it had yesterday, then today’s breakfast has $latex 4(200\,\text{cal})=800\,\text{cal} $ of energy. Two scalar quantities can also be multiplied or divided by each other to form a derived scalar quantity. For example, if a train covers a distance of 100 km in 1.0 h, its speed is 100.0 km/1.0 h = 27.8 m/s, where the speed is a derived scalar quantity obtained by dividing distance by time.
Many physical quantities, however, cannot be described completely by just a single number of physical units. For example, when the U.S. Coast Guard dispatches a ship or a helicopter for a rescue mission, the rescue team must know not only the distance to the distress signal, but also the direction from which the signal is coming so they can get to its origin as quickly as possible. Physical quantities specified completely by giving a number of units (magnitude) and a direction are called vector quantities. Examples of vector quantities include displacement, velocity, position, force, and torque. In the language of mathematics, physical vector quantities are represented by mathematical objects called vectors (Figure). We can add or subtract two vectors, and we can multiply a vector by a scalar or by another vector, but we cannot divide by a vector. The operation of division by a vector is not defined.
Let’s examine vector algebra using a graphical method to be aware of basic terms and to develop a qualitative understanding. In practice, however, when it comes to solving physics problems, we use analytical methods, which we’ll see in the next section. Analytical methods are more simple computationally and more accurate than graphical methods. From now on, to distinguish between a vector and a scalar quantity, we adopt the common convention that a letter in bold type with an arrow above it denotes a vector, and a letter without an arrow denotes a scalar. For example, a distance of 2.0 km, which is a scalar quantity, is denoted by d = 2.0 km, whereas a displacement of 2.0 km in some direction, which is a vector quantity, is denoted by $latex \mathbf{\overset{\to }{d}} $.
Suppose you tell a friend on a camping trip that you have discovered a terrific fishing hole 6 km from your tent. It is unlikely your friend would be able to find the hole easily unless you also communicate the direction in which it can be found with respect to your campsite. You may say, for example, “Walk about 6 km northeast from my tent.” The key concept here is that you have to give not one but two pieces of information—namely, the distance or magnitude (6 km) and the direction (northeast).
Displacement is a general term used to describe a change in position, such as during a trip from the tent to the fishing hole. Displacement is an example of a vector quantity. If you walk from the tent (location A) to the hole (location B), as shown in Figure, the vector $latex \mathbf{\overset{\to }{D}} $, representing your displacement, is drawn as the arrow that originates at point A and ends at point B. The arrowhead marks the end of the vector. The direction of the displacement vector $latex \mathbf{\overset{\to }{D}} $ is the direction of the arrow. The length of the arrow represents the magnitude D of vector $latex \mathbf{\overset{\to }{D}} $. Here, D = 6 km. Since the magnitude of a vector is its length, which is a positive number, the magnitude is also indicated by placing the absolute value notation around the symbol that denotes the vector; so, we can write equivalently that $latex D\equiv |\mathbf{\overset{\to }{D}}| $. To solve a vector problem graphically, we need to draw the vector $latex \mathbf{\overset{\to }{D}} $ to scale. For example, if we assume 1 unit of distance (1 km) is represented in the drawing by a line segment of length u = 2 cm, then the total displacement in this example is represented by a vector of length $latex d=6u=6(2\,\text{cm})=12\,\text{cm} $, as shown in Figure. Notice that here, to avoid confusion, we used $latex D=6\,\text{km} $ to denote the magnitude of the actual displacement and d = 12 cm to denote the length of its representation in the drawing.
Suppose your friend walks from the campsite at A to the fishing pond at B and then walks back: from the fishing pond at B to the campsite at A. The magnitude of the displacement vector $latex {\mathbf{\overset{\to }{D}}}_{AB} $ from A to B is the same as the magnitude of the displacement vector $latex {\mathbf{\overset{\to }{D}}}_{BA} $ from B to A (it equals 6 km in both cases), so we can write $latex {D}_{AB}={D}_{BA} $. However, vector $latex {\mathbf{\overset{\to }{D}}}_{AB} $ is not equal to vector $latex {\mathbf{\overset{\to }{D}}}_{BA} $ because these two vectors have different directions: $latex {\mathbf{\overset{\to }{D}}}_{AB}\ne {\mathbf{\overset{\to }{D}}}_{BA} $. In Figure, vector $latex {\mathbf{\overset{\to }{D}}}_{BA} $ would be represented by a vector with an origin at point B and an end at point A, indicating vector $latex {\mathbf{\overset{\to }{D}}}_{BA} $ points to the southwest, which is exactly $latex 180^\circ $ opposite to the direction of vector $latex {\mathbf{\overset{\to }{D}}}_{AB} $. We say that vector $latex {\mathbf{\overset{\to }{D}}}_{BA} $ is antiparallel to vector $latex {\mathbf{\overset{\to }{D}}}_{AB} $ and write $latex {\mathbf{\overset{\to }{D}}}_{AB}=\text{−}{\mathbf{\overset{\to }{D}}}_{BA} $, where the minus sign indicates the antiparallel direction.
Two vectors that have identical directions are said to be parallel vectors—meaning, they are parallel to each other. Two parallel vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ are equal, denoted by $latex \mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{B}} $, if and only if they have equal magnitudes $latex |\mathbf{\overset{\to }{A}}|=|\mathbf{\overset{\to }{B}}| $. Two vectors with directions perpendicular to each other are said to be orthogonal vectors. These relations between vectors are illustrated in Figure.
Two motorboats named Alice and Bob are moving on a lake. Given the information about their velocity vectors in each of the following situations, indicate whether their velocity vectors are equal or otherwise. (a) Alice moves north at 6 knots and Bob moves west at 6 knots. (b) Alice moves west at 6 knots and Bob moves west at 3 knots. (c) Alice moves northeast at 6 knots and Bob moves south at 3 knots. (d) Alice moves northeast at 6 knots and Bob moves southwest at 6 knots. (e) Alice moves northeast at 2 knots and Bob moves closer to the shore northeast at 2 knots.
a. not equal because they are orthogonal; b. not equal because they have different magnitudes; c. not equal because they have different magnitudes and directions; d. not equal because they are antiparallel; e. equal.
Vectors can be multiplied by scalars, added to other vectors, or subtracted from other vectors. We can illustrate these vector concepts using an example of the fishing trip seen in Figure.
Suppose your friend departs from point A (the campsite) and walks in the direction to point B (the fishing pond), but, along the way, stops to rest at some point C located three-quarters of the distance between A and B, beginning from point A (Figure(a)). What is his displacement vector $latex {\mathbf{\overset{\to }{D}}}_{AC} $ when he reaches point C? We know that if he walks all the way to B, his displacement vector relative to A is $latex {\mathbf{\overset{\to }{D}}}_{AB} $, which has magnitude $latex {D}_{AB}=6\,\text{km} $ and a direction of northeast. If he walks only a 0.75 fraction of the total distance, maintaining the northeasterly direction, at point C he must be $latex 0.75{D}_{AB}=4.5\,\text{km} $ away from the campsite at A. So, his displacement vector at the rest point C has magnitude $latex {D}_{AC}=4.5\,\text{km}=0.75{D}_{AB} $ and is parallel to the displacement vector $latex {\mathbf{\overset{\to }{D}}}_{AB} $. All of this can be stated succinctly in the form of the following vector equation:
In a vector equation, both sides of the equation are vectors. The previous equation is an example of a vector multiplied by a positive scalar (number) $latex \alpha =0.75 $. The result, $latex {\mathbf{\overset{\to }{D}}}_{AC} $, of such a multiplication is a new vector with a direction parallel to the direction of the original vector $latex {\mathbf{\overset{\to }{D}}}_{AB} $.
In general, when a vector $latex \mathbf{\overset{\to }{A}} $ is multiplied by a positive scalar $latex \alpha $, the result is a new vector $latex \mathbf{\overset{\to }{B}} $ that is parallel to $latex \mathbf{\overset{\to }{A}} $:
The magnitude $latex |\mathbf{\overset{\to }{B}}| $ of this new vector is obtained by multiplying the magnitude $latex |\mathbf{\overset{\to }{A}}| $ of the original vector, as expressed by the scalar equation:
In a scalar equation, both sides of the equation are numbers. Figure is a scalar equation because the magnitudes of vectors are scalar quantities (and positive numbers). If the scalar $latex \alpha $ is negative in the vector equation Figure, then the magnitude $latex |\mathbf{\overset{\to }{B}}| $ of the new vector is still given by Figure, but the direction of the new vector $latex \mathbf{\overset{\to }{B}} $ is antiparallel to the direction of $latex \mathbf{\overset{\to }{A}} $. These principles are illustrated in Figure(a) by two examples where the length of vector $latex \mathbf{\overset{\to }{A}} $ is 1.5 units. When $latex \alpha =2 $, the new vector $latex \mathbf{\overset{\to }{B}}=2\mathbf{\overset{\to }{A}} $ has length $latex B=2A=3.0\,\text{units} $ (twice as long as the original vector) and is parallel to the original vector. When $latex \alpha =-2 $, the new vector $latex \mathbf{\overset{\to }{C}}=-2\mathbf{\overset{\to }{A}} $ has length $latex C=|-2|A=3.0\,\text{units} $ (twice as long as the original vector) and is antiparallel to the original vector.
Now suppose your fishing buddy departs from point A (the campsite), walking in the direction to point B (the fishing hole), but he realizes he lost his tackle box when he stopped to rest at point C (located three-quarters of the distance between A and B, beginning from point A). So, he turns back and retraces his steps in the direction toward the campsite and finds the box lying on the path at some point D only 1.2 km away from point C (see Figure(b)). What is his displacement vector $latex {\mathbf{\overset{\to }{D}}}_{AD} $ when he finds the box at point D? What is his displacement vector $latex {\mathbf{\overset{\to }{D}}}_{DB} $ from point D to the hole? We have already established that at rest point C his displacement vector is $latex {\mathbf{\overset{\to }{D}}}_{AC}=0.75{\mathbf{\overset{\to }{D}}}_{AB} $. Starting at point C, he walks southwest (toward the campsite), which means his new displacement vector $latex {\mathbf{\overset{\to }{D}}}_{CD} $ from point C to point D is antiparallel to $latex {\mathbf{\overset{\to }{D}}}_{AB} $. Its magnitude $latex |{\mathbf{\overset{\to }{D}}}_{CD}| $ is $latex {D}_{CD}=1.2\,\text{km}=0.2{D}_{AB} $, so his second displacement vector is $latex {\mathbf{\overset{\to }{D}}}_{CD}=-0.2{\mathbf{\overset{\to }{D}}}_{AB} $. His total displacement $latex {\mathbf{\overset{\to }{D}}}_{AD} $ relative to the campsite is the vector sum of the two displacement vectors: vector $latex {\mathbf{\overset{\to }{D}}}_{AC} $ (from the campsite to the rest point) and vector $latex {\mathbf{\overset{\to }{D}}}_{CD} $ (from the rest point to the point where he finds his box):
The vector sum of two (or more) vectors is called the resultant vector or, for short, the resultant. When the vectors on the right-hand-side of Figure are known, we can find the resultant $latex {\mathbf{\overset{\to }{D}}}_{AD} $ as follows:
When your friend finally reaches the pond at B, his displacement vector $latex {\mathbf{\overset{\to }{D}}}_{AB} $ from point A is the vector sum of his displacement vector $latex {\mathbf{\overset{\to }{D}}}_{AD} $ from point A to point D and his displacement vector $latex {\mathbf{\overset{\to }{D}}}_{DB} $ from point D to the fishing hole: $latex {\mathbf{\overset{\to }{D}}}_{AB}={\mathbf{\overset{\to }{D}}}_{AD}+{\mathbf{\overset{\to }{D}}}_{DB} $ (see Figure(c)). This means his displacement vector $latex {\mathbf{\overset{\to }{D}}}_{DB} $ is the difference of two vectors:
Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second term in Figure is vector $latex \text{−}{\mathbf{\overset{\to }{D}}}_{AD} $ (which is antiparallel to $latex {\mathbf{\overset{\to }{D}}}_{AD}) $. When we substitute Figure into Figure, we obtain the second displacement vector:
This result means your friend walked $latex {D}_{DB}=0.45{D}_{AB}=0.45(6.0\,\text{km})=2.7\,\text{km} $ from the point where he finds his tackle box to the fishing hole.
When vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ lie along a line (that is, in one dimension), such as in the camping example, their resultant $latex \mathbf{\overset{\to }{R}}=\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}} $ and their difference $latex \mathbf{\overset{\to }{D}}=\mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}} $ both lie along the same direction. We can illustrate the addition or subtraction of vectors by drawing the corresponding vectors to scale in one dimension, as shown in Figure.
To illustrate the resultant when $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ are two parallel vectors, we draw them along one line by placing the origin of one vector at the end of the other vector in head-to-tail fashion (see Figure(b)). The magnitude of this resultant is the sum of their magnitudes: R = A + B. The direction of the resultant is parallel to both vectors. When vector $latex \mathbf{\overset{\to }{A}} $ is antiparallel to vector $latex \mathbf{\overset{\to }{B}} $, we draw them along one line in either head-to-head fashion (Figure(c)) or tail-to-tail fashion. The magnitude of the vector difference, then, is the absolute value $latex D=|A-B| $ of the difference of their magnitudes. The direction of the difference vector $latex \mathbf{\overset{\to }{D}} $ is parallel to the direction of the longer vector.
In general, in one dimension—as well as in higher dimensions, such as in a plane or in space—we can add any number of vectors and we can do so in any order because the addition of vectors is commutative,
Moreover, multiplication by a scalar is distributive:
We used the distributive property in Figure and Figure.
When adding many vectors in one dimension, it is convenient to use the concept of a unit vector. A unit vector, which is denoted by a letter symbol with a hat, such as $latex \mathbf{\hat{u}} $, has a magnitude of one and does not have any physical unit so that $latex |\mathbf{\hat{u}}|\equiv u=1 $. The only role of a unit vector is to specify direction. For example, instead of saying vector $latex {\mathbf{\overset{\to }{D}}}_{AB} $ has a magnitude of 6.0 km and a direction of northeast, we can introduce a unit vector $latex \mathbf{\hat{u}} $ that points to the northeast and say succinctly that $latex {\mathbf{\overset{\to }{D}}}_{AB}=(6.0\,\text{km})\mathbf{\hat{u}} $. Then the southwesterly direction is simply given by the unit vector $latex \text{−}\mathbf{\hat{u}} $. In this way, the displacement of 6.0 km in the southwesterly direction is expressed by the vector
The total displacement $latex \mathbf{\overset{\to }{D}} $ is the resultant of all its displacement vectors.
A cave diver enters a long underwater tunnel. When her displacement with respect to the entry point is 20 m, she accidentally drops her camera, but she doesn’t notice it missing until she is some 6 m farther into the tunnel. She swims back 10 m but cannot find the camera, so she decides to end the dive. How far from the entry point is she? Taking the positive direction out of the tunnel, what is her displacement vector relative to the entry point?
16 m; $latex \mathbf{\overset{\to }{D}}=-16\,\text{m}\mathbf{\hat{u}} $
When vectors lie in a plane—that is, when they are in two dimensions—they can be multiplied by scalars, added to other vectors, or subtracted from other vectors in accordance with the general laws expressed by Figure, Figure, Figure, and Figure. However, the addition rule for two vectors in a plane becomes more complicated than the rule for vector addition in one dimension. We have to use the laws of geometry to construct resultant vectors, followed by trigonometry to find vector magnitudes and directions. This geometric approach is commonly used in navigation (Figure). In this section, we need to have at hand two rulers, a triangle, a protractor, a pencil, and an eraser for drawing vectors to scale by geometric constructions.
For a geometric construction of the sum of two vectors in a plane, we follow the parallelogram rule. Suppose two vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ are at the arbitrary positions shown in Figure. Translate either one of them in parallel to the beginning of the other vector, so that after the translation, both vectors have their origins at the same point. Now, at the end of vector $latex \mathbf{\overset{\to }{A}} $ we draw a line parallel to vector $latex \mathbf{\overset{\to }{B}} $ and at the end of vector $latex \mathbf{\overset{\to }{B}} $ we draw a line parallel to vector $latex \mathbf{\overset{\to }{A}} $ (the dashed lines in Figure). In this way, we obtain a parallelogram. From the origin of the two vectors we draw a diagonal that is the resultant $latex \mathbf{\overset{\to }{R}} $ of the two vectors: $latex \mathbf{\overset{\to }{R}}=\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}} $ (Figure(a)). The other diagonal of this parallelogram is the vector difference of the two vectors $latex \mathbf{\overset{\to }{D}}=\mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}} $, as shown in Figure(b). Notice that the end of the difference vector is placed at the end of vector $latex \mathbf{\overset{\to }{A}} $.
It follows from the parallelogram rule that neither the magnitude of the resultant vector nor the magnitude of the difference vector can be expressed as a simple sum or difference of magnitudes A and B, because the length of a diagonal cannot be expressed as a simple sum of side lengths. When using a geometric construction to find magnitudes $latex |\mathbf{\overset{\to }{R}}| $ and $latex |\mathbf{\overset{\to }{D}}| $, we have to use trigonometry laws for triangles, which may lead to complicated algebra. There are two ways to circumvent this algebraic complexity. One way is to use the method of components, which we examine in the next section. The other way is to draw the vectors to scale, as is done in navigation, and read approximate vector lengths and angles (directions) from the graphs. In this section we examine the second approach.
If we need to add three or more vectors, we repeat the parallelogram rule for the pairs of vectors until we find the resultant of all of the resultants. For three vectors, for example, we first find the resultant of vector 1 and vector 2, and then we find the resultant of this resultant and vector 3. The order in which we select the pairs of vectors does not matter because the operation of vector addition is commutative and associative (see Figure and Figure). Before we state a general rule that follows from repetitive applications of the parallelogram rule, let’s look at the following example.
Suppose you plan a vacation trip in Florida. Departing from Tallahassee, the state capital, you plan to visit your uncle Joe in Jacksonville, see your cousin Vinny in Daytona Beach, stop for a little fun in Orlando, see a circus performance in Tampa, and visit the University of Florida in Gainesville. Your route may be represented by five displacement vectors $latex \mathbf{\overset{\to }{A}}, $ $latex \mathbf{\overset{\to }{B}} $, $latex \mathbf{\overset{\to }{C}} $, $latex \mathbf{\overset{\to }{D}} $, and $latex \mathbf{\overset{\to }{E}} $, which are indicated by the red vectors in Figure. What is your total displacement when you reach Gainesville? The total displacement is the vector sum of all five displacement vectors, which may be found by using the parallelogram rule four times. Alternatively, recall that the displacement vector has its beginning at the initial position (Tallahassee) and its end at the final position (Gainesville), so the total displacement vector can be drawn directly as an arrow connecting Tallahassee with Gainesville (see the green vector in Figure). When we use the parallelogram rule four times, the resultant $latex \mathbf{\overset{\to }{R}} $ we obtain is exactly this green vector connecting Tallahassee with Gainesville: $latex \mathbf{\overset{\to }{R}}=\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}+\mathbf{\overset{\to }{C}}+\mathbf{\overset{\to }{D}}+\mathbf{\overset{\to }{E}} $.
Drawing the resultant vector of many vectors can be generalized by using the following tail-to-head geometric construction. Suppose we want to draw the resultant vector $latex \mathbf{\overset{\to }{R}} $ of four vectors $latex \mathbf{\overset{\to }{A}} $, $latex \mathbf{\overset{\to }{B}} $, $latex \mathbf{\overset{\to }{C}} $, and $latex \mathbf{\overset{\to }{D}} $ (Figure(a)). We select any one of the vectors as the first vector and make a parallel translation of a second vector to a position where the origin (“tail”) of the second vector coincides with the end (“head”) of the first vector. Then, we select a third vector and make a parallel translation of the third vector to a position where the origin of the third vector coincides with the end of the second vector. We repeat this procedure until all the vectors are in a head-to-tail arrangement like the one shown in Figure. We draw the resultant vector $latex \mathbf{\overset{\to }{R}} $ by connecting the origin (“tail”) of the first vector with the end (“head”) of the last vector. The end of the resultant vector is at the end of the last vector. Because the addition of vectors is associative and commutative, we obtain the same resultant vector regardless of which vector we choose to be first, second, third, or fourth in this construction.
For parts (a) and (b), we attach the origin of vector $latex \mathbf{\overset{\to }{B}} $ to the origin of vector $latex \mathbf{\overset{\to }{A}} $, as shown in Figure, and construct a parallelogram. The shorter diagonal of this parallelogram is the sum $latex \mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}} $. The longer of the diagonals is the difference $latex \mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}} $. We use a ruler to measure the lengths of the diagonals, and a protractor to measure the angles with the horizontal. For the resultant $latex \mathbf{\overset{\to }{R}} $, we obtain R = 5.8 cm and $latex {\theta }_{R}\approx 0^\circ $. For the difference $latex \mathbf{\overset{\to }{D}} $, we obtain D = 16.2 cm and $latex {\theta }_{D}=49.3^\circ $, which are shown in Figure.
For (c), we can start with vector $latex -3\mathbf{\overset{\to }{B}} $ and draw the remaining vectors tail-to-head as shown in Figure. In vector addition, the order in which we draw the vectors is unimportant, but drawing the vectors to scale is very important. Next, we draw vector $latex \mathbf{\overset{\to }{S}} $ from the origin of the first vector to the end of the last vector and place the arrowhead at the end of $latex \mathbf{\overset{\to }{S}} $. We use a ruler to measure the length of $latex \mathbf{\overset{\to }{S}} $, and find that its magnitude is
Using the three displacement vectors $latex \mathbf{\overset{\to }{A}} $, $latex \mathbf{\overset{\to }{B}} $, and $latex \mathbf{\overset{\to }{F}} $ in Figure, choose a convenient scale, and use a ruler and a protractor to find vector $latex \mathbf{\overset{\to }{G}} $ given by the vector equation $latex \mathbf{\overset{\to }{G}}=\mathbf{\overset{\to }{A}}+2\mathbf{\overset{\to }{B}}-\mathbf{\overset{\to }{F}} $.
G = 28.2 cm, $latex {\theta }_{G}=291^\circ $
Observe the addition of vectors in a plane by visiting this vector calculator and this Phet simulation.
A weather forecast states the temperature is predicted to be $latex -5\,^\circ\text{C} $ the following day. Is this temperature a vector or a scalar quantity? Explain.
scalar
Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the velocity of a fly, the age of Earth, the boiling point of water, the cost of a book, Earth’s population, or the acceleration of gravity?
Give a specific example of a vector, stating its magnitude, units, and direction.
answers may vary
What do vectors and scalars have in common? How do they differ?
Suppose you add two vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $. What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?
parallel, sum of magnitudes, antiparallel, zero
Is it possible to add a scalar quantity to a vector quantity?
Is it possible for two vectors of different magnitudes to add to zero? Is it possible for three vectors of different magnitudes to add to zero? Explain.
no, yes
Does the odometer in an automobile indicate a scalar or a vector quantity?
When a 10,000-m runner competing on a 400-m track crosses the finish line, what is the runner’s net displacement? Can this displacement be zero? Explain.
zero, yes
A vector has zero magnitude. Is it necessary to specify its direction? Explain.
Can a magnitude of a vector be negative?
no
Can the magnitude of a particle’s displacement be greater that the distance traveled?
If two vectors are equal, what can you say about their components? What can you say about their magnitudes? What can you say about their directions?
equal, equal, the same
If three vectors sum up to zero, what geometric condition do they satisfy?
A scuba diver makes a slow descent into the depths of the ocean. His vertical position with respect to a boat on the surface changes several times. He makes the first stop 9.0 m from the boat but has a problem with equalizing the pressure, so he ascends 3.0 m and then continues descending for another 12.0 m to the second stop. From there, he ascends 4 m and then descends for 18.0 m, ascends again for 7 m and descends again for 24.0 m, where he makes a stop, waiting for his buddy. Assuming the positive direction up to the surface, express his net vertical displacement vector in terms of the unit vector. What is his distance to the boat?
$latex \mathbf{\overset{\to }{h}}=-49\,\text{m}\mathbf{\hat{u}} $, 49 m
In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one side or the other. Two students pull with force 196 N each to the right, four students pull with force 98 N each to the left, five students pull with force 62 N each to the left, three students pull with force 150 N each to the right, and one student pulls with force 250 N to the left. Assuming the positive direction to the right, express the net pull on the knot in terms of the unit vector. How big is the net pull on the knot? In what direction?
Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point and what is the compass direction of a line connecting your starting point to your final position? Use a graphical method.
30.8 m, $latex 35.7^\circ $ west of north
For the vectors given in the following figure, use a graphical method to find the following resultants: (a) $latex \mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}} $, (b) $latex \mathbf{\overset{\to }{C}}+\mathbf{\overset{\to }{B}} $, (c) $latex \mathbf{\overset{\to }{D}}+\mathbf{\overset{\to }{F}} $, (d) $latex \mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}} $, (e) $latex \mathbf{\overset{\to }{D}}-\mathbf{\overset{\to }{F}} $, (f) $latex \mathbf{\overset{\to }{A}}+2\mathbf{\overset{\to }{F}} $, (g); and (h) $latex \mathbf{\overset{\to }{A}}-4\mathbf{\overset{\to }{D}}+2\mathbf{\overset{\to }{F}} $.
A delivery man starts at the post office, drives 40 km north, then 20 km west, then 60 km northeast, and finally 50 km north to stop for lunch. Use a graphical method to find his net displacement vector.
134 km, $latex 80^\circ $
An adventurous dog strays from home, runs three blocks east, two blocks north, one block east, one block north, and two blocks west. Assuming that each block is about 100 m, how far from home and in what direction is the dog? Use a graphical method.
In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day and he is blown along the following directions: 2.50 km and $latex 45.0^\circ $ north of west, then 4.70 km and $latex 60.0^\circ $ south of east, then 1.30 km and $latex 25.0^\circ $ south of west, then 5.10 km straight east, then 1.70 km and $latex 5.00^\circ $ east of north, then 7.20 km and $latex 55.0^\circ $ south of west, and finally 2.80 km and $latex 10.0^\circ $ north of east. Use a graphical method to find the castaway’s final position relative to the island.
7.34 km, $latex 63.5^\circ $ south of east
A small plane flies 40.0 km in a direction $latex 60^\circ $ north of east and then flies 30.0 km in a direction $latex 15^\circ $ north of east. Use a graphical method to find the total distance the plane covers from the starting point and the direction of the path to the final position.
A trapper walks a 5.0-km straight-line distance from his cabin to the lake, as shown in the following figure. Use a graphical method (the parallelogram rule) to determine the trapper’s displacement directly to the east and displacement directly to the north that sum up to his resultant displacement vector. If the trapper walked only in directions east and north, zigzagging his way to the lake, how many kilometers would he have to walk to get to the lake?
3.8 km east, 3.2 km north, 7.0 km
A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks 100 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is $latex 35^\circ $. How wide is the river?
A pedestrian walks 6.0 km east and then 13.0 km north. Use a graphical method to find the pedestrian’s resultant displacement and geographic direction.
14.3 km, $latex 65^\circ $
The magnitudes of two displacement vectors are A = 20 m and B = 6 m. What are the largest and the smallest values of the magnitude of the resultant $latex \mathbf{\overset{\to }{R}}=\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}? $
Vectors are usually described in terms of their components in a coordinate system. Even in everyday life we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction $latex 37^\circ $ north of east.
In a rectangular (Cartesian) xy-coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y). In a similar fashion, a vector $latex \mathbf{\overset{\to }{A}} $ in a plane is described by a pair of its vector coordinates. The x-coordinate of vector $latex \mathbf{\overset{\to }{A}} $ is called its x-component and the y-coordinate of vector $latex \mathbf{\overset{\to }{A}} $ is called its y-component. The vector x-component is a vector denoted by $latex {\mathbf{\overset{\to }{A}}}_{x} $. The vector y-component is a vector denoted by $latex {\mathbf{\overset{\to }{A}}}_{y} $. In the Cartesian system, the x and y vector components of a vector are the orthogonal projections of this vector onto the x- and y-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components:
As illustrated in Figure, vector $latex \mathbf{\overset{\to }{A}} $ is the diagonal of the rectangle where the x-component $latex {\mathbf{\overset{\to }{A}}}_{x} $ is the side parallel to the x-axis and the y-component $latex {\mathbf{\overset{\to }{A}}}_{y} $ is the side parallel to the y-axis. Vector component $latex {\mathbf{\overset{\to }{A}}}_{x} $ is orthogonal to vector component $latex {\mathbf{\overset{\to }{A}}}_{y} $.
It is customary to denote the positive direction on the x-axis by the unit vector $latex \mathbf{\hat{i}} $ and the positive direction on the y-axis by the unit vector $latex \mathbf{\hat{j}} $. Unit vectors of the axes, $latex \mathbf{\hat{i}} $ and $latex \mathbf{\hat{j}} $, define two orthogonal directions in the plane. As shown in Figure, the x- and y- components of a vector can now be written in terms of the unit vectors of the axes:
The vectors $latex {\mathbf{\overset{\to }{A}}}_{x} $ and $latex {\mathbf{\overset{\to }{A}}}_{y} $ defined by Figure are the vector components of vector $latex \mathbf{\overset{\to }{A}} $. The numbers $latex {A}_{x} $ and $latex {A}_{y} $ that define the vector components in Figure are the scalar components of vector $latex \mathbf{\overset{\to }{A}} $. Combining Figure with Figure, we obtain the component form of a vector:
If we know the coordinates $latex b({x}_{b},{y}_{b}) $ of the origin point of a vector (where b stands for “beginning”) and the coordinates $latex e({x}_{e},{y}_{e}) $ of the end point of a vector (where e stands for “end”), we can obtain the scalar components of a vector simply by subtracting the origin point coordinates from the end point coordinates:
The vector x-component $latex {\mathbf{\overset{\to }{D}}}_{x}=-4.0\mathbf{\hat{i}}=4.0(\text{−}\mathbf{\hat{i}}) $ of the displacement vector has the magnitude $latex |{\mathbf{\overset{\to }{D}}}_{x}|=|-4.0||\mathbf{\hat{i}}|=4.0 $ because the magnitude of the unit vector is $latex |\mathbf{\hat{i}}|=1 $. Notice, too, that the direction of the x-component is $latex \text{−}\mathbf{\hat{i}} $, which is antiparallel to the direction of the +x-axis; hence, the x-component vector $latex {\mathbf{\overset{\to }{D}}}_{x} $ points to the left, as shown in Figure. The scalar x-component of vector $latex \mathbf{\overset{\to }{D}} $ is $latex {D}_{x}=-4.0 $.
Similarly, the vector y-component $latex {\mathbf{\overset{\to }{D}}}_{y}=+2.9\mathbf{\hat{j}} $ of the displacement vector has magnitude $latex |{\mathbf{\overset{\to }{D}}}_{y}|=|2.9||\mathbf{\hat{j}}|=\,2.9 $ because the magnitude of the unit vector is $latex |\mathbf{\hat{j}}|=1 $. The direction of the y-component is $latex +\mathbf{\hat{j}} $, which is parallel to the direction of the +y-axis. Therefore, the y-component vector $latex {\mathbf{\overset{\to }{D}}}_{y} $ points up, as seen in Figure. The scalar y-component of vector $latex \mathbf{\overset{\to }{D}} $ is $latex {D}_{y}=+2.9 $. The displacement vector $latex \mathbf{\overset{\to }{D}} $ is the resultant of its two vector components.
The vector component form of the displacement vector Figure tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.
A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.
$latex \mathbf{\overset{\to }{D}}=(-5.0\mathbf{\hat{i}}-3.0\mathbf{\hat{j}})\text{cm} $; the fly moved 5.0 cm to the left and 3.0 cm down from its landing site.
When we know the scalar components $latex {A}_{x} $ and $latex {A}_{y} $ of a vector $latex \mathbf{\overset{\to }{A}} $, we can find its magnitude A and its direction angle $latex {\theta }_{A} $. The direction angle—or direction, for short—is the angle the vector forms with the positive direction on the x-axis. The angle $latex {\theta }_{A} $ is measured in the counterclockwise direction from the +x-axis to the vector (Figure). Because the lengths A, $latex {A}_{x} $, and $latex {A}_{y} $ form a right triangle, they are related by the Pythagorean theorem:
This equation works even if the scalar components of a vector are negative. The direction angle $latex {\theta }_{A} $ of a vector is defined via the tangent function of angle $latex {\theta }_{A} $ in the triangle shown in Figure:
When the vector lies either in the first quadrant or in the fourth quadrant, where component $latex {A}_{x} $ is positive (Figure), the angle $latex \theta $ in Figure is identical to the direction angle $latex {\theta }_{A} $. For vectors in the fourth quadrant, angle $latex \theta $ is negative, which means that for these vectors, direction angle $latex {\theta }_{A} $ is measured clockwise from the positive x-axis. Similarly, for vectors in the second quadrant, angle $latex \theta $ is negative. When the vector lies in either the second or third quadrant, where component $latex {A}_{x} $ is negative, the direction angle is $latex {\theta }_{A}=\theta +180^\circ $ (Figure).
If the displacement vector of a blue fly walking on a sheet of graph paper is $latex \mathbf{\overset{\to }{D}}=(-5.00\mathbf{\hat{i}}-3.00\mathbf{\hat{j}})\text{cm} $, find its magnitude and direction.
5.83 cm, $latex 211^\circ $
In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known.
Let us return to the right triangle in Figure. The quotient of the adjacent side $latex {A}_{x} $ to the hypotenuse A is the cosine function of direction angle $latex {\theta }_{A} $, $latex {A}_{x}\text{/}A=\text{cos}\,{\theta }_{A} $, and the quotient of the opposite side $latex {A}_{y} $ to the hypotenuse A is the sine function of $latex {\theta }_{A} $, $latex {A}_{y}\text{/}A=\text{sin}\,{\theta }_{A} $. When magnitude A and direction $latex {\theta }_{A} $ are known, we can solve these relations for the scalar components:
When calculating vector components with Figure, care must be taken with the angle. The direction angle $latex {\theta }_{A} $ of a vector is the angle measured counterclockwise from the positive direction on the x-axis to the vector. The clockwise measurement gives a negative angle.
On the first leg, the displacement magnitude is $latex {L}_{1}=200.0\,\text{m} $ and the direction is southeast. For direction angle $latex {\theta }_{1} $ we can take either $latex 45^\circ $ measured clockwise from the east direction or $latex 45^\circ+270^\circ $ measured counterclockwise from the east direction. With the first choice, $latex {\theta }_{1}=-45^\circ $. With the second choice, $latex {\theta }_{1}=+315^\circ $. We can use either one of these two angles. The components are
The displacement vector of the first leg is
On the second leg of Trooper’s wanderings, the magnitude of the displacement is $latex {L}_{2}=300.0\,\text{m} $ and the direction is north. The direction angle is $latex {\theta }_{2}=+90^\circ $. We obtain the following results:
On the third leg, the displacement magnitude is $latex {L}_{3}=50.0\,\text{m} $ and the direction is $latex 30^\circ $ west of north. The direction angle measured counterclockwise from the eastern direction is $latex {\theta }_{3}=30^\circ+90^\circ=+120^\circ $. This gives the following answers:
On the fourth leg of the excursion, the displacement magnitude is $latex {L}_{4}=80.0\,\text{m} $ and the direction is south. The direction angle can be taken as either $latex {\theta }_{4}=-90^\circ $ or $latex {\theta }_{4}=+270^\circ $. We obtain
If Trooper runs 20 m west before taking a rest, what is his displacement vector?
$latex \mathbf{\overset{\to }{D}}=(-20\,\text{m})\mathbf{\hat{j}} $
To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system these directions are given by unit vectors $latex \mathbf{\hat{i}} $ and $latex \mathbf{\hat{j}} $ along the x-axis and the y-axis, respectively. The Cartesian coordinate system is very convenient to use in describing displacements and velocities of objects and the forces acting on them. However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually work in the polar coordinate system.
In the polar coordinate system, the location of point P in a plane is given by two polar coordinates (Figure). The first polar coordinate is the radial coordinate r, which is the distance of point P from the origin. The second polar coordinate is an angle $latex \phi $ that the radial vector makes with some chosen direction, usually the positive x-direction. In polar coordinates, angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point P. This radial direction is described by a unit radial vector $latex \mathbf{\hat{r}} $. The second unit vector $latex \mathbf{\hat{t}} $ is a vector orthogonal to the radial direction $latex \mathbf{\hat{r}} $. The positive $latex +\mathbf{\hat{t}} $ direction indicates how the angle $latex \phi $ changes in the counterclockwise direction. In this way, a point P that has coordinates (x, y) in the rectangular system can be described equivalently in the polar coordinate system by the two polar coordinates $latex (r,\phi ) $. Figure is valid for any vector, so we can use it to express the x- and y-coordinates of vector $latex \mathbf{\overset{\to }{r}} $. In this way, we obtain the connection between the polar coordinates and rectangular coordinates of point P:
To specify the location of a point in space, we need three coordinates (x, y, z), where coordinates x and y specify locations in a plane, and coordinate z gives a vertical position above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but three unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the x-axis $latex \mathbf{\hat{i}} $ and the unit vector of the y-axis $latex \mathbf{\hat{j}} $. The third unit vector $latex \mathbf{\hat{k}} $ is the direction of the z-axis (Figure). The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order x-y-z, which is equivalent to the order $latex \mathbf{\hat{i}} $ - $latex \mathbf{\hat{j}} $ - $latex \mathbf{\hat{k}} $, defines the standard right-handed coordinate system (positive orientation).
In three-dimensional space, vector $latex \mathbf{\overset{\to }{A}} $ has three vector components: the x-component $latex {\mathbf{\overset{\to }{A}}}_{x}={A}_{x}\mathbf{\hat{i}} $, which is the part of vector $latex \mathbf{\overset{\to }{A}} $ along the x-axis; the y-component $latex {\mathbf{\overset{\to }{A}}}_{y}={A}_{y}\mathbf{\hat{j}} $, which is the part of $latex \mathbf{\overset{\to }{A}} $ along the y-axis; and the z-component $latex {\mathbf{\overset{\to }{A}}}_{z}={A}_{z}\mathbf{\hat{k}} $, which is the part of the vector along the z-axis. A vector in three-dimensional space is the vector sum of its three vector components (Figure):
If we know the coordinates of its origin $latex b({x}_{b},{y}_{b},{z}_{b}) $ and of its end $latex e({x}_{e},{y}_{e},{z}_{e}) $, its scalar components are obtained by taking their differences: $latex {A}_{x} $ and $latex {A}_{y} $ are given by Figure and the z-component is given by
Magnitude A is obtained by generalizing Figure to three dimensions:
This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in Figure, the diagonal in the xy-plane has length $latex \sqrt{{A}_{x}^{2}+{A}_{y}^{2}} $ and its square adds to the square $latex {A}_{z}^{2} $ to give $latex {A}^{2} $. Note that when the z-component is zero, the vector lies entirely in the xy-plane and its description is reduced to two dimensions.
If the average velocity vector of the drone in the displacement in Figure is $latex \mathbf{\overset{\to }{u}}=(15.0\mathbf{\hat{i}}+31.7\mathbf{\hat{j}}+2.5\mathbf{\hat{k}})\text{m}\text{/}\text{s} $, what is the magnitude of the drone’s velocity vector?
35.1 m/s = 126.4 km/h
Give an example of a nonzero vector that has a component of zero.
a unit vector of the x-axis
Explain why a vector cannot have a component greater than its own magnitude.
If two vectors are equal, what can you say about their components?
They are equal.
If vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ are orthogonal, what is the component of $latex \mathbf{\overset{\to }{B}} $ along the direction of $latex \mathbf{\overset{\to }{A}} $? What is the component of $latex \mathbf{\overset{\to }{A}} $ along the direction of $latex \mathbf{\overset{\to }{B}} $?
If one of the two components of a vector is not zero, can the magnitude of the other vector component of this vector be zero?
yes
If two vectors have the same magnitude, do their components have to be the same?
Assuming the +x-axis is horizontal and points to the right, resolve the vectors given in the following figure to their scalar components and express them in vector component form.
a. $latex \mathbf{\overset{\to }{A}}=+8.66\mathbf{\hat{i}}+5.00\mathbf{\hat{j}} $, b. $latex \mathbf{\overset{\to }{B}}=+30.09\mathbf{\hat{i}}+39.93\mathbf{\hat{j}} $, c. $latex \mathbf{\overset{\to }{C}}=+6.00\mathbf{\hat{i}}-10.39\mathbf{\hat{j}} $, d. $latex \mathbf{\overset{\to }{D}}=-15.97\mathbf{\hat{i}}+12.04\mathbf{\hat{j}} $, f. $latex \mathbf{\overset{\to }{F}}=-17.32\mathbf{\hat{i}}-10.00\mathbf{\hat{j}} $
Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +x-axis is horizontal to the right.
You drive 7.50 km in a straight line in a direction $latex 15^\circ $ east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (b) Show that you still arrive at the same point if the east and north legs are reversed in order. Assume the +x-axis is to the east.
a. 1.94 km, 7.24 km; b. proof
A sledge is being pulled by two horses on a flat terrain. The net force on the sledge can be expressed in the Cartesian coordinate system as vector $latex \mathbf{\overset{\to }{F}}=(-2980.0\mathbf{\hat{i}}+8200.0\mathbf{\hat{j}})\text{N} $, where $latex \mathbf{\hat{i}} $ and $latex \mathbf{\hat{j}} $ denote directions to the east and north, respectively. Find the magnitude and direction of the pull.
A trapper walks a 5.0-km straight-line distance from her cabin to the lake, as shown in the following figure. Determine the east and north components of her displacement vector. How many more kilometers would she have to walk if she walked along the component displacements? What is her displacement vector?
3.8 km east, 3.2 km north, 2.0 km, $latex \mathbf{\overset{\to }{D}}=(3.8\mathbf{\hat{i}}+3.2\mathbf{\hat{j}})\text{km} $
The polar coordinates of a point are $latex 4\pi \text{/}3 $ and 5.50 m. What are its Cartesian coordinates?
Two points in a plane have polar coordinates $latex {P}_{1}(2.500\,\text{m},\pi \text{/}6) $ and $latex {P}_{2}(3.800\,\text{m},2\pi \text{/}3) $. Determine their Cartesian coordinates and the distance between them in the Cartesian coordinate system. Round the distance to a nearest centimeter.
$latex {P}_{1}(2.165\,\text{m},1.250\,\text{m}) $, $latex {P}_{2}(-1.900\,\text{m},3.290\,\text{m}) $, 5.27 m
A chameleon is resting quietly on a lanai screen, waiting for an insect to come by. Assume the origin of a Cartesian coordinate system at the lower left-hand corner of the screen and the horizontal direction to the right as the +x-direction. If its coordinates are (2.000 m, 1.000 m), (a) how far is it from the corner of the screen? (b) What is its location in polar coordinates?
Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar coordinates.
8.60 m, $latex A(2\sqrt{5}\,\text{m},0.647\pi ) $, $latex B(3\sqrt{2}\,\text{m},0.75\pi ) $
A fly enters through an open window and zooms around the room. In a Cartesian coordinate system with three axes along three edges of the room, the fly changes its position from point b(4.0 m, 1.5 m, 2.5 m) to point e(1.0 m, 4.5 m, 0.5 m). Find the scalar components of the fly’s displacement vector and express its displacement vector in vector component form. What is its magnitude?
Area
$latex 1\,{\text{cm}}^{2}=0.155\,{\text{in.}}^{2} $
$latex 1\,{\text{m}}^{2}={10}^{4}\,{\text{cm}}^{2}=10.76\,{\text{ft}}^{2} $
$latex 1\,{\text{in.}}^{2}=6.452\,{\text{cm}}^{2} $
$latex 1\,{\text{ft}}^{2}=144\,{\text{in.}}^{2}=0.0929\,{\text{m}}^{2} $
Volume
$latex 1\,\text{liter}=1000\,{\text{cm}}^{3}={10}^{-3}\,{\text{m}}^{3}=0.03531\,{\text{ft}}^{3}=61.02\,{\text{in.}}^{3} $
$latex 1\,{\text{ft}}^{3}=0.02832\,{\text{m}}^{3}=28.32\,\text{liters}=7.477\,\text{gallons} $
$latex 1\,\text{gallon}=3.788\,\text{liters} $
s | min | h | day | yr | |
---|---|---|---|---|---|
1 second | 1 | $latex 1.667\times {10}^{-2} $ | $latex 2.778\times {10}^{-4} $ | $latex 1.157\times {10}^{-5} $ | $latex 3.169\times {10}^{-8} $ |
1 minute | 60 | 1 | $latex 1.667\times {10}^{-2} $ | $latex 6.944\times {10}^{-4} $ | $latex 1.901\times {10}^{-6} $ |
1 hour | 3600 | 60 | 1 | $latex 4.167\times {10}^{-2} $ | $latex 1.141\times {10}^{-4} $ |
1 day | $latex 8.640\times {10}^{4} $ | 1440 | 24 | 1 | $latex 2.738\times {10}^{-3} $ |
1 year | $latex 3.156\times {10}^{7} $ | $latex 5.259\times {10}^{5} $ | $latex 8.766\times {10}^{3} $ | 365.25 | 1 |
m/s | cm/s | ft/s | mi/h | |
---|---|---|---|---|
1 meter/second | 1 | 10^{2} | 3.281 | 2.237 |
1 centimeter/second | 10^{−2} | 1 | $latex 3.281\times {10}^{-2} $ | $latex 2.237\times {10}^{-2} $ |
1 foot/second | 0.3048 | 30.48 | 1 | 0.6818 |
1 mile/hour | 0.4470 | 44.70 | 1.467 | 1 |
Acceleration
$latex 1\,{\text{m/s}}^{2}=100\,{\text{cm/s}}^{2}=3.281\,{\text{ft/s}}^{2} $
$latex 1\,{\text{cm/s}}^{2}=0.01\,{\text{m/s}}^{2}=0.03281\,{\text{ft/s}}^{2} $
$latex 1\,{\text{ft/s}}^{2}=0.3048\,{\text{m/s}}^{2}=30.48\,{\text{cm/s}}^{2} $
$latex 1\,\text{mi/h}\cdot \text{s}=1.467\,{\text{ft/s}}^{2} $
kg | g | slug | u | |
---|---|---|---|---|
1 kilogram | 1 | 10^{3} | $latex 6.852\times {10}^{-2} $ | $latex 6.024\times {10}^{26} $ |
1 gram | 10^{−3} | 1 | $latex 6.852\times {10}^{-5} $ | $latex 6.024\times {10}^{23} $ |
1 slug | 14.59 | $latex 1.459\times {10}^{4} $ | 1 | $latex 8.789\times {10}^{27} $ |
1 atomic mass unit | $latex 1.661\times {10}^{-27} $ | $latex 1.661\times {10}^{-24} $ | $latex 1.138\times {10}^{-28} $ | 1 |
1 metric ton | 1000 |
N | dyne | lb | |
---|---|---|---|
$latex 1 $ newton | 1 | 10^{5} | 0.2248 |
$latex 1 $ dyne | 10^{−5} | 1 | $latex 2.248\times {10}^{-6} $ |
$latex 1 $ pound | 4.448 | $latex 4.448\times {10}^{5} $ | 1 |
Pa | dyne/cm^{2} | atm | cmHg | lb/in.^{2} | |
---|---|---|---|---|---|
*Where the acceleration due to gravity is $latex 9.80665\,{\text{m/s}}^{2} $ and the temperature is $latex 0^\circ\text{C} $ | |||||
1 pascal | 1 | 10 | $latex 9.869\times {10}^{-6} $ | $latex 7.501\times {10}^{-4} $ | $latex 1.450\times {10}^{-4} $ |
1 dyne/centimeter^{2} | 10^{−1} | 1 | $latex 9.869\times {10}^{-7} $ | $latex 7.501\times {10}^{-5} $ | $latex 1.450\times {10}^{-5} $ |
1 atmosphere | $latex 1.013\times {10}^{5} $ | $latex 1.013\times {10}^{6} $ | 1 | 76 | 14.70 |
1 centimeter mercury* | $latex 1.333\times {10}^{3} $ | $latex 1.333\times {10}^{4} $ | $latex 1.316\times {10}^{-2} $ | 1 | 0.1934 |
1 pound/inch^{2} | $latex 6.895\times {10}^{3} $ | $latex 6.895\times {10}^{4} $ | $latex 6.805\times {10}^{-2} $ | 5.171 | 1 |
1 bar | 10^{5} | ||||
1 torr | 1 (mmHg) |
J | erg | ft.lb | |
---|---|---|---|
1 joule | 1 | 10^{7} | 0.7376 |
1 erg | 10^{−7} | 1 | $latex 7.376\times {10}^{-8} $ |
1 foot-pound | 1.356 | $latex 1.356\times {10}^{7} $ | 1 |
1 electron-volt | $latex 1.602\times {10}^{-19} $ | $latex 1.602\times {10}^{\text{-12}} $ | $latex 1.182\times {10}^{-19} $ |
1 calorie | 4.186 | $latex 4.186\times {10}^{7} $ | 3.088 |
1 British thermal unit | $latex 1.055\times {10}^{3} $ | $latex 1.055\times {10}^{10} $ | $latex 7.779\times {10}^{2} $ |
1 kilowatt-hour | $latex 3.600\times {10}^{6} $ | ||
eV | cal | Btu | |
1 joule | $latex 6.242\times {10}^{18} $ | 0.2389 | $latex 9.481\times {10}^{-4} $ |
1 erg | $latex 6.242\times {10}^{11} $ | $latex 2.389\times {10}^{-8} $ | $latex 9.481\times {10}^{-11} $ |
1 foot-pound | $latex 8.464\times {10}^{18} $ | 0.3239 | $latex 1.285\times {10}^{-3} $ |
1 electron-volt | 1 | $latex 3.827\times {10}^{-20} $ | $latex 1.519\times {10}^{-22} $ |
1 calorie | $latex 2.613\times {10}^{19} $ | 1 | $latex 3.968\times {10}^{-3} $ |
1 British thermal unit | $latex 6.585\times {10}^{21} $ | $latex 2.520\times {10}^{2} $ | 1 |
Power
$latex 1\,\text{W}=1\,\text{J/s} $
$latex 1\,\text{hp}=746\,\text{W}=550\,\text{ft}\cdot \text{lb/s} $
$latex 1\,\text{Btu/h}=0.293\,\text{W} $
Angle
$latex 1\,\text{rad}=57.30^\circ=180^\circ\text{/}\text{π} $
$latex 1^\circ=0.01745\,\text{rad}=\text{π}\text{/}180\,\text{rad} $
$latex 1\,\text{revolution}=360^\circ=2\text{π}\,\text{rad} $
$latex 1\,\text{rev/min}(\text{rpm})=0.1047\,\text{rad/s} $
]]>Useful combinations of constants for calculations:
$latex hc=12,400\,\text{eV}\cdot Å=1240\,\text{eV}\cdot \text{nm}=1240\,\text{MeV}\cdot \text{fm} $
$latex \hslash c=1973\,\text{eV}\cdot Å=197.3\,\text{eV}\cdot \text{nm}=197.3\,\text{MeV}\cdot \text{fm} $
$latex {k}_{e}{e}^{2}=14.40\,\text{eV}\cdot Å=1.440\,\text{eV}\cdot \text{nm}=1.440\,\text{MeV}\cdot \text{fm} $
$latex {k}_{\text{B}}T=0.02585\,\text{eV at}\,T=300\,\text{K} $
]]>Other Data:
Mass of Earth: $latex 5.97\times {10}^{24}\,\text{kg} $
Mass of the Moon: $latex 7.36\times {10}^{22}\,\text{kg} $
Mass of the Sun: $latex 1.99\times {10}^{30}\,\text{kg} $
]]>If $latex a{x}^{2}+bx+c=0, $ then $latex x=\frac{\text{−}b±\sqrt{{b}^{2}-4ac}}{2a} $
Triangle of base $latex b $ and height $latex h$ | Area $latex =\frac{1}{2}bh $ | |
---|---|---|
Circle of radius $latex r$ | Circumference $latex =2\pi r $ | Area $latex =\pi {r}^{2} $ |
Sphere of radius $latex r$ | Surface area $latex =4\pi {r}^{2} $ | Volume $latex =\frac{4}{3}\pi {r}^{3} $ |
Cylinder of radius $latex r $ and height $latex h$ | Area of curved surface $latex =2\pi rh $ | Volume $latex =\pi {r}^{2}h $ |
Trigonometry
Trigonometric Identities
Triangles
Series expansions
Derivatives
Integrals
Vectors can be added together and multiplied by scalars. Vector addition is associative (Figure) and commutative (Figure), and vector multiplication by a sum of scalars is distributive (Figure). Also, scalar multiplication by a sum of vectors is distributive:
In this equation, $latex \alpha $ is any number (a scalar). For example, a vector antiparallel to vector $latex \mathbf{\overset{\to }{A}}={A}_{x}\mathbf{\hat{i}}+{A}_{y}\mathbf{\hat{j}}+{A}_{z}\mathbf{\hat{k}} $ can be expressed simply by multiplying $latex \mathbf{\overset{\to }{A}} $ by the scalar $latex \alpha =-1 $:
The generalization of the number zero to vector algebra is called the null vector, denoted by $latex \mathbf{\overset{\to }{0}} $. All components of the null vector are zero, $latex \mathbf{\overset{\to }{0}}=0\mathbf{\hat{i}}+0\mathbf{\hat{j}}+0\mathbf{\hat{k}} $, so the null vector has no length and no direction.
Two vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ are equal vectors if and only if their difference is the null vector:
This vector equation means we must have simultaneously $latex {A}_{x}-{B}_{x}=0 $, $latex {A}_{y}-{B}_{y}=0 $, and $latex {A}_{z}-{B}_{z}=0 $. Hence, we can write $latex \mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{B}} $ if and only if the corresponding components of vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ are equal:
Two vectors are equal when their corresponding scalar components are equal.
Resolving vectors into their scalar components (i.e., finding their scalar components) and expressing them analytically in vector component form (given by Figure) allows us to use vector algebra to find sums or differences of many vectors analytically (i.e., without using graphical methods). For example, to find the resultant of two vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $, we simply add them component by component, as follows:
In this way, using Figure, scalar components of the resultant vector $latex \mathbf{\overset{\to }{R}}={R}_{x}\mathbf{\hat{i}}+{R}_{y}\mathbf{\hat{j}}+{R}_{z}\mathbf{\hat{k}} $ are the sums of corresponding scalar components of vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $:
Analytical methods can be used to find components of a resultant of many vectors. For example, if we are to sum up $latex N $ vectors $latex {\mathbf{\overset{\to }{F}}}_{1},{\mathbf{\overset{\to }{F}}}_{2},{\mathbf{\overset{\to }{F}}}_{3},\dots ,{\mathbf{\overset{\to }{F}}}_{N} $, where each vector is $latex {\mathbf{\overset{\to }{F}}}_{k}={F}_{kx}\mathbf{\hat{i}}+{F}_{ky}\mathbf{\hat{j}}+{F}_{kz}\mathbf{\hat{k}} $, the resultant vector $latex {\mathbf{\overset{\to }{F}}}_{R} $ is
Therefore, scalar components of the resultant vector are
Having found the scalar components, we can write the resultant in vector component form:
Analytical methods for finding the resultant and, in general, for solving vector equations are very important in physics because many physical quantities are vectors. For example, we use this method in kinematics to find resultant displacement vectors and resultant velocity vectors, in mechanics to find resultant force vectors and the resultants of many derived vector quantities, and in electricity and magnetism to find resultant electric or magnetic vector fields.
$latex \begin{array}{l} \left\{\begin{array}{l}{A}_{x}=A\,\text{cos}\,\alpha =(10.0\,\text{cm})\,\text{cos}\,35^\circ=8.19\,\text{cm}\\ {A}_{y}=A\,\text{sin}\,\alpha =(10.0\,\text{cm})\,\text{sin}\,35^\circ=5.73\,\text{cm}\end{array}\right. \\ \left\{\begin{array}{l}{B}_{x}=B\,\text{cos}\,\beta =(7.0\,\text{cm})\,\text{cos}\,(-110^\circ)=-2.39\,\text{cm} \\ {B}_{y}=B\,\text{sin}\,\beta =(7.0\,\text{cm})\,\text{sin}\,(-110^\circ)=-6.58\,\text{cm}\end{array}\right. \\ \left\{\begin{array}{l}{C}_{x}=C\,\text{cos}\,\gamma =(8.0\,\text{cm})\,\text{cos}\,30^\circ=6.93\,\text{cm} \\ {C}_{y}=C\,\text{sin}\,\gamma =(8.0\,\text{cm})\,\text{sin}\,30^\circ=4.00\,\text{cm}\end{array}\right. \end{array}. $
For (a) we may substitute directly into (Figure) to find the scalar components of the resultant:Three displacement vectors $latex \mathbf{\overset{\to }{A}} $, $latex \mathbf{\overset{\to }{B}} $, and $latex \mathbf{\overset{\to }{F}} $ (Figure) are specified by their magnitudes A = 10.00, B = 7.00, and F = 20.00, respectively, and by their respective direction angles with the horizontal direction $latex \alpha =35^\circ $, $latex \beta =-110^\circ $, and $latex \phi =110^\circ $. The physical units of the magnitudes are centimeters. Use the analytical method to find vector $latex \mathbf{\overset{\to }{G}}=\mathbf{\overset{\to }{A}}+2\mathbf{\overset{\to }{B}}-\mathbf{\overset{\to }{F}} $. Verify that
G = 28.15 cm and that $latex {\theta }_{G}=-68.65^\circ $.$latex \mathbf{\overset{\to }{G}}=(10.25\mathbf{\hat{i}}-26.22\mathbf{\hat{j}})\text{cm} $
Suppose that Bing in Figure leaves the game to attend to more important matters, but Ang, Chang, and Dong continue playing. Ang and Chang’s pull on the toy does not change, but Dong runs around and bites on the toy in a different place. With how big a force and in what direction must Dong pull on the toy now to balance out the combined pulls from Chang and Ang? Illustrate this situation by drawing a vector diagram indicating all forces involved.
D = 55.7 N; direction $latex 65.7^\circ $ north of east
In many physical situations, we often need to know the direction of a vector. For example, we may want to know the direction of a magnetic field vector at some point or the direction of motion of an object. We have already said direction is given by a unit vector, which is a dimensionless entity—that is, it has no physical units associated with it. When the vector in question lies along one of the axes in a Cartesian system of coordinates, the answer is simple, because then its unit vector of direction is either parallel or antiparallel to the direction of the unit vector of an axis. For example, the direction of vector $latex \mathbf{\overset{\to }{d}}=-5\,\text{m}\mathbf{\hat{i}} $ is unit vector $latex \mathbf{\hat{d}}=\text{−}\mathbf{\hat{i}} $. The general rule of finding the unit vector $latex \mathbf{\hat{V}} $ of direction for any vector $latex \mathbf{\overset{\to }{V}} $ is to divide it by its magnitude V:
We see from this expression that the unit vector of direction is indeed dimensionless because the numerator and the denominator in Figure have the same physical unit. In this way, Figure allows us to express the unit vector of direction in terms of unit vectors of the axes. The following example illustrates this principle.
Verify that vector $latex \mathbf{\hat{v}} $ obtained in Figure is indeed a unit vector by computing its magnitude. If the convoy in Figure was moving across a desert flatland—that is, if the third component of its velocity was zero—what is the unit vector of its direction of motion? Which geographic direction does it represent?
$latex \mathbf{\hat{v}}=0.8\mathbf{\hat{i}}+0.6\mathbf{\hat{j}} $, $latex 36.87^\circ $ north of east
For vectors $latex \mathbf{\overset{\to }{B}}=\text{−}\mathbf{\hat{i}}-4\mathbf{\hat{j}} $ and $latex \mathbf{\overset{\to }{A}}=-3\mathbf{\hat{i}}-2\mathbf{\hat{j}} $, calculate (a) $latex \mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}} $ and its magnitude and direction angle, and (b) $latex \mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}} $ and its magnitude and direction angle.
a. $latex \mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}=-4\mathbf{\hat{i}}-6\mathbf{\hat{j}} $, $latex |\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}|=7.211,\theta =213.7^\circ $; b. $latex \mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}}=2\mathbf{\hat{i}}-2\mathbf{\hat{j}} $, $latex |\mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}}|=2\sqrt{2},\theta =-45^\circ $
A particle undergoes three consecutive displacements given by vectors $latex {\mathbf{\overset{\to }{D}}}_{1}=(3.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}}-2.0\mathbf{\hat{k}})\text{mm} $, $latex {\mathbf{\overset{\to }{D}}}_{2}=(1.0\mathbf{\hat{i}}-7.0\mathbf{\hat{j}}+4.0\mathbf{\hat{k}})\text{mm} $, and $latex {\mathbf{\overset{\to }{D}}}_{3}=(-7.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}}+1.0\mathbf{\hat{k}})\text{mm} $. (a) Find the resultant displacement vector of the particle. (b) What is the magnitude of the resultant displacement? (c) If all displacements were along one line, how far would the particle travel?
Given two displacement vectors $latex \mathbf{\overset{\to }{A}}=(3.00\mathbf{\hat{i}}-4.00\mathbf{\hat{j}}+4.00\mathbf{\hat{k}})\text{m} $ and $latex \mathbf{\overset{\to }{B}}=(2.00\mathbf{\hat{i}}+3.00\mathbf{\hat{j}}-7.00\mathbf{\hat{k}})\text{m} $, find the displacements and their magnitudes for (a) $latex \mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}} $ and (b) $latex \mathbf{\overset{\to }{D}}=2\mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}} $.
a. $latex \mathbf{\overset{\to }{C}}=(5.0\mathbf{\hat{i}}-1.0\mathbf{\hat{j}}-3.0\mathbf{\hat{k}})\text{m},C=5.92\,\text{m} $;
b. $latex \mathbf{\overset{\to }{D}}=(4.0\mathbf{\hat{i}}-11.0\mathbf{\hat{j}}+15.0\mathbf{\hat{k}})\text{m},D=19.03\,\text{m} $A small plane flies $latex 40.0\,\text{km} $ in a direction $latex 60^\circ $ north of east and then flies $latex 30.0\,\text{km} $ in a direction $latex 15^\circ $ north of east. Use the analytical method to find the total distance the plane covers from the starting point, and the geographic direction of its displacement vector. What is its displacement vector?
In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day, and she is blown along the following straight lines: 2.50 km and $latex 45.0^\circ $ north of west, then 4.70 km and $latex 60.0^\circ $ south of east, then 1.30 km and $latex 25.0^\circ $ south of west, then 5.10 km due east, then 1.70 km and $latex 5.00^\circ $ east of north, then 7.20 km and $latex 55.0^\circ $ south of west, and finally 2.80 km and $latex 10.0^\circ $ north of east. Use the analytical method to find the resultant vector of all her displacement vectors. What is its magnitude and direction?
$latex \mathbf{\overset{\to }{D}}=(3.3\mathbf{\hat{i}}-6.6\mathbf{\hat{j}})\text{km} $, $latex \mathbf{\hat{i}} $ is to the east, 7.34 km, $latex -63.5^\circ $
Assuming the +x-axis is horizontal to the right for the vectors given in the following figure, use the analytical method to find the following resultants: (a) $latex \mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}, $ (b) $latex \mathbf{\overset{\to }{C}}+\mathbf{\overset{\to }{B}} $, (c) $latex \mathbf{\overset{\to }{D}}+\mathbf{\overset{\to }{F}} $, (d) $latex \mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}} $, (e) $latex \mathbf{\overset{\to }{D}}-\mathbf{\overset{\to }{F}} $, (f) $latex \mathbf{\overset{\to }{A}}+2\mathbf{\overset{\to }{F}} $, (g) $latex \mathbf{\overset{\to }{C}}-2\mathbf{\overset{\to }{D}}+3\mathbf{\overset{\to }{F}} $, and (h) $latex \mathbf{\overset{\to }{A}}-4\mathbf{\overset{\to }{D}}+2\mathbf{\overset{\to }{F}} $.
Given the vectors in the preceding figure, find vector $latex \mathbf{\overset{\to }{R}} $ that solves equations (a) $latex \mathbf{\overset{\to }{D}}+\mathbf{\overset{\to }{R}}=\mathbf{\overset{\to }{F}} $ and (b) $latex \mathbf{\overset{\to }{C}}-2\mathbf{\overset{\to }{D}}+5\mathbf{\overset{\to }{R}}=3\mathbf{\overset{\to }{F}} $. Assume the +x-axis is horizontal to the right.
a. $latex \mathbf{\overset{\to }{R}}=-1.35\mathbf{\hat{i}}-22.04\mathbf{\hat{j}} $, b. $latex \mathbf{\overset{\to }{R}}=-17.98\mathbf{\hat{i}}+0.89\mathbf{\hat{j}} $
A delivery man starts at the post office, drives 40 km north, then 20 km west, then 60 km northeast, and finally 50 km north to stop for lunch. Use the analytical method to determine the following: (a) Find his net displacement vector. (b) How far is the restaurant from the post office? (c) If he returns directly from the restaurant to the post office, what is his displacement vector on the return trip? (d) What is his compass heading on the return trip? Assume the +x-axis is to the east.
An adventurous dog strays from home, runs three blocks east, two blocks north, and one block east, one block north, and two blocks west. Assuming that each block is about a 100 yd, use the analytical method to find the dog’s net displacement vector, its magnitude, and its direction. Assume the +x-axis is to the east. How would your answer be affected if each block was about 100 m?
$latex \mathbf{\overset{\to }{D}}=(200\mathbf{\hat{i}}+300\mathbf{\hat{j}})\text{yd} $, D = 360.5 yd, $latex 56.3^\circ $ north of east; The numerical answers would stay the same but the physical unit would be meters. The physical meaning and distances would be about the same because 1 yd is comparable with 1 m.
If $latex \mathbf{\overset{\to }{D}}=(6.00\mathbf{\hat{i}}-8.00\mathbf{\hat{j}})\text{m} $, $latex \mathbf{\overset{\to }{B}}=(-8.00\mathbf{\hat{i}}+3.00\mathbf{\hat{j}})\text{m} $, and $latex \mathbf{\overset{\to }{A}}=(26.0\mathbf{\hat{i}}+19.0\mathbf{\hat{j}})\text{m} $, find the unknown constants a and b such that $latex a\mathbf{\overset{\to }{D}}+b\mathbf{\overset{\to }{B}}+\mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{0}} $.
Given the displacement vector $latex \mathbf{\overset{\to }{D}}=(3\mathbf{\hat{i}}-4\mathbf{\hat{j}})\text{m,} $ find the displacement vector $latex \mathbf{\overset{\to }{R}} $ so that $latex \mathbf{\overset{\to }{D}}+\mathbf{\overset{\to }{R}}=-4D\mathbf{\hat{j}} $.
$latex \mathbf{\overset{\to }{R}}=-3\mathbf{\hat{i}}-16\mathbf{\hat{j}} $
Find the unit vector of direction for the following vector quantities: (a) Force $latex \mathbf{\overset{\to }{F}}=(3.0\mathbf{\hat{i}}-2.0\mathbf{\hat{j}})\text{N} $, (b) displacement $latex \mathbf{\overset{\to }{D}}=(-3.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}})\text{m} $, and (c) velocity $latex \mathbf{\overset{\to }{v}}=(-5.00\mathbf{\hat{i}}+4.00\mathbf{\hat{j}})\text{m/s} $.
At one point in space, the direction of the electric field vector is given in the Cartesian system by the unit vector $latex \mathbf{\hat{E}}=1\,\text{/}\sqrt{5}\mathbf{\hat{i}}-2\,\text{/}\sqrt{5}\mathbf{\hat{j}} $. If the magnitude of the electric field vector is E = 400.0 V/m, what are the scalar components $latex {E}_{x} $, $latex {E}_{y} $, and $latex {E}_{z} $ of the electric field vector $latex \mathbf{\overset{\to }{E}} $ at this point? What is the direction angle $latex {\theta }_{E} $ of the electric field vector at this point?
$latex \mathbf{\overset{\to }{E}}=E\mathbf{\hat{E}} $, $latex {E}_{x}=+178.9\text{V}\text{/}\text{m} $, $latex {E}_{y}=-357.8\text{V}\text{/}\text{m} $, $latex {E}_{z}=0.0\text{V}\text{/}\text{m} $, $latex {\theta }_{E}=\text{−}{\text{tan}}^{-1}(2) $
A barge is pulled by the two tugboats shown in the following figure. One tugboat pulls on the barge with a force of magnitude 4000 units of force at $latex 15^\circ $ above the line AB (see the figure and the other tugboat pulls on the barge with a force of magnitude 5000 units of force at $latex 12^\circ $ below the line AB. Resolve the pulling forces to their scalar components and find the components of the resultant force pulling on the barge. What is the magnitude of the resultant pull? What is its direction relative to the line AB?
In the control tower at a regional airport, an air traffic controller monitors two aircraft as their positions change with respect to the control tower. One plane is a cargo carrier Boeing 747 and the other plane is a Douglas DC-3. The Boeing is at an altitude of 2500 m, climbing at $latex 10^\circ $ above the horizontal, and moving $latex 30^\circ $ north of west. The DC-3 is at an altitude of 3000 m, climbing at $latex 5^\circ $ above the horizontal, and cruising directly west. (a) Find the position vectors of the planes relative to the control tower. (b) What is the distance between the planes at the moment the air traffic controller makes a note about their positions?
a. $latex {\mathbf{\overset{\to }{R}}}_{B}=(12.278\mathbf{\hat{i}}+7.089\mathbf{\hat{j}}+2.500\mathbf{\hat{k}})\text{km} $, $latex {\mathbf{\overset{\to }{R}}}_{D}=(-0.262\mathbf{\hat{i}}+3.000\mathbf{\hat{k}})\text{km} $; b. $latex |{\mathbf{\overset{\to }{R}}}_{B}-{\mathbf{\overset{\to }{R}}}_{D}|=14.414\,\text{km} $
A vector can be multiplied by another vector but may not be divided by another vector. There are two kinds of products of vectors used broadly in physics and engineering. One kind of multiplication is a scalar multiplication of two vectors. Taking a scalar product of two vectors results in a number (a scalar), as its name indicates. Scalar products are used to define work and energy relations. For example, the work that a force (a vector) performs on an object while causing its displacement (a vector) is defined as a scalar product of the force vector with the displacement vector. A quite different kind of multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, as its name suggests. Vector products are used to define other derived vector quantities. For example, in describing rotations, a vector quantity called torque is defined as a vector product of an applied force (a vector) and its distance from pivot to force (a vector). It is important to distinguish between these two kinds of vector multiplications because the scalar product is a scalar quantity and a vector product is a vector quantity.
Scalar multiplication of two vectors yields a scalar product.
The scalar product $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}} $ of two vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ is a number defined by the equation
where $latex \phi $ is the angle between the vectors (shown in Figure). The scalar product is also called the dot product because of the dot notation that indicates it.
In the definition of the dot product, the direction of angle $latex \phi $ does not matter, and $latex \phi $ can be measured from either of the two vectors to the other because $latex \text{cos}\,\phi =\text{cos}\,(\text{−}\phi )=\text{cos}\,(2\pi -\phi ) $. The dot product is a negative number when $latex 90^\circ \lt \phi \le 180^\circ $ and is a positive number when $latex 0^\circ\le \phi \lt 90^\circ $. Moreover, the dot product of two parallel vectors is $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}=AB\,\text{cos}\,0^\circ=AB $, and the dot product of two antiparallel vectors is $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}=AB\,\text{cos}\,180^\circ=\text{−}AB $. The scalar product of two orthogonal vectors vanishes: $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}=AB\,\text{cos}\,90^\circ=0 $. The scalar product of a vector with itself is the square of its magnitude:
For the vectors given in Figure, find the scalar products $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}} $ and $latex \mathbf{\overset{\to }{F}}\cdot \mathbf{\overset{\to }{C}} $.
$latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}=-57.3 $, $latex \mathbf{\overset{\to }{F}}\cdot \mathbf{\overset{\to }{C}}=27.8 $
In the Cartesian coordinate system, scalar products of the unit vector of an axis with other unit vectors of axes always vanish because these unit vectors are orthogonal:
In these equations, we use the fact that the magnitudes of all unit vectors are one: $latex |\mathbf{\hat{i}}|=|\mathbf{\hat{j}}|=|\mathbf{\hat{k}}|=1 $. For unit vectors of the axes, Figure gives the following identities:
The scalar product $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}} $ can also be interpreted as either the product of B with the orthogonal projection $latex {A}_{\perp } $ of vector $latex \mathbf{\overset{\to }{A}} $ onto the direction of vector $latex \mathbf{\overset{\to }{B}} $ (Figure(b)) or the product of A with the orthogonal projection $latex {B}_{\perp } $ of vector $latex \mathbf{\overset{\to }{B}} $ onto the direction of vector $latex \mathbf{\overset{\to }{A}} $ (Figure(c)):
For example, in the rectangular coordinate system in a plane, the scalar x-component of a vector is its dot product with the unit vector $latex \mathbf{\hat{i}} $, and the scalar y-component of a vector is its dot product with the unit vector $latex \mathbf{\hat{j}} $:
Scalar multiplication of vectors is commutative,
and obeys the distributive law:
We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot product of two vectors in terms of their scalar components.
For vector $latex \mathbf{\overset{\to }{A}}={A}_{x}\mathbf{\hat{i}}+{A}_{y}\mathbf{\hat{j}}+{A}_{z}\mathbf{\hat{k}} $ in a rectangular coordinate system, use Figure through Figure to show that $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\hat{i}}={A}_{x} $ $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\hat{j}}={A}_{y} $ and $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\hat{k}}={A}_{z} $.
When the vectors in Figure are given in their vector component forms,
we can compute their scalar product as follows:
Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselves give one (see Figure and Figure), there are only three nonzero terms in this expression. Thus, the scalar product simplifies to
We can use Figure for the scalar product in terms of scalar components of vectors to find the angle between two vectors. When we divide Figure by AB, we obtain the equation for $latex \text{cos}\,\phi $, into which we substitute Figure:
Angle $latex \phi $ between vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ is obtained by taking the inverse cosine of the expression in Figure.
$latex {F}_{1}=\sqrt{{F}_{1x}^{2}+{F}_{1y}^{2}+{F}_{1z}^{2}}=\sqrt{{10.0}^{2}+{20.4}^{2}+{2.0}^{2}}\,\text{N}=22.8\,\text{N} $
and$latex {F}_{2}=\sqrt{{F}_{2x}^{2}+{F}_{2y}^{2}+{F}_{2z}^{2}}=\sqrt{{15.0}^{2}+{6.2}^{2}}\,\text{N}=16.2\,\text{N}. $
Substituting the scalar components into (Figure) yields the scalar product$latex \begin{array}{cc}\hfill {\mathbf{\overset{\to }{F}}}_{1}\cdot {\mathbf{\overset{\to }{F}}}_{2}& ={F}_{1x}{F}_{2x}+{F}_{1y}{F}_{2y}+{F}_{1z}{F}_{2z}\hfill \\ & =(10.0\,\text{N})(-15.0\,\text{N})+(-20.4\,\text{N})(0.0\,\text{N})+(2.0\,\text{N})(-6.2\,\text{N})\hfill \\ & =-162.4\,{\text{N}}^{2}.\hfill \end{array} $
Finally, substituting everything into (Figure) gives the angle$latex \text{cos}\,\phi =\frac{{\mathbf{\overset{\to }{F}}}_{1}\cdot {\mathbf{\overset{\to }{F}}}_{2}}{{F}_{1}{F}_{2}}=\frac{-162.4\,{\text{N}}^{2}}{(22.8\,\text{N})(16.2\,\text{N})}=-0.439\Rightarrow \enspace\phi ={\text{cos}}^{-1}(-0.439)=116.0^\circ. $
Find the angle between forces $latex {\mathbf{\overset{\to }{F}}}_{1} $ and $latex {\mathbf{\overset{\to }{F}}}_{3} $ in Figure.
$latex 131.9^\circ $
$latex \begin{array}{cc}\hfill {W}_{3}& ={\mathbf{\overset{\to }{F}}}_{3}\cdot \mathbf{\overset{\to }{D}}={F}_{3x}{D}_{x}+{F}_{3y}{D}_{y}+{F}_{3z}{D}_{z}\hfill \\ & =(5.0\,\text{N})(0.0\,\text{cm})+(12.5\,\text{N})(-7.9\,\text{cm})+(0.0\,\text{N})(-4.2\,\text{cm})\hfill \\ & =-98.7\,\text{N}\cdot \text{cm}.\hfill \end{array} $
Vector multiplication of two vectors yields a vector product.
The vector product of two vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ is denoted by $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $ and is often referred to as a cross product. The vector product is a vector that has its direction perpendicular to both vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $. In other words, vector $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $ is perpendicular to the plane that contains vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $, as shown in Figure. The magnitude of the vector product is defined as
where angle $latex \phi $, between the two vectors, is measured from vector $latex \mathbf{\overset{\to }{A}} $ (first vector in the product) to vector $latex \mathbf{\overset{\to }{B}} $ (second vector in the product), as indicated in Figure, and is between $latex 0^\circ $ and $latex 180^\circ $.
According to Figure, the vector product vanishes for pairs of vectors that are either parallel $latex (\phi =0^\circ) $ or antiparallel $latex (\phi =180^\circ) $ because $latex \text{sin}\,0^\circ=\text{sin}\,180^\circ=0 $.
On the line perpendicular to the plane that contains vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ there are two alternative directions—either up or down, as shown in Figure—and the direction of the vector product may be either one of them. In the standard right-handed orientation, where the angle between vectors is measured counterclockwise from the first vector, vector $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $ points upward, as seen in Figure(a). If we reverse the order of multiplication, so that now $latex \mathbf{\overset{\to }{B}} $ comes first in the product, then vector $latex \mathbf{\overset{\to }{B}}\times \mathbf{\overset{\to }{A}} $ must point downward, as seen in Figure(b). This means that vectors $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $ and $latex \mathbf{\overset{\to }{B}}\times \mathbf{\overset{\to }{A}} $ are antiparallel to each other and that vector multiplication is not commutative but anticommutative. The anticommutative property means the vector product reverses the sign when the order of multiplication is reversed:
The corkscrew right-hand rule is a common mnemonic used to determine the direction of the vector product. As shown in Figure, a corkscrew is placed in a direction perpendicular to the plane that contains vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $, and its handle is turned in the direction from the first to the second vector in the product. The direction of the cross product is given by the progression of the corkscrew.
To loosen a rusty nut, a 20.00-N force is applied to the wrench handle at angle $latex \phi =40^\circ $ and at a distance of 0.25 m from the nut, as shown in Figure(a). Find the magnitude and direction of the torque applied to the nut. What would the magnitude and direction of the torque be if the force were applied at angle $latex \phi =45^\circ $, as shown in Figure(b)? For what value of angle $latex \phi $ does the torque have the largest magnitude?
$latex \tau \,=|\mathbf{\overset{\to }{R}}\times \mathbf{\overset{\to }{F}}|=\,RF\,\text{sin}\,\phi =(0.25\,\text{m})(20.00\,\text{N})\,\text{sin}\,40^\circ=3.21\,\text{N}\cdot \text{m}. $
For the situation in (b), the corkscrew rule gives the direction of $latex \mathbf{\overset{\to }{R}}\times \mathbf{\overset{\to }{F}} $ in the negative direction of the z-axis. Physically, it means the vector $latex \mathbf{\overset{\to }{\tau }} $ points into the page, perpendicular to the wrench handle. The magnitude of this torque is$latex \tau \,=|\mathbf{\overset{\to }{R}}\times \mathbf{\overset{\to }{F}}|=\,RF\,\text{sin}\,\phi =(0.25\,\text{m})(20.00\,\text{N})\,\text{sin}\,45^\circ=3.53\,\text{N}\cdot \text{m}. $
The torque has the largest value when $latex \text{sin}\,\phi =1 $, which happens when $latex \phi =90^\circ $. Physically, it means the wrench is most effective—giving us the best mechanical advantage—when we apply the force perpendicular to the wrench handle. For the situation in this example, this best-torque value is $latex {\tau }_{\text{best}}=RF=(0.25\,\text{m})(20.00\,\text{N})=5.00\,\text{N}\cdot \text{m} $.In this equation, the number that multiplies $latex \mathbf{\hat{k}} $ is the scalar z-component of the vector $latex \mathbf{\overset{\to }{R}}\times \mathbf{\overset{\to }{F}} $. In the computation of this component, care must be taken that the angle $latex \phi $ is measured counterclockwise from $latex \mathbf{\overset{\to }{R}} $ (first vector) to $latex \mathbf{\overset{\to }{F}} $ (second vector). Following this principle for the angles, we obtain $latex RF\,\text{sin}\,(+40^\circ)=+3.2\,\text{N}\cdot \text{m} $ for the situation in (a), and we obtain $latex RF\,\text{sin}\,(-45^\circ)=-3.5\,\text{N}\cdot \text{m} $ for the situation in (b). In the latter case, the angle is negative because the graph in Figure indicates the angle is measured clockwise; but, the same result is obtained when this angle is measured counterclockwise because $latex +(360^\circ-45^\circ)=+315^\circ $ and $latex \text{sin}\,(+315^\circ)=\text{sin}\,(-45^\circ) $. In this way, we obtain the solution without reference to the corkscrew rule. For the situation in (a), the solution is $latex \mathbf{\overset{\to }{R}}\times \mathbf{\overset{\to }{F}}=+3.2\,\text{N}\cdot \text{m}\mathbf{\hat{k}} $; for the situation in (b), the solution is $latex \mathbf{\overset{\to }{R}}\times \mathbf{\overset{\to }{F}}=-3.5\,\text{N}\cdot \text{m}\mathbf{\hat{k}} $.
For the vectors given in Figure, find the vector products $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $ and $latex \mathbf{\overset{\to }{C}}\times \mathbf{\overset{\to }{F}} $.
$latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}}=-40.1\mathbf{\hat{k}} $ or, equivalently, $latex |\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}}|=40.1 $, and the direction is into the page; $latex \mathbf{\overset{\to }{C}}\times \mathbf{\overset{\to }{F}}=+157.6\mathbf{\hat{k}} $ or, equivalently, $latex |\mathbf{\overset{\to }{C}}\times \mathbf{\overset{\to }{F}}|=157.6 $, and the direction is out of the page.
Similar to the dot product (Figure), the cross product has the following distributive property:
The distributive property is applied frequently when vectors are expressed in their component forms, in terms of unit vectors of Cartesian axes.
When we apply the definition of the cross product, Figure, to unit vectors $latex \mathbf{\hat{i}} $, $latex \mathbf{\hat{j}} $, and $latex \mathbf{\hat{k}} $ that define the positive x-, y-, and z-directions in space, we find that
All other cross products of these three unit vectors must be vectors of unit magnitudes because $latex \mathbf{\hat{i}} $, $latex \mathbf{\hat{j}} $, and $latex \mathbf{\hat{k}} $ are orthogonal. For example, for the pair $latex \mathbf{\hat{i}} $ and $latex \mathbf{\hat{j}} $, the magnitude is $latex |\mathbf{\hat{i}}\times \mathbf{\hat{j}}|=ij\,\text{sin}\,90^\circ=(1)(1)(1)=1 $. The direction of the vector product $latex \mathbf{\hat{i}}\times \mathbf{\hat{j}} $ must be orthogonal to the xy-plane, which means it must be along the z-axis. The only unit vectors along the z-axis are $latex \text{−}\mathbf{\hat{k}} $ or $latex +\mathbf{\hat{k}} $. By the corkscrew rule, the direction of vector $latex \mathbf{\hat{i}}\times \mathbf{\hat{j}} $ must be parallel to the positive z-axis. Therefore, the result of the multiplication $latex \mathbf{\hat{i}}\times \mathbf{\hat{j}} $ is identical to $latex +\mathbf{\hat{k}} $. We can repeat similar reasoning for the remaining pairs of unit vectors. The results of these multiplications are
Notice that in Figure, the three unit vectors $latex \mathbf{\hat{i}} $, $latex \mathbf{\hat{j}} $, and $latex \mathbf{\hat{k}} $ appear in the cyclic order shown in a diagram in Figure(a). The cyclic order means that in the product formula, $latex \mathbf{\hat{i}} $ follows $latex \mathbf{\hat{k}} $ and comes before $latex \mathbf{\hat{j}} $, or $latex \mathbf{\hat{k}} $ follows $latex \mathbf{\hat{j}} $ and comes before $latex \mathbf{\hat{i}} $, or $latex \mathbf{\hat{j}} $ follows $latex \mathbf{\hat{i}} $ and comes before $latex \mathbf{\hat{k}} $. The cross product of two different unit vectors is always a third unit vector. When two unit vectors in the cross product appear in the cyclic order, the result of such a multiplication is the remaining unit vector, as illustrated in Figure(b). When unit vectors in the cross product appear in a different order, the result is a unit vector that is antiparallel to the remaining unit vector (i.e., the result is with the minus sign, as shown by the examples in Figure(c) and Figure(d). In practice, when the task is to find cross products of vectors that are given in vector component form, this rule for the cross-multiplication of unit vectors is very useful.
Suppose we want to find the cross product $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $ for vectors $latex \mathbf{\overset{\to }{A}}={A}_{x}\mathbf{\hat{i}}+{A}_{y}\mathbf{\hat{j}}+{A}_{z}\mathbf{\hat{k}} $ and $latex \mathbf{\overset{\to }{B}}={B}_{x}\mathbf{\hat{i}}+{B}_{y}\mathbf{\hat{j}}+{B}_{z}\mathbf{\hat{k}} $. We can use the distributive property (Figure), the anticommutative property (Figure), and the results in Figure and Figure for unit vectors to perform the following algebra:
When performing algebraic operations involving the cross product, be very careful about keeping the correct order of multiplication because the cross product is anticommutative. The last two steps that we still have to do to complete our task are, first, grouping the terms that contain a common unit vector and, second, factoring. In this way we obtain the following very useful expression for the computation of the cross product:
In this expression, the scalar components of the cross-product vector are
When finding the cross product, in practice, we can use either Figure or Figure, depending on which one of them seems to be less complex computationally. They both lead to the same final result. One way to make sure if the final result is correct is to use them both.
A particle moving in space with velocity vector $latex \mathbf{\overset{\to }{u}}=-5.0\mathbf{\hat{i}}-2.0\mathbf{\hat{j}}+3.5\mathbf{\hat{k}} $ enters a region with a magnetic field and experiences a magnetic force. Find the magnetic force $latex \mathbf{\overset{\to }{F}} $ on this particle at the entry point to the region where the magnetic field vector is (a) $latex \mathbf{\overset{\to }{B}}=7.2\mathbf{\hat{i}}-\mathbf{\hat{j}}-2.4\mathbf{\hat{k}} $ and (b) $latex \mathbf{\overset{\to }{B}}=4.5\mathbf{\hat{k}} $. In each case, find magnitude F of the magnetic force and angle $latex \theta $ the force vector $latex \mathbf{\overset{\to }{F}} $ makes with the given magnetic field vector $latex \mathbf{\overset{\to }{B}} $.
$latex \left\{\begin{array}{l}{F}_{x}=\zeta ({u}_{y}{B}_{z}-{u}_{z}{B}_{y})=\zeta [(-2.0)(-2.4)-(3.5)(-1.0)]=8.3\zeta \\ {F}_{y}=\zeta ({u}_{z}{B}_{x}-{u}_{x}{B}_{z})=\zeta [(3.5)(7.2)-(-5.0)(-2.4)]=13.2\zeta \\ {F}_{z}=\zeta ({u}_{x}{B}_{y}-{u}_{y}{B}_{x})=\zeta [(-5.0)(-1.0)-(-2.0)(7.2)]=19.4\zeta \end{array}.\right. $
Thus, the magnetic force is $latex \mathbf{\overset{\to }{F}}=\zeta (8.3\mathbf{\hat{i}}+13.2\mathbf{\hat{j}}+19.4\mathbf{\hat{k}}) $ and its magnitude is$latex F=\sqrt{{F}_{x}^{2}+{F}_{y}^{2}+{F}_{z}^{2}}=\zeta \sqrt{{(8.3)}^{2}+{(13.2)}^{2}+{(19.4)}^{2}}=24.9\zeta . $
To compute angle $latex \theta $, we may need to find the magnitude of the magnetic field vector,$latex B=\sqrt{{B}_{x}^{2}+{B}_{y}^{2}+{B}_{z}^{2}}=\sqrt{{(7.2)}^{2}+{(-1.0)}^{2}+{(-2.4)}^{2}}=7.6, $
and the scalar product $latex \mathbf{\overset{\to }{F}}\cdot \mathbf{\overset{\to }{B}} $:$latex \mathbf{\overset{\to }{F}}\cdot \mathbf{\overset{\to }{B}}={F}_{x}{B}_{x}+{F}_{y}{B}_{y}+{F}_{z}{B}_{z}=(8.3\zeta )(7.2)+(13.2\zeta )(-1.0)+(19.4\zeta )(-2.4)=0. $
Now, substituting into (Figure) gives angle $latex \theta $:$latex \text{cos}\,\theta =\frac{\mathbf{\overset{\to }{F}}\cdot \mathbf{\overset{\to }{B}}}{FB}=\frac{0}{(18.2\zeta )(7.6)}=0\,\Rightarrow \enspace\theta =90^\circ. $
Hence, the magnetic force vector is perpendicular to the magnetic field vector. (We could have saved some time if we had computed the scalar product earlier.) (b) Because vector $latex \mathbf{\overset{\to }{B}}=4.5\mathbf{\hat{k}} $ has only one component, we can perform the algebra quickly and find the vector product directly:$latex \begin{array}{ll}\hfill \mathbf{\overset{\to }{F}}& =\zeta \mathbf{\overset{\to }{u}}\times \mathbf{\overset{\to }{B}}=\zeta (-5.0\mathbf{\hat{i}}-2.0\mathbf{\hat{j}}+3.5\mathbf{\hat{k}})\times (4.5\mathbf{\hat{k}})\hfill \\ & =\zeta [(-5.0)(4.5)\mathbf{\hat{i}}\times \mathbf{\hat{k}}+(-2.0)(4.5)\mathbf{\hat{j}}\times \mathbf{\hat{k}}+(3.5)(4.5)\mathbf{\hat{k}}\times \mathbf{\hat{k}}]\hfill \\ & =\zeta [-22.5(\text{−}\mathbf{\hat{j}})-9.0(+\mathbf{\hat{i}})+0]=\zeta (-9.0\mathbf{\hat{i}}+22.5\mathbf{\hat{j}}).\hfill \end{array} $
The magnitude of the magnetic force is$latex F=\sqrt{{F}_{x}^{2}+{F}_{y}^{2}+{F}_{z}^{2}}=\zeta \sqrt{{(-9.0)}^{2}+{(22.5)}^{2}+{(0.0)}^{2}}=24.2\zeta . $
Because the scalar product is$latex \mathbf{\overset{\to }{F}}\cdot \mathbf{\overset{\to }{B}}={F}_{x}{B}_{x}+{F}_{y}{B}_{y}+{F}_{z}{B}_{z}=(-9.0\zeta )(0)+(22.5\zeta )(0)+(0)(4.5)=0, $
the magnetic force vector $latex \mathbf{\overset{\to }{F}} $ is perpendicular to the magnetic field vector $latex \mathbf{\overset{\to }{B}} $.Given two vectors $latex \mathbf{\overset{\to }{A}}=\text{−}\mathbf{\hat{i}}+\mathbf{\hat{j}} $ and $latex \mathbf{\overset{\to }{B}}=3\mathbf{\hat{i}}-\mathbf{\hat{j}} $, find (a) $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $, (b) $latex |\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}}| $, (c) the angle between $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $, and (d) the angle between $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $ and vector $latex \mathbf{\overset{\to }{C}}=\mathbf{\hat{i}}+\mathbf{\hat{k}} $.
a. $latex -2\mathbf{\hat{k}} $, b. 2, c. $latex 153.4^\circ $, d. $latex 135^\circ $
In conclusion to this section, we want to stress that “dot product” and “cross product” are entirely different mathematical objects that have different meanings. The dot product is a scalar; the cross product is a vector. Later chapters use the terms dot product and scalar product interchangeably. Similarly, the terms cross product and vector product are used interchangeably.
Multiplication by a scalar (vector equation) | $latex \mathbf{\overset{\to }{B}}=\alpha \mathbf{\overset{\to }{A}} $ |
Multiplication by a scalar (scalar equation for magnitudes) | $latex B=|\alpha |A $ |
Resultant of two vectors | $latex {\mathbf{\overset{\to }{D}}}_{AD}={\mathbf{\overset{\to }{D}}}_{AC}+{\mathbf{\overset{\to }{D}}}_{CD} $ |
Commutative law | $latex \mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}=\mathbf{\overset{\to }{B}}+\mathbf{\overset{\to }{A}} $ |
Associative law | $latex (\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}})+\mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}+(\mathbf{\overset{\to }{B}}+\mathbf{\overset{\to }{C}}) $ |
Distributive law | $latex {\alpha }_{1}\mathbf{\overset{\to }{A}}+{\alpha }_{2}\mathbf{\overset{\to }{A}}=({\alpha }_{1}+{\alpha }_{2})\mathbf{\overset{\to }{A}} $ |
The component form of a vector in two dimensions | $latex \mathbf{\overset{\to }{A}}={A}_{x}\mathbf{\hat{i}}+{A}_{y}\mathbf{\hat{j}} $ |
Scalar components of a vector in two dimensions | $latex \left\{\begin{array}{c}{A}_{x}={x}_{e}-{x}_{b}\hfill \\ {A}_{y}={y}_{e}-{y}_{b}\hfill \end{array}\right. $ |
Magnitude of a vector in a plane | $latex A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}} $ |
The direction angle of a vector in a plane | $latex {\theta }_{A}={\text{tan}}^{-1}(\frac{{A}_{y}}{{A}_{x}}) $ |
Scalar components of a vector in a plane | $latex \left\{\begin{array}{c}{A}_{x}=A\,\text{cos}\,{\theta }_{A}\hfill \\ {A}_{y}=A\,\text{sin}\,{\theta }_{A}\hfill \end{array}\right. $ |
Polar coordinates in a plane | $latex \left\{\begin{array}{c}x=r\,\text{cos}\,\phi \hfill \\ y=r\,\text{sin}\,\phi \hfill \end{array}\right. $ |
The component form of a vector in three dimensions | $latex \mathbf{\overset{\to }{A}}={A}_{x}\mathbf{\hat{i}}+{A}_{y}\mathbf{\hat{j}}+{A}_{z}\mathbf{\hat{k}} $ |
The scalar z-component of a vector in three dimensions | $latex {A}_{z}={z}_{e}-{z}_{b} $ |
Magnitude of a vector in three dimensions | $latex A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}} $ |
Distributive property | $latex \alpha (\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}})=\alpha \mathbf{\overset{\to }{A}}+\alpha \mathbf{\overset{\to }{B}} $ |
Antiparallel vector to $latex \mathbf{\overset{\to }{A}} $ | $latex \text{−}\mathbf{\overset{\to }{A}}=\text{−}{A}_{x}\mathbf{\hat{i}}-{A}_{y}\mathbf{\hat{j}}-{A}_{z}\mathbf{\hat{k}} $ |
Equal vectors | $latex \mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{B}}\enspace\Leftrightarrow\enspace\left\{\begin{array}{c}{A}_{x}={B}_{x}\hfill \\ {A}_{y}={B}_{y}\hfill \\ {A}_{z}={B}_{z}\hfill \end{array}\right. $ |
Components of the resultant of N vectors | $latex \left\{\begin{array}{c}{F}_{Rx}=\sum _{k=1}^{N}{F}_{kx}={F}_{1x}+{F}_{2x}+\dots+{F}_{Nx}\hfill \\ {F}_{Ry}=\sum _{k=1}^{N}{F}_{ky}={F}_{1y}+{F}_{2y}+\dots+{F}_{Ny}\hfill \\ {F}_{Rz}=\sum _{k=1}^{N}{F}_{kz}={F}_{1z}+{F}_{2z}+\dots+{F}_{Nz}\hfill \end{array}\right. $ |
General unit vector | $latex \mathbf{\hat{V}}=\frac{\mathbf{\overset{\to }{V}}}{V} $ |
Definition of the scalar product | $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}=AB\,\text{cos}\,\phi $ |
Commutative property of the scalar product | $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}=\mathbf{\overset{\to }{B}}\cdot \mathbf{\overset{\to }{A}} $ |
Distributive property of the scalar product | $latex \mathbf{\overset{\to }{A}}\cdot (\mathbf{\overset{\to }{B}}+\mathbf{\overset{\to }{C}})=\mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}+\mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{C}} $ |
Scalar product in terms of scalar components of vectors | $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}={A}_{x}{B}_{x}+{A}_{y}{B}_{y}+{A}_{z}{B}_{z} $ |
Cosine of the angle between two vectors | $latex \text{cos}\,\phi =\frac{\mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}}{AB} $ |
Dot products of unit vectors | $latex \mathbf{\hat{i}}\cdot \mathbf{\hat{j}}=\mathbf{\hat{j}}\cdot \mathbf{\hat{k}}=\mathbf{\hat{k}}\cdot \mathbf{\hat{i}}=0 $ |
Magnitude of the vector product (definition) | $latex |\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}}|=AB\,\text{sin}\,\phi $ |
Anticommutative property of the vector product | $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}}=\text{−}\mathbf{\overset{\to }{B}}\times \mathbf{\overset{\to }{A}} $ |
Distributive property of the vector product | $latex \mathbf{\overset{\to }{A}}\times (\mathbf{\overset{\to }{B}}+\mathbf{\overset{\to }{C}})=\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}}+\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{C}} $ |
Cross products of unit vectors | $latex \left\{\begin{array}{l}\mathbf{\hat{i}}\times \mathbf{\hat{j}}=+\mathbf{\hat{k}},\hfill \\ \mathbf{\hat{j}}\times \mathbf{\hat{k}}=+\mathbf{\hat{i}},\hfill \\ \mathbf{\hat{k}}\times \mathbf{\hat{i}}=+\mathbf{\hat{j}}.\hfill \end{array}\right. $ |
The cross product in terms of scalar components of vectors | $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}}=({A}_{y}{B}_{z}-{A}_{z}{B}_{y})\mathbf{\hat{i}}+({A}_{z}{B}_{x}-{A}_{x}{B}_{z})\mathbf{\hat{j}}+({A}_{x}{B}_{y}-{A}_{y}{B}_{x})\mathbf{\hat{k}} $ |
What is wrong with the following expressions? How can you correct them? (a) $latex C=\mathbf{\overset{\to }{A}}\mathbf{\overset{\to }{B}} $, (b) $latex \mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}\mathbf{\overset{\to }{B}} $, (c) $latex C=\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $, (d) $latex C=A\mathbf{\overset{\to }{B}} $, (e) $latex C+2\mathbf{\overset{\to }{A}}=B $, (f) $latex \mathbf{\overset{\to }{C}}=A\times \mathbf{\overset{\to }{B}} $, (g) $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}}=\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $, (h) $latex \mathbf{\overset{\to }{C}}=2\mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}} $, (i) $latex C=\mathbf{\overset{\to }{A}}\text{/}\mathbf{\overset{\to }{B}} $, and (j) $latex C=\mathbf{\overset{\to }{A}}\text{/}B $.
a. $latex C=\mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{B}} $, b. $latex \mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $ or $latex \mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}} $, c. $latex \mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $, d. $latex \mathbf{\overset{\to }{C}}=A\mathbf{\overset{\to }{B}} $, e. $latex \mathbf{\overset{\to }{C}}+2\mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{B}} $, f. $latex \mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $, g. left side is a scalar and right side is a vector, h. $latex \mathbf{\overset{\to }{C}}=2\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{B}} $, i. $latex \mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}\text{/}B $, j. $latex \mathbf{\overset{\to }{C}}=\mathbf{\overset{\to }{A}}\text{/}B $
If the cross product of two vectors vanishes, what can you say about their directions?
If the dot product of two vectors vanishes, what can you say about their directions?
They are orthogonal.
What is the dot product of a vector with the cross product that this vector has with another vector?
Assuming the +x-axis is horizontal to the right for the vectors in the following figure, find the following scalar products: (a) $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{C}} $, (b) $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{F}} $, (c) $latex \mathbf{\overset{\to }{D}}\cdot \mathbf{\overset{\to }{C}} $, (d) $latex \mathbf{\overset{\to }{A}}\cdot (\mathbf{\overset{\to }{F}}+2\mathbf{\overset{\to }{C}}) $, (e) $latex \mathbf{\hat{i}}\cdot \mathbf{\overset{\to }{B}} $, (f) $latex \mathbf{\hat{j}}\cdot \mathbf{\overset{\to }{B}} $, (g) $latex (3\mathbf{\hat{i}}-\mathbf{\hat{j}})\cdot \mathbf{\overset{\to }{B}} $, and (h) $latex \mathbf{\hat{B}}\cdot \mathbf{\overset{\to }{B}} $.
Assuming the +x-axis is horizontal to the right for the vectors in the preceding figure, find (a) the component of vector $latex \mathbf{\overset{\to }{A}} $ along vector $latex \mathbf{\overset{\to }{C}} $, (b) the component of vector $latex \mathbf{\overset{\to }{C}} $ along vector $latex \mathbf{\overset{\to }{A}} $, (c) the component of vector $latex \mathbf{\hat{i}} $ along vector $latex \mathbf{\overset{\to }{F}} $, and (d) the component of vector $latex \mathbf{\overset{\to }{F}} $ along vector $latex \mathbf{\hat{i}} $.
a. 8.66, b. 10.39, c. 0.866, d. 17.32
Find the angle between vectors for (a) $latex \mathbf{\overset{\to }{D}}=(-3.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}})\text{m} $ and $latex \mathbf{\overset{\to }{A}}=(-3.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}})\text{m} $ and (b) $latex \mathbf{\overset{\to }{D}}=(2.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}}+\mathbf{\hat{k}})\text{m} $ and $latex \mathbf{\overset{\to }{B}}=(-2.0\mathbf{\hat{i}}+3.0\mathbf{\hat{j}}+2.0\mathbf{\hat{k}})\text{m} $.
Find the angles that vector $latex \mathbf{\overset{\to }{D}}=(2.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}}+\mathbf{\hat{k}})\text{m} $ makes with the x-, y-, and z- axes.
$latex {\theta }_{i}=64.12^\circ,{\theta }_{j}=150.79^\circ,{\theta }_{k}=77.39^\circ $
Show that the force vector $latex \mathbf{\overset{\to }{D}}=(2.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}}+\mathbf{\hat{k}})\text{N} $ is orthogonal to the force vector $latex \mathbf{\overset{\to }{G}}=(3.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}}+10.0\mathbf{\hat{k}})\text{N} $.
Assuming the +x-axis is horizontal to the right for the vectors in the previous figure, find the following vector products: (a) $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{C}} $, (b) $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{F}} $, (c) $latex \mathbf{\overset{\to }{D}}\times \mathbf{\overset{\to }{C}} $, (d) $latex \mathbf{\overset{\to }{A}}\times (\mathbf{\overset{\to }{F}}+2\mathbf{\overset{\to }{C}}) $, (e) $latex \mathbf{\hat{i}}\times \mathbf{\overset{\to }{B}} $, (f) $latex \mathbf{\hat{j}}\times \mathbf{\overset{\to }{B}} $, (g) $latex (3\mathbf{\hat{i}}-\mathbf{\hat{j}})\times \mathbf{\overset{\to }{B}} $, and (h) $latex \mathbf{\hat{B}}\times \mathbf{\overset{\to }{B}} $.
a. $latex -119.98\mathbf{\hat{k}} $, b. $latex -173.2\mathbf{\hat{k}} $, c. $latex +93.69\mathbf{\hat{k}} $, d. $latex -413.2\mathbf{\hat{k}} $, e. $latex +39.93\mathbf{\hat{k}} $, f. $latex -30.09\mathbf{\hat{k}} $, g. $latex +149.9\mathbf{\hat{k}} $, h. 0
Find the cross product $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{C}} $ for (a) $latex \mathbf{\overset{\to }{A}}=2.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}}+\mathbf{\hat{k}} $ and $latex \mathbf{\overset{\to }{C}}=3.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}}+10.0\mathbf{\hat{k}} $, (b) $latex \mathbf{\overset{\to }{A}}=3.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}}+10.0\mathbf{\hat{k}} $ and $latex \mathbf{\overset{\to }{C}}=2.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}}+\mathbf{\hat{k}} $, (c) $latex \mathbf{\overset{\to }{A}}=-3.0\mathbf{\hat{i}}-4.0\mathbf{\hat{j}} $ and $latex \mathbf{\overset{\to }{C}}=-3.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}} $, and (d) $latex \mathbf{\overset{\to }{C}}=-2.0\mathbf{\hat{i}}+3.0\mathbf{\hat{j}}+2.0\mathbf{\hat{k}} $ and $latex \mathbf{\overset{\to }{A}}=-9.0\mathbf{\hat{j}} $.
For the vectors in the earlier figure, find (a) $latex (\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{F}})\cdot \mathbf{\overset{\to }{D}} $, (b) $latex (\mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{F}})\cdot (\mathbf{\overset{\to }{D}}\times \mathbf{\overset{\to }{B}}) $, and (c) $latex (\mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{F}})(\mathbf{\overset{\to }{D}}\times \mathbf{\overset{\to }{B}}) $.
a. 0, b. 173,194, c. $latex +199,993\mathbf{\hat{k}} $
(a) If $latex \mathbf{\overset{\to }{A}}\times \mathbf{\overset{\to }{F}}=\mathbf{\overset{\to }{B}}\times \mathbf{\overset{\to }{F}} $, can we conclude $latex \mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{B}} $? (b) If $latex \mathbf{\overset{\to }{A}}\cdot \mathbf{\overset{\to }{F}}=\mathbf{\overset{\to }{B}}\cdot \mathbf{\overset{\to }{F}} $, can we conclude $latex \mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{B}} $? (c) If $latex F\mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{B}}F $, can we conclude $latex \mathbf{\overset{\to }{A}}=\mathbf{\overset{\to }{B}} $? Why or why not?
You fly $latex 32.0\,\text{km} $ in a straight line in still air in the direction $latex 35.0^\circ $ south of west. (a) Find the distances you would have to fly due south and then due west to arrive at the same point. (b) Find the distances you would have to fly first in a direction $latex 45.0^\circ $ south of west and then in a direction $latex 45.0^\circ $ west of north. Note these are the components of the displacement along a different set of axes—namely, the one rotated by $latex 45^\circ $ with respect to the axes in (a).
a. 18.4 km and 26.2 km, b. 31.5 km and 5.56 km
Rectangular coordinates of a point are given by (2, y) and its polar coordinates are given by $latex (r,\pi \text{/}6) $. Find y and r.
If the polar coordinates of a point are $latex (r,\phi ) $ and its rectangular coordinates are $latex (x,y)$, determine the polar coordinates of the following points: (a) (−x, y), (b) (−2x, −2y), and (c) (3x, −3y).
a. $latex (r,\phi +\pi \text{/}2)$, b. $latex (2r,\phi +2\pi ) $, (c) $latex (3r,\text{−}\phi ) $
Vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ have identical magnitudes of 5.0 units. Find the angle between them if $latex \mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}=5\sqrt{2}\mathbf{\hat{j}} $.
Starting at the island of Moi in an unknown archipelago, a fishing boat makes a round trip with two stops at the islands of Noi and Poi. It sails from Moi for 4.76 nautical miles (nmi) in a direction $latex 37^\circ $ north of east to Noi. From Noi, it sails $latex 69^\circ $ west of north to Poi. On its return leg from Poi, it sails $latex 28^\circ $ east of south. What distance does the boat sail between Noi and Poi? What distance does it sail between Moi and Poi? Express your answer both in nautical miles and in kilometers. Note: 1 nmi = 1852 m.
$latex {d}_{\text{PM}}=33.12\,\text{nmi}=61.34\,\text{km},\enspace{d}_{\text{NP}}=35.47\,\text{nmi}=65.69\,\text{km} $
An air traffic controller notices two signals from two planes on the radar monitor. One plane is at altitude 800 m and in a 19.2-km horizontal distance to the tower in a direction $latex 25^\circ $ south of west. The second plane is at altitude 1100 m and its horizontal distance is 17.6 km and $latex 20^\circ $ south of west. What is the distance between these planes?
Show that when $latex \mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}=\mathbf{\overset{\to }{C}} $, then $latex {C}^{2}={A}^{2}+{B}^{2}+2AB\,\text{cos}\,\phi $, where $latex \phi $ is the angle between vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $.
proof
Four force vectors each have the same magnitude f. What is the largest magnitude the resultant force vector may have when these forces are added? What is the smallest magnitude of the resultant? Make a graph of both situations.
A skater glides along a circular path of radius 5.00 m in clockwise direction. When he coasts around one-half of the circle, starting from the west point, find (a) the magnitude of his displacement vector and (b) how far he actually skated. (c) What is the magnitude of his displacement vector when he skates all the way around the circle and comes back to the west point?
a. 10.00 m, b. $latex 5\pi \,\text{m} $, c. 0
A stubborn dog is being walked on a leash by its owner. At one point, the dog encounters an interesting scent at some spot on the ground and wants to explore it in detail, but the owner gets impatient and pulls on the leash with force $latex \mathbf{\overset{\to }{F}}=(98.0\mathbf{\hat{i}}+132.0\mathbf{\hat{j}}+32.0\mathbf{\hat{k}})\text{N} $ along the leash. (a) What is the magnitude of the pulling force? (b) What angle does the leash make with the vertical?
If the velocity vector of a polar bear is $latex \mathbf{\overset{\to }{u}}=(-18.0\mathbf{\hat{i}}-13.0\mathbf{\hat{j}})\text{km}\text{/}\text{h} $, how fast and in what geographic direction is it heading? Here, $latex \mathbf{\hat{i}} $ and $latex \mathbf{\hat{j}} $ are directions to geographic east and north, respectively.
22.2 km/h, $latex 35.8^\circ $ south of west
Find the scalar components of three-dimensional vectors $latex \mathbf{\overset{\to }{G}} $ and $latex \mathbf{\overset{\to }{H}} $ in the following figure and write the vectors in vector component form in terms of the unit vectors of the axes.
A diver explores a shallow reef off the coast of Belize. She initially swims 90.0 m north, makes a turn to the east and continues for 200.0 m, then follows a big grouper for 80.0 m in the direction $latex 30^\circ $ north of east. In the meantime, a local current displaces her by 150.0 m south. Assuming the current is no longer present, in what direction and how far should she now swim to come back to the point where she started?
240.2 m, $latex 2.2^\circ $ south of west
A force vector $latex \mathbf{\overset{\to }{A}} $ has x- and y-components, respectively, of −8.80 units of force and 15.00 units of force. The x- and y-components of force vector $latex \mathbf{\overset{\to }{B}} $ are, respectively, 13.20 units of force and −6.60 units of force. Find the components of force vector $latex \mathbf{\overset{\to }{C}} $ that satisfies the vector equation $latex \mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}}+3\mathbf{\overset{\to }{C}}=0 $.
Vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $ are two orthogonal vectors in the xy-plane and they have identical magnitudes. If $latex \mathbf{\overset{\to }{A}}=3.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}} $, find $latex \mathbf{\overset{\to }{B}} $.
$latex \mathbf{\overset{\to }{B}}=-4.0\mathbf{\hat{i}}+3.0\mathbf{\hat{j}} $ or $latex \mathbf{\overset{\to }{B}}=4.0\mathbf{\hat{i}}-3.0\mathbf{\hat{j}} $
For the three-dimensional vectors in the following figure, find (a) $latex \mathbf{\overset{\to }{G}}\times \mathbf{\overset{\to }{H}} $, (b) $latex |\mathbf{\overset{\to }{G}}\times \mathbf{\overset{\to }{H}}| $, and (c) $latex \mathbf{\overset{\to }{G}}\cdot \mathbf{\overset{\to }{H}} $.
Show that $latex (\mathbf{\overset{\to }{B}}\times \mathbf{\overset{\to }{C}})\cdot \mathbf{\overset{\to }{A}} $ is the volume of the parallelepiped, with edges formed by the three vectors in the following figure.
proof
Vector $latex \mathbf{\overset{\to }{B}} $ is 5.0 cm long and vector $latex \mathbf{\overset{\to }{A}} $ is 4.0 cm long. Find the angle between these two vectors when $latex |\mathbf{\overset{\to }{A}}+\mathbf{\overset{\to }{B}}|=\,3.0\,\text{cm} $ and $latex |\mathbf{\overset{\to }{A}}-\mathbf{\overset{\to }{B}}|=\,3.0\,\text{cm} $.
What is the component of the force vector $latex \mathbf{\overset{\to }{G}}=(3.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}}+10.0\mathbf{\hat{k}})\text{N} $ along the force vector $latex \mathbf{\overset{\to }{H}}=(1.0\mathbf{\hat{i}}+4.0\mathbf{\hat{j}})\text{N} $?
$latex {G}_{\perp }=2375\sqrt{17}\approx 9792 $
The following figure shows a triangle formed by the three vectors $latex \mathbf{\overset{\to }{A}} $, $latex \mathbf{\overset{\to }{B}} $, and $latex \mathbf{\overset{\to }{C}} $. If vector $latex {\mathbf{\overset{\to }{C}}}^{\prime } $ is drawn between the midpoints of vectors $latex \mathbf{\overset{\to }{A}} $ and $latex \mathbf{\overset{\to }{B}} $, show that $latex {\mathbf{\overset{\to }{C}}}^{\prime }=\mathbf{\overset{\to }{C}}\text{/}2 $.
Distances between points in a plane do not change when a coordinate system is rotated. In other words, the magnitude of a vector is invariant under rotations of the coordinate system. Suppose a coordinate system S is rotated about its origin by angle $latex \phi $ to become a new coordinate system $latex {\text{S}}^{\prime } $, as shown in the following figure. A point in a plane has coordinates (x, y) in S and coordinates $latex ({x}^{\prime },{y}^{\prime }) $ in $latex {\text{S}}^{\prime } $.
(a) Show that, during the transformation of rotation, the coordinates in $latex {\text{S}}^{\prime } $ are expressed in terms of the coordinates in S by the following relations:
(b) Show that the distance of point P to the origin is invariant under rotations of the coordinate system. Here, you have to show that
(c) Show that the distance between points P and Q is invariant under rotations of the coordinate system. Here, you have to show that
proof
Our universe is full of objects in motion. From the stars, planets, and galaxies; to the motion of people and animals; down to the microscopic scale of atoms and molecules—everything in our universe is in motion. We can describe motion using the two disciplines of kinematics and dynamics. We study dynamics, which is concerned with the causes of motion, in Newton’s Laws of Motion; but, there is much to be learned about motion without referring to what causes it, and this is the study of kinematics. Kinematics involves describing motion through properties such as position, time, velocity, and acceleration.
A full treatment of kinematics considers motion in two and three dimensions. For now, we discuss motion in one dimension, which provides us with the tools necessary to study multidimensional motion. A good example of an object undergoing one-dimensional motion is the maglev (magnetic levitation) train depicted at the beginning of this chapter. As it travels, say, from Tokyo to Kyoto, it is at different positions along the track at various times in its journey, and therefore has displacements, or changes in position. It also has a variety of velocities along its path and it undergoes accelerations (changes in velocity). With the skills learned in this chapter we can calculate these quantities and average velocity. All these quantities can be described using kinematics, without knowing the train’s mass or the forces involved.
]]>When you’re in motion, the basic questions to ask are: Where are you? Where are you going? How fast are you getting there? The answers to these questions require that you specify your position, your displacement, and your average velocity—the terms we define in this section.
To describe the motion of an object, you must first be able to describe its position (x): where it is at any particular time. More precisely, we need to specify its position relative to a convenient frame of reference. A frame of reference is an arbitrary set of axes from which the position and motion of an object are described. Earth is often used as a frame of reference, and we often describe the position of an object as it relates to stationary objects on Earth. For example, a rocket launch could be described in terms of the position of the rocket with respect to Earth as a whole, whereas a cyclist’s position could be described in terms of where she is in relation to the buildings she passes Figure. In other cases, we use reference frames that are not stationary but are in motion relative to Earth. To describe the position of a person in an airplane, for example, we use the airplane, not Earth, as the reference frame. To describe the position of an object undergoing one-dimensional motion, we often use the variable x. Later in the chapter, during the discussion of free fall, we use the variable y.
If an object moves relative to a frame of reference—for example, if a professor moves to the right relative to a whiteboard Figure—then the object’s position changes. This change in position is called displacement. The word displacement implies that an object has moved, or has been displaced. Although position is the numerical value of x along a straight line where an object might be located, displacement gives the change in position along this line. Since displacement indicates direction, it is a vector and can be either positive or negative, depending on the choice of positive direction. Also, an analysis of motion can have many displacements embedded in it. If right is positive and an object moves 2 m to the right, then 4 m to the left, the individual displacements are 2 m and $latex -4 $ m, respectively.
Displacement $latex \Delta x $ is the change in position of an object:
where $latex \Delta x $ is displacement, $latex {x}_{\text{f}} $ is the final position, and $latex {x}_{0} $ is the initial position.
We use the uppercase Greek letter delta (Δ) to mean “change in” whatever quantity follows it; thus, $latex \Delta x $ means change in position (final position less initial position). We always solve for displacement by subtracting initial position $latex {x}_{0} $ from final position $latex {x}_{\text{f}} $. Note that the SI unit for displacement is the meter, but sometimes we use kilometers or other units of length. Keep in mind that when units other than meters are used in a problem, you may need to convert them to meters to complete the calculation (see Conversion Factors).
Objects in motion can also have a series of displacements. In the previous example of the pacing professor, the individual displacements are 2 m and $latex -4 $ m, giving a total displacement of −2 m. We define total displacement $latex \Delta {x}_{\text{Total}} $, as the sum of the individual displacements, and express this mathematically with the equation
where $latex \Delta {x}_{i} $ are the individual displacements. In the earlier example,
Similarly,
Thus,
The total displacement is 2 − 4 = −2 m to the left, or in the negative direction. It is also useful to calculate the magnitude of the displacement, or its size. The magnitude of the displacement is always positive. This is the absolute value of the displacement, because displacement is a vector and cannot have a negative value of magnitude. In our example, the magnitude of the total displacement is 2 m, whereas the magnitudes of the individual displacements are 2 m and 4 m.
The magnitude of the total displacement should not be confused with the distance traveled. Distance traveled $latex {x}_{\text{Total}} $, is the total length of the path traveled between two positions. In the previous problem, the distance traveled is the sum of the magnitudes of the individual displacements:
To calculate the other physical quantities in kinematics we must introduce the time variable. The time variable allows us not only to state where the object is (its position) during its motion, but also how fast it is moving. How fast an object is moving is given by the rate at which the position changes with time.
For each position $latex {x}_{\text{i}} $, we assign a particular time $latex {t}_{\text{i}} $. If the details of the motion at each instant are not important, the rate is usually expressed as the average velocity $latex \overset{\text{–}}{v} $. This vector quantity is simply the total displacement between two points divided by the time taken to travel between them. The time taken to travel between two points is called the elapsed time $latex \Delta t $.
If $latex {x}_{1} $ and $latex {x}_{2} $ are the positions of an object at times $latex {t}_{1} $ and $latex {t}_{2} $, respectively, then
It is important to note that the average velocity is a vector and can be negative, depending on positions $latex {x}_{1} $ and $latex {x}_{2} $.
A sketch of Jill’s movements is shown in Figure.
Time t_{i} (min) | Position $latex {x}_{i} $ (km) | Displacement $latex \Delta {x}_{\text{i}} $ (km) |
---|---|---|
$latex {t}_{0}=0 $ | $latex {x}_{0}=0 $ | $latex \Delta {x}_{0}=0 $ |
$latex {t}_{1}=9 $ | $latex {x}_{1}=0.5 $ | $latex \Delta {x}_{1}={x}_{1}-{x}_{0}=0.5 $ |
$latex {t}_{2}=18 $ | $latex {x}_{2}=0 $ | $latex \Delta {x}_{2}={x}_{2}-{x}_{1}=-0.5 $ |
$latex {t}_{3}=33 $ | $latex {x}_{3}=1.0 $ | $latex \Delta {x}_{3}={x}_{3}-{x}_{2}=1.0 $ |
$latex {t}_{4}=58 $ | $latex {x}_{4}=-0.75 $ | $latex \Delta {x}_{4}={x}_{4}-{x}_{3}=-1.75 $ |
$latex \sum \Delta {x}_{\text{i}}=0.5-0.5+1.0-1.75\,\text{km}=-0.75\,\text{km}\text{.} $
$latex {x}_{\text{Total}}=\sum |\Delta {x}_{\text{i}}|=0.5+0.5+1.0+1.75\,\text{km}=3.75\,\text{km} $.
A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is his displacement? (b) What is the distance traveled? (c) What is the magnitude of his displacement?
Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Identify each quantity in your example specifically.
You drive your car into town and return to drive past your house to a friend’s house.
Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of displacement and displacement are exactly the same?
Bacteria move back and forth using their flagella (structures that look like little tails). Speeds of up to 50 μm/s (50 × 10^{−6} m/s) have been observed. The total distance traveled by a bacterium is large for its size, whereas its displacement is small. Why is this?
If the bacteria are moving back and forth, then the displacements are canceling each other and the final displacement is small.
Give an example of a device used to measure time and identify what change in that device indicates a change in time.
Does a car’s odometer measure distance traveled or displacement?
Distance traveled
During a given time interval the average velocity of an object is zero. What can you say conclude about its displacement over the time interval?
Consider a coordinate system in which the positive x axis is directed upward vertically. What are the positions of a particle (a) 5.0 m directly above the origin and (b) 2.0 m below the origin?
A car is 2.0 km west of a traffic light at t = 0 and 5.0 km east of the light at t = 6.0 min. Assume the origin of the coordinate system is the light and the positive x direction is eastward. (a) What are the car’s position vectors at these two times? (b) What is the car’s displacement between 0 min and 6.0 min?
a. $latex {\mathbf{\overset{\to }{x}}}_{1}=(-2.0\,\text{m})\mathbf{\hat{i}} $, $latex {\mathbf{\overset{\to }{x}}}_{2}=(5.0\,\text{m})\mathbf{\hat{i}} $; b. 7.0 m east
The Shanghai maglev train connects Longyang Road to Pudong International Airport, a distance of 30 km. The journey takes 8 minutes on average. What is the maglev train’s average velocity?
The position of a particle moving along the x-axis is given by $latex x(t)=4.0-2.0t $ m. (a) At what time does the particle cross the origin? (b) What is the displacement of the particle between $latex \text{t}=3.0\,\text{s} $ and $latex \text{t}=6.0\,\text{s}? $
a. $latex t=2.0 $ s; b. $latex x(6.0)-x(3.0)=-8.0-(-2.0)=-6.0\,\text{m} $
A cyclist rides 8.0 km east for 20 minutes, then he turns and heads west for 8 minutes and 3.2 km. Finally, he rides east for 16 km, which takes 40 minutes. (a) What is the final displacement of the cyclist? (b) What is his average velocity?
On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth’s atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity of the blast wave? b) Compare this with the speed of sound, which is 343 m/s at sea level.
a. 150.0 s, $latex \overset{\text{–}}{v}=156.7\,\text{m/s} $; b. 45.7% the speed of sound at sea level
We have now seen how to calculate the average velocity between two positions. However, since objects in the real world move continuously through space and time, we would like to find the velocity of an object at any single point. We can find the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us better insight into the physics of motion and will be useful in later chapters.
The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usually called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position x as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation is $latex \overset{\text{–}}{v}=\frac{x({t}_{2})-x({t}_{1})}{{t}_{2}-{t}_{1}} $. To find the instantaneous velocity at any position, we let $latex {t}_{1}=t $ and $latex {t}_{2}=t+\Delta t $. After inserting these expressions into the equation for the average velocity and taking the limit as $latex \Delta t\to 0 $, we find the expression for the instantaneous velocity:
The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t:
Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point $latex {t}_{0} $ is the rate of change of the position function, which is the slope of the position function $latex x(t) $ at $latex {t}_{0} $. Figure shows how the average velocity $latex \overset{\text{–}}{v}=\frac{\Delta x}{\Delta t} $ between two times approaches the instantaneous velocity at $latex {t}_{0}. $ The instantaneous velocity is shown at time $latex {t}_{0} $, which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, $latex {t}_{1},{t}_{2} $, and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.
In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not have the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.
We can calculate the average speed by finding the total distance traveled divided by the elapsed time:
Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time, then we would be interested in our average speed.
However, we can calculate the instantaneous speed from the magnitude of the instantaneous velocity:
If a particle is moving along the x-axis at +7.0 m/s and another particle is moving along the same axis at −7.0 m/s, they have different velocities, but both have the same speed of 7.0 m/s. Some typical speeds are shown in the following table.
Speed | m/s | mi/h |
---|---|---|
Continental drift | $latex {10}^{-7} $ | $latex 2\times {10}^{-7} $ |
Brisk walk | 1.7 | 3.9 |
Cyclist | 4.4 | 10 |
Sprint runner | 12.2 | 27 |
Rural speed limit | 24.6 | 56 |
Official land speed record | 341.1 | 763 |
Speed of sound at sea level | 343 | 768 |
Space shuttle on reentry | 7800 | 17,500 |
Escape velocity of Earth* | 11,200 | 25,000 |
Orbital speed of Earth around the Sun | 29,783 | 66,623 |
Speed of light in a vacuum | 299,792,458 | 670,616,629 |
When calculating instantaneous velocity, we need to specify the explicit form of the position function x(t). For the moment, let’s use polynomials $latex x(t)=A{t}^{n} $, because they are easily differentiated using the power rule of calculus:
The following example illustrates the use of Figure.
StrategyFigure gives the instantaneous velocity of the particle as the derivative of the position function. Looking at the form of the position function given, we see that it is a polynomial in t. Therefore, we can use Figure, the power rule from calculus, to find the solution. We use Figure to calculate the average velocity of the particle.
The position of an object as a function of time is $latex x(t)=-3{t}^{2}\,\text{m}$. (a) What is the velocity of the object as a function of time? (b) Is the velocity ever positive? (c) What are the velocity and speed at t = 1.0 s?
(a) Taking the derivative of x(t) gives v(t) = −6t m/s. (b) No, because time can never be negative. (c) The velocity is v(1.0 s) = −6 m/s and the speed is $latex |v(1.0\,\text{s})|=6\,\text{m/s}$.
There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.
Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement/Elapsed time, your average velocity is zero.
Does the speedometer of a car measure speed or velocity?
If you divide the total distance traveled on a car trip (as determined by the odometer) by the elapsed time of the trip, are you calculating average speed or magnitude of average velocity? Under what circumstances are these two quantities the same?
Average speed. They are the same if the car doesn’t reverse direction.
How are instantaneous velocity and instantaneous speed related to one another? How do they differ?
A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the woodchuck? (b) What is its average speed?
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Given the following velocity-versus-time graph, sketch the position-versus-time graph.
An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.
A particle moves along the x-axis according to $latex x(t)=10t-2{t}^{2}\,\text{m}$. (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between t = 2 s and t = 3 s?
Unreasonable results. A particle moves along the x-axis according to $latex x(t)=3{t}^{3}+5t\text{} $. At what time is the velocity of the particle equal to zero? Is this reasonable?
The importance of understanding acceleration spans our day-to-day experience, as well as the vast reaches of outer space and the tiny world of subatomic physics. In everyday conversation, to accelerate means to speed up; applying the brake pedal causes a vehicle to slow down. We are familiar with the acceleration of our car, for example. The greater the acceleration, the greater the change in velocity over a given time. Acceleration is widely seen in experimental physics. In linear particle accelerator experiments, for example, subatomic particles are accelerated to very high velocities in collision experiments, which tell us information about the structure of the subatomic world as well as the origin of the universe. In space, cosmic rays are subatomic particles that have been accelerated to very high energies in supernovas (exploding massive stars) and active galactic nuclei. It is important to understand the processes that accelerate cosmic rays because these rays contain highly penetrating radiation that can damage electronics flown on spacecraft, for example.
The formal definition of acceleration is consistent with these notions just described, but is more inclusive.
Average acceleration is the rate at which velocity changes:
where $latex \overset{\text{−}}{a} $ is average acceleration, v is velocity, and t is time. (The bar over the a means average acceleration.)
Because acceleration is velocity in meters divided by time in seconds, the SI units for acceleration are often abbreviated m/s^{2}—that is, meters per second squared or meters per second per second. This literally means by how many meters per second the velocity changes every second. Recall that velocity is a vector—it has both magnitude and direction—which means that a change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if a runner traveling at 10 km/h due east slows to a stop, reverses direction, continues her run at 10 km/h due west, her velocity has changed as a result of the change in direction, although the magnitude of the velocity is the same in both directions. Thus, acceleration occurs when velocity changes in magnitude (an increase or decrease in speed) or in direction, or both.
Acceleration is a vector in the same direction as the change in velocity, $latex \Delta v $. Since velocity is a vector, it can change in magnitude or in direction, or both. Acceleration is, therefore, a change in speed or direction, or both.
Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction of motion. When an object slows down, its acceleration is opposite to the direction of its motion. Although this is commonly referred to as deceleration Figure, we say the train is accelerating in a direction opposite to its direction of motion.
The term deceleration can cause confusion in our analysis because it is not a vector and it does not point to a specific direction with respect to a coordinate system, so we do not use it. Acceleration is a vector, so we must choose the appropriate sign for it in our chosen coordinate system. In the case of the train in Figure, acceleration is in the negative direction in the chosen coordinate system, so we say the train is undergoing negative acceleration.
If an object in motion has a velocity in the positive direction with respect to a chosen origin and it acquires a constant negative acceleration, the object eventually comes to a rest and reverses direction. If we wait long enough, the object passes through the origin going in the opposite direction. This is illustrated in Figure.
We can solve this problem by identifying $latex \Delta v\,\text{and}\,\Delta t $ from the given information, and then calculating the average acceleration directly from the equation $latex \overset{\text{–}}{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{\text{f}}-{v}_{0}}{{t}_{\text{f}}-{t}_{0}} $.
Second, find the change in velocity. Since the horse is going from zero to –15.0 m/s, its change in velocity equals its final velocity:
Last, substitute the known values ($latex \Delta v\,\text{and}\,\Delta t $) and solve for the unknown $latex \overset{\text{–}}{a} $:
Protons in a linear accelerator are accelerated from rest to $latex 2.0\times {10}^{7}\,\text{m/s} $ in 10^{–4} s. What is the average acceleration of the protons?
Inserting the knowns, we have
$latex \overset{\text{–}}{a}=\frac{\Delta v}{\Delta t}=\frac{2.0\times {10}^{7}\,\text{m/s}-0}{{10}^{-4}\,\text{s}-0}=2.0\times {10}^{11}{\text{m/s}}^{2}. $Instantaneous acceleration a, or acceleration at a specific instant in time, is obtained using the same process discussed for instantaneous velocity. That is, we calculate the average velocity between two points in time separated by $latex \Delta t $ and let $latex \Delta t $ approach zero. The result is the derivative of the velocity function v(t), which is instantaneous acceleration and is expressed mathematically as
Thus, similar to velocity being the derivative of the position function, instantaneous acceleration is the derivative of the velocity function. We can show this graphically in the same way as instantaneous velocity. In Figure, instantaneous acceleration at time t_{0} is the slope of the tangent line to the velocity-versus-time graph at time t_{0}. We see that average acceleration $latex \overset{\text{–}}{a}=\frac{\Delta v}{\Delta t} $ approaches instantaneous acceleration as $latex \Delta t $ approaches zero. Also in part (a) of the figure, we see that velocity has a maximum when its slope is zero. This time corresponds to the zero of the acceleration function. In part (b), instantaneous acceleration at the minimum velocity is shown, which is also zero, since the slope of the curve is zero there, too. Thus, for a given velocity function, the zeros of the acceleration function give either the minimum or the maximum velocity.
To illustrate this concept, let’s look at two examples. First, a simple example is shown using Figure(b), the velocity-versus-time graph of Figure, to find acceleration graphically. This graph is depicted in Figure(a), which is a straight line. The corresponding graph of acceleration versus time is found from the slope of velocity and is shown in Figure(b). In this example, the velocity function is a straight line with a constant slope, thus acceleration is a constant. In the next example, the velocity function has a more complicated functional dependence on time.
If we know the functional form of velocity, v(t), we can calculate instantaneous acceleration a(t) at any time point in the motion using Figure.
At t = 2 s, velocity has increased to$latex v(2\,\text{s)}=20\,\text{m/s} $, where it is maximum, which corresponds to the time when the acceleration is zero. We see that the maximum velocity occurs when the slope of the velocity function is zero, which is just the zero of the acceleration function.
At t = 3 s, velocity is $latex v(3\,\text{s)}=15\,\text{m/s} $ and acceleration is negative. The particle has reduced its velocity and the acceleration vector is negative. The particle is slowing down.
At t = 5 s, velocity is $latex v(5\,\text{s)}=-25\,\text{m/s} $ and acceleration is increasingly negative. Between the times t = 3 s and t = 5 s the particle has decreased its velocity to zero and then become negative, thus reversing its direction. The particle is now speeding up again, but in the opposite direction.
We can see these results graphically in Figure.
An airplane lands on a runway traveling east. Describe its acceleration.
If we take east to be positive, then the airplane has negative acceleration because it is accelerating toward the west. It is also decelerating; its acceleration is opposite in direction to its velocity.
You are probably used to experiencing acceleration when you step into an elevator, or step on the gas pedal in your car. However, acceleration is happening to many other objects in our universe with which we don’t have direct contact. Figure presents the acceleration of various objects. We can see the magnitudes of the accelerations extend over many orders of magnitude.
Acceleration | Value (m/s^{2}) |
---|---|
High-speed train | 0.25 |
Elevator | 2 |
Cheetah | 5 |
Object in a free fall without air resistance near the surface of Earth | 9.8 |
Space shuttle maximum during launch | 29 |
Parachutist peak during normal opening of parachute | 59 |
F16 aircraft pulling out of a dive | 79 |
Explosive seat ejection from aircraft | 147 |
Sprint missile | 982 |
Fastest rocket sled peak acceleration | 1540 |
Jumping flea | 3200 |
Baseball struck by a bat | 30,000 |
Closing jaws of a trap-jaw ant | 1,000,000 |
Proton in the large Hadron collider | $latex 1.9\times {10}^{9} $ |
In this table, we see that typical accelerations vary widely with different objects and have nothing to do with object size or how massive it is. Acceleration can also vary widely with time during the motion of an object. A drag racer has a large acceleration just after its start, but then it tapers off as the vehicle reaches a constant velocity. Its average acceleration can be quite different from its instantaneous acceleration at a particular time during its motion. Figure compares graphically average acceleration with instantaneous acceleration for two very different motions.
Learn about position, velocity, and acceleration graphs. Move the little man back and forth with a mouse and plot his motion. Set the position, velocity, or acceleration and let the simulation move the man for you. Visit this link to use the moving man simulation.
Is it possible for speed to be constant while acceleration is not zero?
No, in one dimension constant speed requires zero acceleration.
Is it possible for velocity to be constant while acceleration is not zero? Explain.
Give an example in which velocity is zero yet acceleration is not.
A ball is thrown into the air and its velocity is zero at the apex of the throw, but acceleration is not zero.
If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceleration? Is the acceleration positive or negative?
Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?
Plus, minus
A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?
$latex a=4.29{\text{m/s}}^{2} $
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s and was brought jarringly back to rest in only 1.40 s. Calculate his (a) acceleration in his direction of motion and (b) acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s^{2}) by taking its ratio to the acceleration of gravity.
Sketch the acceleration-versus-time graph from the following velocity-versus-time graph.
A commuter backs her car out of her garage with an acceleration of 1.40 m/s^{2}. (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her acceleration?
Assume an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in meters per second and in multiples of g (9.80 m/s^{2})?
$latex a=11.1g $
An airplane, starting from rest, moves down the runway at constant acceleration for 18 s and then takes off at a speed of 60 m/s. What is the average acceleration of the plane?
You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car’s displacement in a given time. But, we have not developed a specific equation that relates acceleration and displacement. In this section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate the motion of two objects, called two-body pursuit problems.
First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is $latex \Delta t={t}_{\text{f}}-{t}_{0} $, taking $latex {t}_{0}=0 $ means that$latex \Delta t={t}_{\text{f}} $, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, $latex {x}_{0} $ is the initial position and $latex {v}_{0} $ is the initial velocity. We put no subscripts on the final values. That is, t is the final time, x is the final position, and v is the final velocity. This gives a simpler expression for elapsed time, $latex \Delta t=t $. It also simplifies the expression for x displacement, which is now $latex \Delta x=x-{x}_{0} $. Also, it simplifies the expression for change in velocity, which is now $latex \Delta v=v-{v}_{0} $. To summarize, using the simplified notation, with the initial time taken to be zero,
where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.
We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—that is,Thus, we can use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor does it degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can describe motion accurately by assuming a constant acceleration equal to the average acceleration for that motion. Lastly, for motion during which acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, motion can be considered in separate parts, each of which has its own constant acceleration.
To get our first two equations, we start with the definition of average velocity:
Substituting the simplified notation for $latex \Delta x $ and $latex \Delta t $ yields
Solving for x gives us
where the average velocity is
The equation $latex \overset{\text{–}}{v}=\frac{{v}_{0}+v}{2} $ reflects the fact that when acceleration is constant, v is just the simple average of the initial and final velocities. Figure illustrates this concept graphically. In part (a) of the figure, acceleration is constant, with velocity increasing at a constant rate. The average velocity during the 1-h interval from 40 km/h to 80 km/h is 60 km/h:
In part (b), acceleration is not constant. During the 1-h interval, velocity is closer to 80 km/h than 40 km/h. Thus, the average velocity is greater than in part (a).
We can derive another useful equation by manipulating the definition of acceleration:
Substituting the simplified notation for $latex \Delta v $ and $latex \Delta t $ gives us
Solving for v yields
Second, we identify the unknown; in this case, it is final velocity $latex {v}_{\text{f}} $.
Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknown in terms of the knowns. We calculate the final velocity using Figure, $latex v={v}_{0}+at $.
In addition to being useful in problem solving, the equation $latex v={v}_{0}+at $ gives us insight into the relationships among velocity, acceleration, and time. We can see, for example, that
All these observations fit our intuition. Note that it is always useful to examine basic equations in light of our intuition and experience to check that they do indeed describe nature accurately.
Adding $latex {v}_{0} $ to each side of this equation and dividing by 2 gives
Now we substitute this expression for $latex \overset{\text{–}}{v} $ into the equation for displacement, $latex x={x}_{0}+\overset{\text{–}}{v}t $, yielding
$latex x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}. $
Since the initial position and velocity are both zero, this equation simplifies to$latex x=\frac{1}{2}a{t}^{2}. $
Substituting the identified values of a and t gives$latex x=\frac{1}{2}(26.0{\,\text{m/s}}^{2}){(5.56\,\text{s})}^{2}=402\,\text{m}\text{.} $
What else can we learn by examining the equation $latex x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}? $ We can see the following relationships:
A fourth useful equation can be obtained from another algebraic manipulation of previous equations. If we solve $latex v={v}_{0}+at $ for t, we get
Substituting this and $latex \overset{\text{–}}{v}=\frac{{v}_{0}+v}{2} $ into $latex x={x}_{0}+\overset{\text{–}}{v}t $, we get
$latex {v}^{2}=0+2(26.0{\,\text{m/s}}^{2})(402\,\text{m}). $
Thus,$latex \begin{array}{ccc}{v}^{2}\hfill & =\hfill & 2.09\times {10}^{4}\,{\text{m}}^{2}{\text{/s}}^{2}\hfill \\ \\ v\hfill & =\hfill & \sqrt{2.09\times {10}^{4}{\,\text{m}}^{2}{\text{/s}}^{2}}=145\,\text{m/s}\text{.}\hfill \end{array} $
An examination of the equation $latex {v}^{2}={v}_{0}^{2}+2a(x-{x}_{0}) $ can produce additional insights into the general relationships among physical quantities:
In the following examples, we continue to explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The note that follows is provided for easy reference to the equations needed. Be aware that these equations are not independent. In many situations we have two unknowns and need two equations from the set to solve for the unknowns. We need as many equations as there are unknowns to solve a given situation.
Before we get into the examples, let’s look at some of the equations more closely to see the behavior of acceleration at extreme values. Rearranging Figure, we have
From this we see that, for a finite time, if the difference between the initial and final velocities is small, the acceleration is small, approaching zero in the limit that the initial and final velocities are equal. On the contrary, in the limit $latex t\to 0 $ for a finite difference between the initial and final velocities, acceleration becomes infinite.
Similarly, rearranging Figure, we can express acceleration in terms of velocities and displacement:
Thus, for a finite difference between the initial and final velocities acceleration becomes infinite in the limit the displacement approaches zero. Acceleration approaches zero in the limit the difference in initial and final velocities approaches zero for a finite displacement.
$latex 200\,\text{m}=0\,\text{m}+(10.0\,\text{m/s})t+\frac{1}{2}(2.00\,{\text{m/s}}^{2}){t}^{2}. $
We then simplify the equation. The units of meters cancel because they are in each term. We can get the units of seconds to cancel by taking t = t s, where t is the magnitude of time and s is the unit. Doing so leaves$latex 200=10t+{t}^{2}. $
We then use the quadratic formula to solve for t,$latex \begin{array}{c}{t}^{2}+10t-200=0\\ \\ \\ t=\frac{\text{−}b\pm\sqrt{{b}^{2}-4ac}}{2a},\end{array} $
which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would mean the event happened 20 s before the motion began. We can discard that solution. Thus,$latex t=10.0\,\text{s}\text{.} $
A manned rocket accelerates at a rate of 20 m/s^{2} during launch. How long does it take the rocket to reach a velocity of 400 m/s?
To answer this, choose an equation that allows us to solve for time t, given only a , v_{0} , and v:
$latex v={v}_{0}+at. $ Rearrange to solve for t: $latex t=\frac{v-{v}_{0}}{a}=\frac{400\,\text{m/s}-0\,\text{m/s}}{20{\,\text{m/s}}^{2}}=20\,\text{s}\text{.} $With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. The next level of complexity in our kinematics problems involves the motion of two interrelated bodies, called two-body pursuit problems.
Up until this point we have looked at examples of motion involving a single body. Even for the problem with two cars and the stopping distances on wet and dry roads, we divided this problem into two separate problems to find the answers. In a two-body pursuit problem, the motions of the objects are coupled—meaning, the unknown we seek depends on the motion of both objects. To solve these problems we write the equations of motion for each object and then solve them simultaneously to find the unknown. This is illustrated in Figure.
The time and distance required for car 1 to catch car 2 depends on the initial distance car 1 is from car 2 as well as the velocities of both cars and the acceleration of car 1. The kinematic equations describing the motion of both cars must be solved to find these unknowns.
Consider the following example.
A bicycle has a constant velocity of 10 m/s. A person starts from rest and runs to catch up to the bicycle in 30 s. What is the acceleration of the person?
$latex a=\frac{2}{3}{\,\text{m/s}}^{2} $.
When analyzing the motion of a single object, what is the required number of known physical variables that are needed to solve for the unknown quantities using the kinematic equations?
State two scenarios of the kinematics of single object where three known quantities require two kinematic equations to solve for the unknowns.
If the acceleration, time, and displacement are the knowns, and the initial and final velocities are the unknowns, then two kinematic equations must be solved simultaneously. Also if the final velocity, time, and displacement are the knowns then two kinematic equations must be solved for the initial velocity and acceleration.
A particle moves in a straight line at a constant velocity of 30 m/s. What is its displacement between t = 0 and t = 5.0 s?
150 m
A particle moves in a straight line with an initial velocity of 30 m/s and a constant acceleration of 30 m/s^{2}. If at $latex t=0,x=0 $ and $latex v=0 $, what is the particle’s position at t = 5 s?
A particle moves in a straight line with an initial velocity of 30 m/s and constant acceleration 30 m/s^{2}. (a) What is its displacement at t = 5 s? (b) What is its velocity at this same time?
a. 525 m;
b. $latex v=180\,\text{m/s} $(a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in the following figure. (b) Identify the time or times (t_{a}, t_{b}, t_{c}, etc.) at which the instantaneous velocity has the greatest positive value. (c) At which times is it zero? (d) At which times is it negative?
(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in the following figure. (b) Identify the time or times (t_{a}, t_{b}, t_{c}, etc.) at which the acceleration has the greatest positive value. (c) At which times is it zero? (d) At which times is it negative?
a.
b. The acceleration has the greatest positive value at $latex {t}_{a} $ c. The acceleration is zero at $latex {t}_{e}\,\text{and}\,{t}_{h} $ d. The acceleration is negative at $latex {t}_{i}\text{,}{t}_{j}\text{,}{t}_{k}\text{,}{t}_{l} $A particle has a constant acceleration of 6.0 m/s^{2}. (a) If its initial velocity is 2.0 m/s, at what time is its displacement 5.0 m? (b) What is its velocity at that time?
At t = 10 s, a particle is moving from left to right with a speed of 5.0 m/s. At t = 20 s, the particle is moving right to left with a speed of 8.0 m/s. Assuming the particle’s acceleration is constant, determine (a) its acceleration, (b) its initial velocity, and (c) the instant when its velocity is zero.
a. $latex a=-1.3{\,\text{m/s}}^{2} $;
b. $latex {v}_{0}=18\,\text{m/s} $; c. $latex t=13.8\,\text{s} $A well-thrown ball is caught in a well-padded mitt. If the acceleration of the ball is$latex 2.10\times {10}^{4}{\,\text{m/s}}^{2} $, and 1.85 ms $latex (1\,\text{ms}={10}^{-3}\,\text{s}) $ elapses from the time the ball first touches the mitt until it stops, what is the initial velocity of the ball?
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of $latex 6.20\times {10}^{5}{\,\text{m/s}}^{2} $ for $latex 8.10\times {10}^{\text{−}4}\,\text{s} $. What is its muzzle velocity (that is, its final velocity)?
$latex v=502.20\,\text{m/s} $
(a) A light-rail commuter train accelerates at a rate of 1.35 m/s^{2}. How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of 1.65 m/s^{2}. How long does it take to come to a stop from its top speed? (c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency acceleration in meters per second squared?
While entering a freeway, a car accelerates from rest at a rate of 2.04 m/s^{2} for 12.0 s. (a) Draw a sketch of the situation. (b) List the knowns in this problem. (c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, then indicate how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. (d) What is the car’s final velocity? Solve for this unknown in the same manner as in (c), showing all steps explicitly.
a.
b. Knowns: $latex a=2.40\,{\text{m/s}}^{2},t=12.0\,\text{s,}\,{v}_{0}=0\,\text{m/s} $, and $latex {x}_{0}=0\,\text{m} $; c. $latex x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}=\frac{1}{2}a{t}^{2}=2.40\,\text{m/}{\text{s}}^{2}{(12.0\,\text{s})}^{2}=172.80\,\text{m} $, the answer seems reasonable at about 172.8 m; d. $latex v=28.8\,\text{m/s} $Unreasonable results At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s^{2}. (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense?
Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?
a.
b. Knowns: $latex v=30.0\,\text{cm}\text{/}\text{s,}\,x=1.80\,\text{cm} $; c. $latex a=250\,\text{cm/}{\text{s}}^{2},\enspace t=0.12\,\text{s} $; d. yesDuring a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes $latex 3.33\times {10}^{\text{−}2}\,\text{s} $, what is the distance over which the puck accelerates?
A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time?
a. 6.87 s^{2}; b. $latex x=52.26\,\text{m} $
Freight trains can produce only relatively small accelerations. (a) What is the final velocity of a freight train that accelerates at a rate of $latex 0.0500\,{\text{m/s}}^{2} $ for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of $latex 0.550\,{\text{m/s}}^{2} $, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?
A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m. (a) Calculate the acceleration. (b) How long did the acceleration last?
a. $latex a=8450\,\text{m/}{\text{s}}^{2} $;
b. $latex t=0.0077\,\text{s} $A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of $latex 0.35\,{\text{m/s}}^{2} $, how far will it travel before becoming airborne? (b) How long does this take?
A woodpecker’s brain is specially protected from large accelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm. (a) Find the acceleration in meters per second squared and in multiples of g, where g = 9.80 m/s^{2}. (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less acceleration of the brain). What is the brain’s acceleration, expressed in multiples of g?
a. $latex a=9.18\,g; $
b. $latex t=6.67\times {10}^{-3}\,\text{s} $; c. $latex \begin{array}{cc} a=\text{−}40.0\,{\text{m/s}}^{2}\hfill \\ a=4.08\,g\hfill \end{array} $An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his acceleration? (b) How long does the collision last?
A care package is dropped out of a cargo plane and lands in the forest. If we assume the care package speed on impact is 54 m/s (123 mph), then what is its acceleration? Assume the trees and snow stops it over a distance of 3.0 m.
Knowns: $latex x=3\,\text{m,}\,v=0\,\text{m/s,}\enspace{v}_{0}=54\,\text{m/s} $. We want a, so we can use this equation: $latex a=\text{−}486\,{\text{m/s}}^{2} $.
An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of $latex 0.150\,{\text{m/s}}^{2} $ as it goes through. The station is 210.0 m long. (a) How fast is it going when the nose leaves the station? (b) How long is the nose of the train in the station? (c) If the train is 130 m long, what is the velocity of the end of the train as it leaves? (d) When does the end of the train leave the station?
Unreasonable results Dragsters can actually reach a top speed of 145.0 m/s in only 4.45 s. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402.0 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? (Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.)
a. $latex a=32.58\,\text{m/}{\text{s}}^{2} $;
b. $latex v=161.85\,\text{m/s} $; c. $latex v \gt {v}_{\text{max}} $, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have a greater acceleration in first gear than second gear than third gear, and so on. The acceleration would be greatest at the beginning, so it would not be accelerating at $latex 32.6{\,\text{m/s}}^{2} $ during the last few meters, but substantially less, and the final velocity would be less than $latex 162\,\text{m/s} $.An interesting application of Figure through Figure is called free fall, which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.
The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass. This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure.
In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.
For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall. The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity. Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.
Acceleration due to gravity is so important that its magnitude is given its own symbol, g. It is constant at any given location on Earth and has the average value
Although g varies from 9.78 m/s^{2} to 9.83 m/s^{2}, depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s^{2} rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value +g or −g depends on how we define our coordinate system. If we define the upward direction as positive, then $latex a=\text{−}g=-9.8\,{\text{m/s}}^{2}, $ and if we define the downward direction as positive, then $latex a=g=9.8\,{\text{m/s}}^{2} $.
We assume here that acceleration equals −g (with the positive direction upward).
$latex \begin{array}{cc} y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}\hfill \\ -98.0\,\text{m}=0-(4.9\,\text{m/s})t-\frac{1}{2}(9.8\,{\text{m/s}}^{2}){t}^{2}.\hfill \end{array} $
This simplifies to$latex {t}^{2}+t-20=0. $
This is a quadratic equation with roots $latex t=-5.0\mathrm{s}\,\text{and}\,t=4.0\mathrm{s} $. The positive root is the one we are interested in, since time $latex t=0 $ is the time when the ball is released at the top of the building. (The time $latex t=-5.0\mathrm{s} $ represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)$latex v={v}_{0}-gt=-4.9\,\text{m/s}-(9.8{\text{m/s}}^{2})(4.0\,\text{s})=-44.1\,\text{m/s}\text{.} $
A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?
It takes 2.47 s to hit the water. The quantity distance traveled increases faster.
Visit this site to learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y = bx) to see how they add to generate the polynomial curve.
What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down? Assume there is no air resistance.
An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration have the same sign on the way up as on the way down?
a. at the top of its trajectory; b. yes, at the top of its trajectory; c. yes
Suppose you throw a rock nearly straight up at a coconut in a palm tree and the rock just misses the coconut on the way up but hits the coconut on the way down. Neglecting air resistance and the slight horizontal variation in motion to account for the hit and miss of the coconut, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain.
The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the same, how many times higher could a safe fall on the Moon than on Earth (gravitational acceleration on the Moon is about one-sixth that of the Earth)?
Earth $latex v={v}_{0}-gt=\text{−}gt $; Moon $latex {v}^{\prime }=\frac{g}{6}{t}^{\prime }v={v}^{\prime }\enspace -gt=-\frac{g}{6}{t}^{\prime }\enspace{t}^{\prime }=6t $; Earth $latex y=-\frac{1}{2}g{t}^{2} $ Moon $latex {y}^{\prime }=-\frac{1}{2}\,\frac{g}{6}{(6t)}^{2}=-\frac{1}{2}g6{t}^{2}=-6(\frac{1}{2}g{t}^{2})=-6y $
How many times higher could an astronaut jump on the Moon than on Earth if her takeoff speed is the same in both locations (gravitational acceleration on the Moon is about on-sixth of that on Earth)?
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be $latex {y}_{0}=0 $.
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
a. $latex \begin{array}{cc} y=-8.23\,\text{m}\hfill \\ {v}_{1}=\text{−}18.9\,\text{m/s}\hfill \end{array} $;
b. $latex \begin{array}{cc} y=-18.9\,\text{m}\hfill \\ {v}_{2}=23.8\,\text{m/s}\hfill \end{array} $; c. $latex \begin{array}{cc} y=-32.0\,\text{m}\hfill \\ {v}_{3}=\text{−}28.7\,\text{m/s}\hfill \end{array} $; d. $latex \begin{array}{cc} y=-47.6\,\text{m}\hfill \\ {v}_{4}=\text{−}33.6\,\text{m/s}\hfill \end{array} $; e. $latex \begin{array}{cc} y=-65.6\,\text{m}\hfill \\ {v}_{5}=\text{−}38.5\,\text{m/s}\hfill \end{array} $A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.
a. Knowns: $latex a=\text{−}9.8\,{\text{m/s}}^{2}\enspace{v}_{0}=-1.4\,\text{m/s}t=1.8\,\text{s}\enspace{y}_{0}=0\,\text{m} $;
b. $latex y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y={v}_{0}t-\frac{1}{2}gt=-1.4\,\text{m}\text{/}\text{s}(1.8\,\text{sec})-\frac{1}{2}(9.8){(1.8\,\text{s})}^{2}=-18.4\,\text{m} $ and the origin is at the rescuers, who are 18.4 m above the water.Unreasonable results A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known, and identify its value. Then, identify the unknown and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long a time is the dolphin in the air? Neglect any effects resulting from his size or orientation.
A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?
a. $latex {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0v=0y=\frac{{v}_{0}^{2}}{2g}=\frac{{(4.0\,\text{m}\text{/}\text{s})}^{2}}{2(9.80)}=0.82\,\text{m} $; b. to the apex $latex v=0.41\,\text{s} $ times 2 to the board = 0.82 s from the board to the water $latex y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-1.80\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=4.0\,\text{m}\text{/}\text{s} $ $latex -1.8=4.0t-4.9{t}^{2}\enspace 4.9{t}^{2}-4.0t-1.80=0 $, solution to quadratic equation gives 1.13 s; c. $latex \begin{array}{c}{v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0\enspace\enspace{v}_{0}=4.0\,\text{m}\text{/}\text{s}y=-1.80\,\text{m}\hfill \\ v=7.16\,\text{m}\text{/}\text{s}\hfill \end{array} $
(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long a time would it take to reach the ground if it is thrown straight down with the same speed?
A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long a time does he have to get out of the way if the shot was released at a height of 2.20 m and he is 1.80 m tall?
Time to the apex: $latex t=1.12\,\text{s} $ times 2 equals 2.24 s to a height of 2.20 m. To 1.80 m in height is an additional 0.40 m. $latex \begin{array}{cc} y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-0.40\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=-11.0\,\text{m}\text{/}\text{s}\hfill \\ y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-0.40\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=-11.0\,\text{m}\text{/}\text{s}\hfill \\ -0.40=-11.0t-4.9{t}^{2}\enspace\text{or}\enspace 4.9{t}^{2}+11.0t-0.40=0\hfill \end{array} $.
Take the positive root, so the time to go the additional 0.4 m is 0.04 s. Total time is $latex 2.24\,\text{s}\,+0.04\,\text{s}\,=2.28\,\text{s} $.You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.0 m. How much additional time elapses before the ball passes the tree branch on the way back down?
A kangaroo can jump over an object 2.50 m high. (a) Considering just its vertical motion, calculate its vertical speed when it leaves the ground. (b) How long a time is it in the air?
a. $latex \begin{array}{cc} {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0v=0y=2.50\,\text{m}\hfill \\ {v}_{0}^{2}=2gy\Rightarrow {v}_{0}=\sqrt{2(9.80)(2.50)}=7.0\,\text{m}\text{/}\text{s}\hfill \end{array} $; b. $latex t=0.72\,\text{s} $ times 2 gives 1.44 s in the air
There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long a time will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335.0 m/s on this day.
a. $latex v=70.0\,\text{m}\text{/}\text{s} $; b. time heard after rock begins to fall: 0.75 s, time to reach the ground: 6.09 s
This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.
Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,
we can take the indefinite integral of both sides, finding
where C_{1} is a constant of integration. Since $latex \int \frac{d}{dt}v(t)dt=v(t) $, the velocity is given by
Similarly, the time derivative of the position function is the velocity function,
Thus, we can use the same mathematical manipulations we just used and find
where C_{2} is a second constant of integration.
We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in Figure, we find
If the initial velocity is v(0) = v_{0}, then
Then, C_{1} = v_{0} and
which is (Equation). Substituting this expression into Figure gives
Doing the integration, we find
If x(0) = x_{0}, we have
so, C_{2} = x_{0}. Substituting back into the equation for x(t), we finally have
which is (Equation).
We take t = 0 to be the time when the boat starts to decelerate.
$latex v(t)=\int a(t)dt+{C}_{1}=\int -\frac{1}{4}tdt+{C}_{1}=-\frac{1}{8}{t}^{2}+{C}_{1}. $
At t = 0 we have v(0) = 5.0 m/s = 0 + C1, so C1 = 5.0 m/s or $latex v(t)=5.0\,\text{m/}\text{s}-\frac{1}{8}{t}^{2} $.$latex v(t)=0=5.0\,\text{m/}\text{s}-\frac{1}{8}{t}^{2}\Rightarrow t=6.3\,\text{s} $
$latex x(t)=\int v(t)dt+{C}_{2}=\int (5.0-\frac{1}{8}{t}^{2})dt+{C}_{2}=5.0t-\frac{1}{24}{t}^{3}+{C}_{2}. $
At t = 0, we set x(0) = 0 = x0, since we are only interested in the displacement from when the boat starts to decelerate. We have$latex x(0)=0={C}_{2}. $
Therefore, the equation for the position is$latex x(t)=5.0t-\frac{1}{24}{t}^{3}. $
$latex x(6.3)=5.0(6.3)-\frac{1}{24}{(6.3)}^{3}=21.1\,\text{m}\text{.} $
A particle starts from rest and has an acceleration function $latex 5-10t{\text{m/s}}^{2} $. (a) What is the velocity function? (b) What is the position function? (c) When is the velocity zero?
Displacement | $latex \Delta x={x}_{\text{f}}-{x}_{\text{i}} $ |
Total displacement | $latex \Delta {x}_{\text{Total}}=\sum \Delta {x}_{\text{i}} $ |
Average velocity | $latex \overset{\text{–}}{v}=\frac{\Delta x}{\Delta t}=\frac{{x}_{2}-{x}_{1}}{{t}_{2}-{t}_{1}} $ |
Instantaneous velocity | $latex v(t)=\frac{dx(t)}{dt} $ |
Average speed | $latex \text{Average speed}=\overset{\text{–}}{s}=\frac{\text{Total distance}}{\text{Elapsed time}} $ |
Instantaneous speed | $latex \text{Instantaneous speed}=|v(t)| $ |
Average acceleration | $latex \overset{\text{–}}{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}} $ |
Instantaneous acceleration | $latex a(t)=\frac{dv(t)}{dt} $ |
Position from average velocity | $latex x={x}_{0}+\overset{\text{–}}{v}t $ |
Average velocity | $latex \overset{\text{–}}{v}=\frac{{v}_{0}+v}{2} $ |
Velocity from acceleration | $latex v={v}_{0}+at\enspace(\text{constant}\,a\text{)} $ |
Position from velocity and acceleration | $latex x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}\enspace(\text{constant}\,a\text{)} $ |
Velocity from distance | $latex {v}^{2}={v}_{0}^{2}+2a(x-{x}_{0})\enspace(\text{constant}\,a\text{)} $ |
Velocity of free fall | $latex v={v}_{0}-gt\,\text{(positive upward)} $ |
Height of free fall | $latex y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2} $ |
Velocity of free fall from height | $latex {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0}) $ |
Velocity from acceleration | $latex v(t)=\int a(t)dt+{C}_{1} $ |
Position from velocity | $latex x(t)=\int v(t)dt+{C}_{2} $ |
When given the acceleration function, what additional information is needed to find the velocity function and position function?
The acceleration of a particle varies with time according to the equation $latex a(t)=p{t}^{2}-q{t}^{3} $. Initially, the velocity and position are zero. (a) What is the velocity as a function of time? (b) What is the position as a function of time?
Between t = 0 and t = t_{0}, a rocket moves straight upward with an acceleration given by $latex a(t)=A-B{t}^{1\,\text{/}2} $, where A and B are constants. (a) If x is in meters and t is in seconds, what are the units of A and B? (b) If the rocket starts from rest, how does the velocity vary between t = 0 and t = t_{0}? (c) If its initial position is zero, what is the rocket’s position as a function of time during this same time interval?
a. $latex A={\text{m/s}}^{2}\enspace B={\text{m/s}}^{5\,\text{/}2} $;
b. $latex \begin{array}{cc} v(t)=\int a(t)dt+{C}_{1}=\int (A-B{t}^{1\,\text{/}2})dt+{C}_{1}=At-\frac{2}{3}B{t}^{3\,\text{/}2}+{C}_{1}\hfill \\ v(0)=0={C}_{1}\enspace\text{so}\enspace v({t}_{0})=A{t}_{0}-\frac{2}{3}B{t}_{0}^{\text{3/2}}\hfill \end{array} $; c. $latex \begin{array}{cc} x(t)=\int v(t)dt+{C}_{2}=\int (At-\frac{2}{3}B{t}^{3\,\text{/}2})dt+{C}_{2}=\frac{1}{2}A{t}^{2}-\frac{4}{15}B{t}^{5\,\text{/}2}+{C}_{2}\hfill \\ x(0)=0={C}_{2}\enspace\text{so}\enspace x({t}_{0})=\frac{1}{2}A{t}_{0}^{2}-\frac{4}{15}B{t}_{0}^{\text{5/2}}\hfill \end{array} $The velocity of a particle moving along the x-axis varies with time according to $latex v(t)=A+B{t}^{-1} $, where A = 2 m/s, B = 0.25 m, and $latex 1.0\,\text{s}\le t\le 8.0\,\text{s} $. Determine the acceleration and position of the particle at t = 2.0 s and t = 5.0 s. Assume that $latex x(t=1\,\text{s})=0 $.
A particle at rest leaves the origin with its velocity increasing with time according to v(t) = 3.2t m/s. At 5.0 s, the particle’s velocity starts decreasing according to [16.0 – 1.5(t – 5.0)] m/s. This decrease continues until t = 11.0 s, after which the particle’s velocity remains constant at 7.0 m/s. (a) What is the acceleration of the particle as a function of time? (b) What is the position of the particle at t = 2.0 s, t = 7.0 s, and t = 12.0 s?
a. $latex \begin{array}{cc} a(t)=3.2{\text{m/s}}^{2}\enspace t\le 5.0\,\text{s}\hfill \\ a(t)=1.5{\text{m/s}}^{2}\enspace 5.0\,\text{s}\le t\le 11.0\,\text{s}\hfill \\ a(t)=0{\text{m/s}}^{2}\enspace t \gt 11.0\,\text{s}\hfill \end{array} $;
b. $latex \begin{array}{cc} x(t)=\int v(t)dt+{C}_{2}=\int 3.2tdt+{C}_{2}=1.6{t}^{2}+{C}_{2}\hfill \\ \quad t\le 5.0\,\text{s}\hfill \\ x(0)=0\Rightarrow {C}_{2}=0\enspace\text{therefore,}\,x(2.0\,\text{s})=6.4\,\text{m}\hfill \\ x(t)=\int v(t)dt+{C}_{2}=\int [16.0-1.5(t-5.0)]dt+{C}_{2}=16t-1.5(\frac{{t}^{2}}{2}-5.0t)+{C}_{2}\hfill \\ 5.0\le t\le 11.0\,\text{s}\hfill \\ x(5\,\text{s})=1.6{(5.0)}^{2}=40\,\text{m}=16(5.0\,\text{s})-1.5(\frac{{5}^{2}}{2}-5.0(5.0))+{C}_{2}\hfill \\ \quad 40=98.75+{C}_{2}\Rightarrow {C}_{2}=-58.75\hfill \\ x(7.0\,\text{s})=16(7.0)-1.5(\frac{{7}^{2}}{2}-5.0(7))-58.75=69\,\text{m}\hfill \\ \quad x(t)=\int 7.0dt+{C}_{2}=7t+{C}_{2}\hfill \\ \phantom{\rule{1.5em}{0ex}}t\ge 11.0\,\text{s}\hfill \\ x(11.0\,\text{s})=16(11)-1.5(\frac{{11}^{2}}{2}-5.0(11))-58.75=109=7(11.0\,\text{s})+{C}_{2}\Rightarrow {C}_{2}=32\,\text{m}\hfill \\ \quad x(t)=7t+32\,\text{m}\hfill \\ \phantom{\rule{1.5em}{0ex}}x\ge 11.0\,\text{s}\Rightarrow x(12.0\,\text{s})=7(12)+32=116\,\text{m}\hfill \end{array} $Professional baseball player Nolan Ryan could pitch a baseball at approximately 160.0 km/h. At that average velocity, how long did it take a ball thrown by Ryan to reach home plate, which is 18.4 m from the pitcher’s mound? Compare this with the average reaction time of a human to a visual stimulus, which is 0.25 s.
An airplane leaves Chicago and makes the 3000-km trip to Los Angeles in 5.0 h. A second plane leaves Chicago one-half hour later and arrives in Los Angeles at the same time. Compare the average velocities of the two planes. Ignore the curvature of Earth and the difference in altitude between the two cities.
Take west to be the positive direction.
1st plane: $latex \overset{\text{–}}{\nu }=600\,\text{km/h} $ 2nd plane $latex \overset{\text{–}}{\nu }=667.0\,\text{km/h} $Unreasonable Results A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. If his average velocity is 24 km/h, how long did it take him to complete the trip? Is this a reasonable time?
An object has an acceleration of $latex +1.2\,{\text{cm/s}}^{2} $. At $latex t=4.0\,\text{s} $, its velocity is $latex -3.4\,\text{cm/s} $. Determine the object’s velocities at $latex t=1.0\,\text{s} $ and $latex t=6.0\,\text{s} $.
$latex a=\frac{v-{v}_{0}}{t-{t}_{0}} $, $latex t=0,\,a=\frac{-3.4\,\text{cm/s}-{v}_{0}}{4\,\text{s}}=1.2\,{\text{cm/s}}^{2}\Rightarrow {v}_{0}=-8.2\,\text{cm/s} $ $latex v={v}_{0}+at=-8.2+1.2\,t $; $latex v=-7.0\,\text{cm/s}\enspace v=-1.0\,\text{cm/s} $
A particle moves along the x-axis according to the equation $latex x(t)=2.0-4.0{t}^{2} $ m. What are the velocity and acceleration at $latex t=2.0 $ s and $latex t=5.0 $ s?
A particle moving at constant acceleration has velocities of $latex 2.0\,\text{m/s} $ at $latex t=2.0 $ s and $latex -7.6\,\text{m/s} $ at $latex t=5.2 $ s. What is the acceleration of the particle?
$latex a=-3\,{\text{m/s}}^{2} $
A train is moving up a steep grade at constant velocity (see following figure) when its caboose breaks loose and starts rolling freely along the track. After 5.0 s, the caboose is 30 m behind the train. What is the acceleration of the caboose?
An electron is moving in a straight line with a velocity of $latex 4.0\times {10}^{5} $ m/s. It enters a region 5.0 cm long where it undergoes an acceleration of $latex 6.0\times {10}^{12}\,{\text{m/s}}^{2} $ along the same straight line. (a) What is the electron’s velocity when it emerges from this region? b) How long does the electron take to cross the region?
a.
$latex v=8.7\times {10}^{5}\,\text{m/s} $; b. $latex t=7.8\times {10}^{-8}\,\text{s} $An ambulance driver is rushing a patient to the hospital. While traveling at 72 km/h, she notices the traffic light at the upcoming intersections has turned amber. To reach the intersection before the light turns red, she must travel 50 m in 2.0 s. (a) What minimum acceleration must the ambulance have to reach the intersection before the light turns red? (b) What is the speed of the ambulance when it reaches the intersection?
A motorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. Calculate (a) the acceleration of the motorcycle and (b) its velocity at the beginning and end of the 2-km trip.
$latex 1\,\text{km}={v}_{0}(80.0\,\text{s})+\frac{1}{2}a{(80.0)}^{2} $; $latex 2\,\text{km}={v}_{0}(200.0)+\frac{1}{2}a{(200.0)}^{2} $ solve simultaneously to get $latex a=-\frac{0.1}{2400.0}{\text{km/s}}^{2} $ and $latex {v}_{0}=0.014167\,\text{km/s} $, which is $latex 51.0\,\text{km/h} $. Velocity at the end of the trip is $latex v=21.0\,\text{km/h} $.
A cyclist travels from point A to point B in 10 min. During the first 2.0 min of her trip, she maintains a uniform acceleration of $latex 0.090\,{\text{m/s}}^{2} $. She then travels at constant velocity for the next 5.0 min. Next, she decelerates at a constant rate so that she comes to a rest at point B 3.0 min later. (a) Sketch the velocity-versus-time graph for the trip. (b) What is the acceleration during the last 3 min? (c) How far does the cyclist travel?
Two trains are moving at 30 m/s in opposite directions on the same track. The engineers see simultaneously that they are on a collision course and apply the brakes when they are 1000 m apart. Assuming both trains have the same acceleration, what must this acceleration be if the trains are to stop just short of colliding?
$latex a=-0.9\,{\text{m/s}}^{2} $
A 10.0-m-long truck moving with a constant velocity of 97.0 km/h passes a 3.0-m-long car moving with a constant velocity of 80.0 km/h. How much time elapses between the moment the front of the truck is even with the back of the car and the moment the back of the truck is even with the front of the car?
A police car waits in hiding slightly off the highway. A speeding car is spotted by the police car doing 40 m/s. At the instant the speeding car passes the police car, the police car accelerates from rest at 4 m/s^{2} to catch the speeding car. How long does it take the police car to catch the speeding car?
Equation for the speeding car: This car has a constant velocity, which is the average velocity, and is not accelerating, so use the equation for displacement with $latex {x}_{0}=0 $:$latex x={x}_{0}+\overset{\text{–}}{v}t=\overset{\text{–}}{v}t $; Equation for the police car: This car is accelerating, so use the equation for displacement with $latex {x}_{0}=0 $ and $latex {v}_{0}=0 $, since the police car starts from rest: $latex x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}=\frac{1}{2}a{t}^{2} $; Now we have an equation of motion for each car with a common parameter, which can be eliminated to find the solution. In this case, we solve for $latex t$. Step 1, eliminating $latex x$: $latex x=\overset{\text{–}}{v}t=\frac{1}{2}a{t}^{2} $; Step 2, solving for $latex t$: $latex t=\frac{2\overset{\text{–}}{v}}{a} $. The speeding car has a constant velocity of 40 m/s, which is its average velocity. The acceleration of the police car is 4 m/s^{2}. Evaluating t, the time for the police car to reach the speeding car, we have $latex t=\frac{2\overset{\text{–}}{v}}{a}=\frac{2(40)}{4}=20\,\text{s} $.
Pablo is running in a half marathon at a velocity of 3 m/s. Another runner, Jacob, is 50 meters behind Pablo with the same velocity. Jacob begins to accelerate at 0.05 m/s^{2}. (a) How long does it take Jacob to catch Pablo? (b) What is the distance covered by Jacob? (c) What is the final velocity of Jacob?
Unreasonable results A runner approaches the finish line and is 75 m away; her average speed at this position is 8 m/s. She decelerates at this point at 0.5 m/s^{2}. How long does it take her to cross the finish line from 75 m away? Is this reasonable?
At this acceleration she comes to a full stop in $latex t=\frac{-{v}_{0}}{a}=\frac{8}{0.5}=16\,\text{s} $, but the distance covered is $latex x=8\,\text{m/s(16}\,\text{s)}-\frac{1}{2}(0.5){(16\,\text{s})}^{2}=64\,\text{m} $, which is less than the distance she is away from the finish line, so she never finishes the race.
An airplane accelerates at 5.0 m/s^{2} for 30.0 s. During this time, it covers a distance of 10.0 km. What are the initial and final velocities of the airplane?
Compare the distance traveled of an object that undergoes a change in velocity that is twice its initial velocity with an object that changes its velocity by four times its initial velocity over the same time period. The accelerations of both objects are constant.
$latex {x}_{1}=\frac{3}{2}{v}_{0}t $
$latex {x}_{2}=\frac{5}{3}{x}_{1} $An object is moving east with a constant velocity and is at position $latex {x}_{0}\,\text{at}\,\text{time}\,{t}_{0}=0 $. (a) With what acceleration must the object have for its total displacement to be zero at a later time t ? (b) What is the physical interpretation of the solution in the case for $latex t\to \infty $?
A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball’s initial velocity?
$latex {v}_{0}=7.9\,\text{m/s} $ velocity at the bottom of the window.
$latex v=7.9\,\text{m/s} $ $latex {v}_{0}=14.1\,\text{m/s} $A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms $latex (3.50\times {10}^{-3}\,\text{s}) $ (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
a. $latex v=5.42\,\text{m/s} $;
b. $latex v=4.64\,\text{m/s} $; c. $latex a=2874.28\,{\text{m/s}}^{2} $; d. $latex (x-{x}_{0})=5.11\times {10}^{-3}\,\text{m} $Unreasonable results. A raindrop falls from a cloud 100 m above the ground. Neglect air resistance. What is the speed of the raindrop when it hits the ground? Is this a reasonable number?
Compare the time in the air of a basketball player who jumps 1.0 m vertically off the floor with that of a player who jumps 0.3 m vertically.
Consider the players fall from rest at the height 1.0 m and 0.3 m.
0.9 s 0.5 sSuppose that a person takes 0.5 s to react and move his hand to catch an object he has dropped. (a) How far does the object fall on Earth, where $latex g=9.8\,{\text{m/s}}^{2}? $ (b) How far does the object fall on the Moon, where the acceleration due to gravity is 1/6 of that on Earth?
A hot-air balloon rises from ground level at a constant velocity of 3.0 m/s. One minute after liftoff, a sandbag is dropped accidentally from the balloon. Calculate (a) the time it takes for the sandbag to reach the ground and (b) the velocity of the sandbag when it hits the ground.
a. $latex t=6.37\,\text{s} $ taking the positive root;
b. $latex v=59.5\,\text{m/s} $(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.
a. $latex y=4.9\,\text{m} $;
b. $latex v=38.3\,\text{m/s} $; c. $latex -33.3\,\text{m} $A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms $latex (8.00\times {10}^{-5}\,\text{s}) $ (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
An object is dropped from a roof of a building of height h. During the last second of its descent, it drops a distance h/3. Calculate the height of the building.
$latex h=\frac{1}{2}g{t}^{2} $, h = total height and time to drop to ground
$latex \frac{2}{3}h=\frac{1}{2}g{(t-1)}^{2} $ in t – 1 seconds it drops 2/3h $latex \frac{2}{3}(\frac{1}{2}g{t}^{2})=\frac{1}{2}g{(t-1)}^{2} $ or $latex \frac{{t}^{2}}{3}=\frac{1}{2}{(t-1)}^{2} $ $latex 0={t}^{2}-6t+3 $ $latex t=\frac{6\pm\sqrt{{6}^{2}-4\cdot 3}}{2}=3\pm\frac{\sqrt{24}}{2} $ t = 5.45 s and h = 145.5 m. Other root is less than 1 s. Check for t = 4.45 s $latex h=\frac{1}{2}g{t}^{2}=97.0 $ m $latex =\frac{2}{3}(145.5) $In a 100-m race, the winner is timed at 11.2 s. The second-place finisher’s time is 11.6 s. How far is the second-place finisher behind the winner when she crosses the finish line? Assume the velocity of each runner is constant throughout the race.
The position of a particle moving along the x-axis varies with time according to $latex x(t)=5.0{t}^{2}-4.0{t}^{3} $ m. Find (a) the velocity and acceleration of the particle as functions of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position.
a. $latex v(t)=10t-12{t}^{2}\text{m/s,}\,a(t)=10-24t\,{\text{m/s}}^{2} $;
b. $latex v(2\,\text{s})=-28\,\text{m/s,}\,a(2\,\text{s})=-38{\text{m/s}}^{2} $; c. The slope of the position function is zero or the velocity is zero. There are two possible solutions: t = 0, which gives x = 0, or t = 10.0/12.0 = 0.83 s, which gives x = 1.16 m. The second answer is the correct choice; d. 0.83 s (e) 1.16 mA cyclist sprints at the end of a race to clinch a victory. She has an initial velocity of 11.5 m/s and accelerates at a rate of 0.500 m/s^{2} for 7.00 s. (a) What is her final velocity? (b) The cyclist continues at this velocity to the finish line. If she is 300 m from the finish line when she starts to accelerate, how much time did she save? (c) The second-place winner was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. What was the difference in finish time in seconds between the winner and runner-up? How far back was the runner-up when the winner crossed the finish line?
In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 295.38 km/h. The one-way course was 8.00 km long. Acceleration rates are often described by the time it takes to reach 96.0 km/h from rest. If this time was 4.00 s and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?
$latex 96\,\text{km/h}=26.67\,\text{m/s,}\,a=\frac{26.67\,\text{m/s}}{4.0\,\text{s}}=6.67{\text{m/s}}^{2} $, 295.38 km/h = 82.05 m/s, $latex t=12.3\,\text{s} $ time to accelerate to maximum speed
$latex x=504.55\,\text{m} $ distance covered during acceleration $latex 7495.44\,\text{m} $ at a constant speed $latex \frac{7495.44\,\text{m}}{82.05\,\text{m/s}}=91.35\,\text{s} $ so total time is $latex 91.35\,\text{s}+12.3\,\text{s}=103.65\,\text{s} $.Displacement | $latex \Delta x={x}_{\text{f}}-{x}_{\text{i}} $ |
Total displacement | $latex \Delta {x}_{\text{Total}}=\sum \Delta {x}_{\text{i}} $ |
Average velocity | $latex \overset{\text{–}}{v}=\frac{\Delta x}{\Delta t}=\frac{{x}_{2}-{x}_{1}}{{t}_{2}-{t}_{1}} $ |
Instantaneous velocity | $latex v(t)=\frac{dx(t)}{dt} $ |
Average speed | $latex \text{Average speed}=\overset{\text{–}}{s}=\frac{\text{Total distance}}{\text{Elapsed time}} $ |
Instantaneous speed | $latex \text{Instantaneous speed}=|v(t)| $ |
Average acceleration | $latex \overset{\text{–}}{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}} $ |
Instantaneous acceleration | $latex a(t)=\frac{dv(t)}{dt} $ |
Position from average velocity | $latex x={x}_{0}+\overset{\text{–}}{v}t $ |
Average velocity | $latex \overset{\text{–}}{v}=\frac{{v}_{0}+v}{2} $ |
Velocity from acceleration | $latex v={v}_{0}+at\enspace(\text{constant}\,a\text{)} $ |
Position from velocity and acceleration | $latex x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}\enspace(\text{constant}\,a\text{)} $ |
Velocity from distance | $latex {v}^{2}={v}_{0}^{2}+2a(x-{x}_{0})\enspace(\text{constant}\,a\text{)} $ |
Velocity of free fall | $latex v={v}_{0}-gt\,\text{(positive upward)} $ |
Height of free fall | $latex y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2} $ |
Velocity of free fall from height | $latex {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0}) $ |
Velocity from acceleration | $latex v(t)=\int a(t)dt+{C}_{1} $ |
Position from velocity | $latex x(t)=\int v(t)dt+{C}_{2} $ |
Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Identify each quantity in your example specifically.
Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of displacement and displacement are exactly the same?
Bacteria move back and forth using their flagella (structures that look like little tails). Speeds of up to 50 μm/s (50 × 10^{−6}m/s) have been observed. The total distance traveled by a bacterium is large for its size, whereas its displacement is small. Why is this?
Give an example of a device used to measure time and identify what change in that device indicates a change in time.
During a given time interval the average velocity of an object is zero. What can you say conclude about its displacement over the time interval?
There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.
Does the speedometer of a car measure speed or velocity?
If you divide the total distance traveled on a car trip (as determined by the odometer) by the elapsed time of the trip, are you calculating average speed or magnitude of average velocity? Under what circumstances are these two quantities the same?
How are instantaneous velocity and instantaneous speed related to one another? How do they differ?
Is it possible for velocity to be constant while acceleration is not zero? Explain.
If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceleration? Is the acceleration positive or negative?
Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?
When analyzing the motion of a single object, what is the required number of known physical variables that are needed to solve for the unknown quantities using the kinematic equations?
State two scenarios of the kinematics of single object where three known quantities require two kinematic equations to solve for the unknowns.
What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down? Assume there is no air resistance.
An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration have the same sign on the way up as on the way down?
Suppose you throw a rock nearly straight up at a coconut in a palm tree and the rock just misses the coconut on the way up but hits the coconut on the way down. Neglecting air resistance and the slight horizontal variation in motion to account for the hit and miss of the coconut, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain.
The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the same, how many times higher could a safe fall on the Moon than on Earth (gravitational acceleration on the Moon is about one-sixth that of the Earth)?
How many times higher could an astronaut jump on the Moon than on Earth if her takeoff speed is the same in both locations (gravitational acceleration on the Moon is about on-sixth of that on Earth)?
When given the acceleration function, what additional information is needed to find the velocity function and position function?
Consider a coordinate system in which the positive x axis is directed upward vertically. What are the positions of a particle (a) 5.0 m directly above the origin and (b) 2.0 m below the origin?
A car is 2.0 km west of a traffic light at t = 0 and 5.0 km east of the light at t = 6.0 min. Assume the origin of the coordinate system is the light and the positive x direction is eastward. (a) What are the car’s position vectors at these two times? (b) What is the car’s displacement between 0 min and 6.0 min?
The Shanghai maglev train connects Longyang Road to Pudong International Airport, a distance of 30 km. The journey takes 8 minutes on average. What is the maglev train’s average velocity?
The position of a particle moving along the x-axis is given by x(t)=4.0−2.0tx(t)=4.0−2.0t m. (a) At what time does the particle cross the origin? (b) What is the displacement of the particle between t=3.0st=3.0s and t=6.0s?t=6.0s?
A cyclist rides 8.0 km east for 20 minutes, then he turns and heads west for 8 minutes and 3.2 km. Finally, he rides east for 16 km, which takes 40 minutes. (a) What is the final displacement of the cyclist? (b) What is his average velocity?
On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth’s atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity of the blast wave? b) Compare this with the speed of sound, which is 343 m/s at sea level.
A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the woodchuck? (b) What is its average speed?
Sketch the velocity-versus-time graph from the following position-versus-time graph.
An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.