Main Fundamentals of Physics
Fundamentals of PhysicsDavid Halliday, Robert Resnick, Jearl Walker
This book arms engineers with the tools to apply key physics concepts in the field. A number of the key figures in the new edition are revised to provide a more inviting and informative treatment. The figures are broken into component parts with supporting commentary so that they can more readily see the key ideas. Material from The Flying Circus is incorporated into the chapter opener puzzlers, sample problems, examples and end-of-chapter problems to make the subject more engaging. Checkpoints enable them to check their understanding of a question with some reasoning based on the narrative or sample problem they just read. Sample Problems also demonstrate how engineers can solve problems with reasoned solutions.
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P U Z Z L E R For thousands of years the spinning Earth provided a natural standard for our measurements of time. However, since 1972 we have added more than 20 “leap seconds” to our clocks to keep them synchronized to the Earth. Why are such adjustments needed? What does it take to be a good standard? (Don Mason/The Stock Market and NASA) c h a p t e r Physics and Measurement Chapter Outline 1.1 Standards of Length, Mass, and Time 1.2 The Building Blocks of Matter 1.3 Density 1.4 Dimensional Analysis 2 1.5 Conversion of Units 1.6 Estimates and Order-of-Magnitude Calculations 1.7 Signiﬁcant Figures L ike all other sciences, physics is based on experimental observations and quantitative measurements. The main objective of physics is to ﬁnd the limited number of fundamental laws that govern natural phenomena and to use them to develop theories that can predict the results of future experiments. The fundamental laws used in developing theories are expressed in the language of mathematics, the tool that provides a bridge between theory and experiment. When a discrepancy between theory and experiment arises, new theories must be formulated to remove the discrepancy. Many times a theory is satisfactory only under limited conditions; a more general theory might be satisfactory without such limitations. For example, the laws of motion discovered by Isaac Newton (1642 – 1727) in the 17th century accurately describe the motion of bodies at normal speeds but do not apply to objects moving at speeds comparable with the speed of light. In contrast, the special theory of relativity developed by Albert Einstein (1879 – 1955) in the early 1900s gives the same results as Newton’s laws at low speeds but also correctly describes motion at speeds approaching the speed of light. Hence, Einstein’s is a more general theory of motion. Classical physics, which means all of the physics developed before 1900, includes the theories, concepts, laws, and experiments in classical mechanics, thermodynamics, and electromagnetism. Important contributions to classical physics were provided by Newton, who developed classical mechanics as a systematic theory and was one of the originators of calculus as a mathematical tool. Major developments in mechanics continued in the 18th century, but the ﬁelds of thermodynamics and electricity and magnetism were not developed until the latter part of the 19th century, principally because before that time the apparatus for controlled experiments was either too crude or unavailable. A new era in physics, usually referred to as modern physics, began near the end of the 19th century. Modern physics developed mainly because of the discovery that many physical phenomena could not be explained by classical physics. The two most important developments in modern physics were the theories of relativity and quantum mechanics. Einstein’s theory of relativity revolutionized the traditional concepts of space, time, and energy; quantum mechanics, which applies to both the microscopic and macroscopic worlds, was originally formulated by a number of distinguished scientists to provide descriptions of physical phenomena at the atomic level. Scientists constantly work at improving our understanding of phenomena and fundamental laws, and new discoveries are made every day. In many research areas, a great deal of overlap exists between physics, chemistry, geology, and biology, as well as engineering. Some of the most notable developments are (1) numerous space missions and the landing of astronauts on the Moon, (2) microcircuitry and high-speed computers, and (3) sophisticated imaging techniques used in scientiﬁc research and medicine. The impact such developments and discoveries have had on our society has indeed been great, and it is very likely that future discoveries and developments will be just as exciting and challenging and of great beneﬁt to humanity. 1.1 STANDARDS OF LENGTH, MASS, AND TIME The laws of physics are expressed in terms of basic quantities that require a clear definition. In mechanics, the three basic quantities are length (L), mass (M), and time (T). All other quantities in mechanics can be expressed in terms of these three. 3 4 CHAPTER 1 Physics and Measurements If we are to report the results of a measurement to someone who wishes to reproduce this measurement, a standard must be deﬁned. It would be meaningless if a visitor from another planet were to talk to us about a length of 8 “glitches” if we do not know the meaning of the unit glitch. On the other hand, if someone familiar with our system of measurement reports that a wall is 2 meters high and our unit of length is deﬁned to be 1 meter, we know that the height of the wall is twice our basic length unit. Likewise, if we are told that a person has a mass of 75 kilograms and our unit of mass is deﬁned to be 1 kilogram, then that person is 75 times as massive as our basic unit.1 Whatever is chosen as a standard must be readily accessible and possess some property that can be measured reliably — measurements taken by different people in different places must yield the same result. In 1960, an international committee established a set of standards for length, mass, and other basic quantities. The system established is an adaptation of the metric system, and it is called the SI system of units. (The abbreviation SI comes from the system’s French name “Système International.”) In this system, the units of length, mass, and time are the meter, kilogram, and second, respectively. Other SI standards established by the committee are those for temperature (the kelvin), electric current (the ampere), luminous intensity (the candela), and the amount of substance (the mole). In our study of mechanics we shall be concerned only with the units of length, mass, and time. Length In A.D. 1120 the king of England decreed that the standard of length in his country would be named the yard and would be precisely equal to the distance from the tip of his nose to the end of his outstretched arm. Similarly, the original standard for the foot adopted by the French was the length of the royal foot of King Louis XIV. This standard prevailed until 1799, when the legal standard of length in France became the meter, deﬁned as one ten-millionth the distance from the equator to the North Pole along one particular longitudinal line that passes through Paris. Many other systems for measuring length have been developed over the years, but the advantages of the French system have caused it to prevail in almost all countries and in scientiﬁc circles everywhere. As recently as 1960, the length of the meter was deﬁned as the distance between two lines on a speciﬁc platinum – iridium bar stored under controlled conditions in France. This standard was abandoned for several reasons, a principal one being that the limited accuracy with which the separation between the lines on the bar can be determined does not meet the current requirements of science and technology. In the 1960s and 1970s, the meter was deﬁned as 1 650 763.73 wavelengths of orange-red light emitted from a krypton-86 lamp. However, in October 1983, the meter (m) was redeﬁned as the distance traveled by light in vacuum during a time of 1/299 792 458 second. In effect, this latest deﬁnition establishes that the speed of light in vacuum is precisely 299 792 458 m per second. Table 1.1 lists approximate values of some measured lengths. 1 The need for assigning numerical values to various measured physical quantities was expressed by Lord Kelvin (William Thomson) as follows: “I often say that when you can measure what you are speaking about, and express it in numbers, you should know something about it, but when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind. It may be the beginning of knowledge but you have scarcely in your thoughts advanced to the state of science.” 1.1 5 Standards of Length, Mass, and Time TABLE 1.1 Approximate Values of Some Measured Lengths Length (m) Distance from the Earth to most remote known quasar Distance from the Earth to most remote known normal galaxies Distance from the Earth to nearest large galaxy (M 31, the Andromeda galaxy) Distance from the Sun to nearest star (Proxima Centauri) One lightyear Mean orbit radius of the Earth about the Sun Mean distance from the Earth to the Moon Distance from the equator to the North Pole Mean radius of the Earth Typical altitude (above the surface) of a satellite orbiting the Earth Length of a football ﬁeld Length of a houseﬂy Size of smallest dust particles Size of cells of most living organisms Diameter of a hydrogen atom Diameter of an atomic nucleus Diameter of a proton 1.4 ⫻ 1026 9 ⫻ 1025 2 ⫻ 1022 4 ⫻ 1016 9.46 ⫻ 1015 1.50 ⫻ 1011 3.84 ⫻ 108 1.00 ⫻ 107 6.37 ⫻ 106 2 ⫻ 105 9.1 ⫻ 101 5 ⫻ 10⫺3 ⬃ 10⫺4 ⬃ 10⫺5 ⬃ 10⫺10 ⬃ 10⫺14 ⬃ 10⫺15 Mass The basic SI unit of mass, the kilogram (kg), is deﬁned as the mass of a speciﬁc platinum – iridium alloy cylinder kept at the International Bureau of Weights and Measures at Sèvres, France. This mass standard was established in 1887 and has not been changed since that time because platinum – iridium is an unusually stable alloy (Fig. 1.1a). A duplicate of the Sèvres cylinder is kept at the National Institute of Standards and Technology (NIST) in Gaithersburg, Maryland. Table 1.2 lists approximate values of the masses of various objects. web Visit the Bureau at www.bipm.fr or the National Institute of Standards at www.NIST.gov TABLE 1.2 Time Before 1960, the standard of time was deﬁned in terms of the mean solar day for the 1 1 1 )(60 )(24 ) of a mean year 1900.2 The mean solar second was originally deﬁned as (60 solar day. The rotation of the Earth is now known to vary slightly with time, however, and therefore this motion is not a good one to use for deﬁning a standard. In 1967, consequently, the second was redeﬁned to take advantage of the high precision obtainable in a device known as an atomic clock (Fig. 1.1b). In this device, the frequencies associated with certain atomic transitions can be measured to a precision of one part in 1012. This is equivalent to an uncertainty of less than one second every 30 000 years. Thus, in 1967 the SI unit of time, the second, was redeﬁned using the characteristic frequency of a particular kind of cesium atom as the “reference clock.” The basic SI unit of time, the second (s), is deﬁned as 9 192 631 770 times the period of vibration of radiation from the cesium-133 atom.3 To keep these atomic clocks — and therefore all common clocks and 2 One solar day is the time interval between successive appearances of the Sun at the highest point it reaches in the sky each day. 3 Period is deﬁned as the time interval needed for one complete vibration. Masses of Various Bodies (Approximate Values) Body Visible Universe Milky Way galaxy Sun Earth Moon Horse Human Frog Mosquito Bacterium Hydrogen atom Electron Mass (kg) ⬃ 1052 7 ⫻ 1041 1.99 ⫻ 1030 5.98 ⫻ 1024 7.36 ⫻ 1022 ⬃ 103 ⬃ 102 ⬃ 10⫺1 ⬃ 10⫺5 ⬃ 10⫺15 1.67 ⫻ 10⫺27 9.11 ⫻ 10⫺31 6 CHAPTER 1 Physics and Measurements Figure 1.1 (Top) The National Standard Kilogram No. 20, an accurate copy of the International Standard Kilogram kept at Sèvres, France, is housed under a double bell jar in a vault at the National Institute of Standards and Technology (NIST). (Bottom) The primary frequency standard (an atomic clock) at the NIST. This device keeps time with an accuracy of about 3 millionths of a second per year. (Courtesy of National Institute of Standards and Technology, U.S. Department of Commerce) watches that are set to them — synchronized, it has sometimes been necessary to add leap seconds to our clocks. This is not a new idea. In 46 B.C. Julius Caesar began the practice of adding extra days to the calendar during leap years so that the seasons occurred at about the same date each year. Since Einstein’s discovery of the linkage between space and time, precise measurement of time intervals requires that we know both the state of motion of the clock used to measure the interval and, in some cases, the location of the clock as well. Otherwise, for example, global positioning system satellites might be unable to pinpoint your location with sufﬁcient accuracy, should you need rescuing. Approximate values of time intervals are presented in Table 1.3. In addition to SI, another system of units, the British engineering system (sometimes called the conventional system), is still used in the United States despite acceptance of SI by the rest of the world. In this system, the units of length, mass, and 1.1 Standards of Length, Mass, and Time TABLE 1.3 Approximate Values of Some Time Intervals Interval (s) Age of the Universe Age of the Earth Average age of a college student One year One day (time for one rotation of the Earth about its axis) Time between normal heartbeats Period of audible sound waves Period of typical radio waves Period of vibration of an atom in a solid Period of visible light waves Duration of a nuclear collision Time for light to cross a proton 5 ⫻ 1017 1.3 ⫻ 1017 6.3 ⫻ 108 3.16 ⫻ 107 8.64 ⫻ 104 8 ⫻ 10⫺1 ⬃ 10⫺3 ⬃ 10⫺6 ⬃ 10⫺13 ⬃ 10⫺15 ⬃ 10⫺22 ⬃ 10⫺24 time are the foot (ft), slug, and second, respectively. In this text we shall use SI units because they are almost universally accepted in science and industry. We shall make some limited use of British engineering units in the study of classical mechanics. In addition to the basic SI units of meter, kilogram, and second, we can also use other units, such as millimeters and nanoseconds, where the preﬁxes milli- and nano- denote various powers of ten. Some of the most frequently used preﬁxes for the various powers of ten and their abbreviations are listed in Table 1.4. For TABLE 1.4 Preﬁxes for SI Units Power Preﬁx Abbreviation 10⫺24 10⫺21 10⫺18 10⫺15 10⫺12 10⫺9 10⫺6 10⫺3 10⫺2 10⫺1 101 103 106 109 1012 1015 1018 1021 1024 yocto zepto atto femto pico nano micro milli centi deci deka kilo mega giga tera peta exa zetta yotta y z a f p n m c d da k M G T P E Z Y 7 8 CHAPTER 1 Physics and Measurements example, 10⫺3 m is equivalent to 1 millimeter (mm), and 103 m corresponds to 1 kilometer (km). Likewise, 1 kg is 103 grams (g), and 1 megavolt (MV) is 106 volts (V). u u 1.2 d Quark composition of a proton Proton Neutron Gold nucleus Nucleus Gold atoms Gold cube Figure 1.2 Levels of organization in matter. Ordinary matter consists of atoms, and at the center of each atom is a compact nucleus consisting of protons and neutrons. Protons and neutrons are composed of quarks. The quark composition of a proton is shown. THE BUILDING BLOCKS OF MATTER A 1-kg cube of solid gold has a length of 3.73 cm on a side. Is this cube nothing but wall-to-wall gold, with no empty space? If the cube is cut in half, the two pieces still retain their chemical identity as solid gold. But what if the pieces are cut again and again, indeﬁnitely? Will the smaller and smaller pieces always be gold? Questions such as these can be traced back to early Greek philosophers. Two of them — Leucippus and his student Democritus — could not accept the idea that such cuttings could go on forever. They speculated that the process ultimately must end when it produces a particle that can no longer be cut. In Greek, atomos means “not sliceable.” From this comes our English word atom. Let us review brieﬂy what is known about the structure of matter. All ordinary matter consists of atoms, and each atom is made up of electrons surrounding a central nucleus. Following the discovery of the nucleus in 1911, the question arose: Does it have structure? That is, is the nucleus a single particle or a collection of particles? The exact composition of the nucleus is not known completely even today, but by the early 1930s a model evolved that helped us understand how the nucleus behaves. Speciﬁcally, scientists determined that occupying the nucleus are two basic entities, protons and neutrons. The proton carries a positive charge, and a speciﬁc element is identiﬁed by the number of protons in its nucleus. This number is called the atomic number of the element. For instance, the nucleus of a hydrogen atom contains one proton (and so the atomic number of hydrogen is 1), the nucleus of a helium atom contains two protons (atomic number 2), and the nucleus of a uranium atom contains 92 protons (atomic number 92). In addition to atomic number, there is a second number characterizing atoms — mass number, deﬁned as the number of protons plus neutrons in a nucleus. As we shall see, the atomic number of an element never varies (i.e., the number of protons does not vary) but the mass number can vary (i.e., the number of neutrons varies). Two or more atoms of the same element having different mass numbers are isotopes of one another. The existence of neutrons was veriﬁed conclusively in 1932. A neutron has no charge and a mass that is about equal to that of a proton. One of its primary purposes is to act as a “glue” that holds the nucleus together. If neutrons were not present in the nucleus, the repulsive force between the positively charged particles would cause the nucleus to come apart. But is this where the breaking down stops? Protons, neutrons, and a host of other exotic particles are now known to be composed of six different varieties of particles called quarks, which have been given the names of up, down, strange, charm, bottom, and top. The up, charm, and top quarks have charges of ⫹ 32 that of the proton, whereas the down, strange, and bottom quarks have charges of ⫺ 13 that of the proton. The proton consists of two up quarks and one down quark (Fig. 1.2), which you can easily show leads to the correct charge for the proton. Likewise, the neutron consists of two down quarks and one up quark, giving a net charge of zero. 1.3 1.3 9 Density DENSITY A property of any substance is its density (Greek letter rho), deﬁned as the amount of mass contained in a unit volume, which we usually express as mass per unit volume: ⬅ m V (1.1) For example, aluminum has a density of 2.70 g/cm3, and lead has a density of 11.3 g/cm3. Therefore, a piece of aluminum of volume 10.0 cm3 has a mass of 27.0 g, whereas an equivalent volume of lead has a mass of 113 g. A list of densities for various substances is given Table 1.5. The difference in density between aluminum and lead is due, in part, to their different atomic masses. The atomic mass of an element is the average mass of one atom in a sample of the element that contains all the element’s isotopes, where the relative amounts of isotopes are the same as the relative amounts found in nature. The unit for atomic mass is the atomic mass unit (u), where 1 u ⫽ 1.660 540 2 ⫻ 10⫺27 kg. The atomic mass of lead is 207 u, and that of aluminum is 27.0 u. However, the ratio of atomic masses, 207 u/27.0 u ⫽ 7.67, does not correspond to the ratio of densities, (11.3 g/cm3)/(2.70 g/cm3) ⫽ 4.19. The discrepancy is due to the difference in atomic separations and atomic arrangements in the crystal structure of these two substances. The mass of a nucleus is measured relative to the mass of the nucleus of the carbon-12 isotope, often written as 12C. (This isotope of carbon has six protons and six neutrons. Other carbon isotopes have six protons but different numbers of neutrons.) Practically all of the mass of an atom is contained within the nucleus. Because the atomic mass of 12C is deﬁned to be exactly 12 u, the proton and neutron each have a mass of about 1 u. One mole (mol) of a substance is that amount of the substance that contains as many particles (atoms, molecules, or other particles) as there are atoms in 12 g of the carbon-12 isotope. One mole of substance A contains the same number of particles as there are in 1 mol of any other substance B. For example, 1 mol of aluminum contains the same number of atoms as 1 mol of lead. TABLE 1.5 Densities of Various Substances Substance Gold Uranium Lead Copper Iron Aluminum Magnesium Water Air Density (103 kg/m3) 19.3 18.7 11.3 8.92 7.86 2.70 1.75 1.00 0.0012 A table of the letters in the Greek alphabet is provided on the back endsheet of this textbook. 10 CHAPTER 1 Physics and Measurements Experiments have shown that this number, known as Avogadro’s number, NA , is NA ⫽ 6.022 137 ⫻ 10 23 particles/mol Avogadro’s number is deﬁned so that 1 mol of carbon-12 atoms has a mass of exactly 12 g. In general, the mass in 1 mol of any element is the element’s atomic mass expressed in grams. For example, 1 mol of iron (atomic mass ⫽ 55.85 u) has a mass of 55.85 g (we say its molar mass is 55.85 g/mol), and 1 mol of lead (atomic mass ⫽ 207 u) has a mass of 207 g (its molar mass is 207 g/mol). Because there are 6.02 ⫻ 1023 particles in 1 mol of any element, the mass per atom for a given element is m atom ⫽ molar mass NA (1.2) For example, the mass of an iron atom is m Fe ⫽ EXAMPLE 1.1 55.85 g/mol ⫽ 9.28 ⫻ 10 ⫺23 g/atom 6.02 ⫻ 10 23 atoms/mol How Many Atoms in the Cube? A solid cube of aluminum (density 2.7 g/cm3) has a volume of 0.20 cm3. How many aluminum atoms are contained in the cube? minum (27 g) contains 6.02 ⫻ 1023 atoms: NA N ⫽ 27 g 0.54 g Solution Since density equals mass per unit volume, the mass m of the cube is m ⫽ V ⫽ (2.7 g/cm3)(0.20 cm3) ⫽ 0.54 g To ﬁnd the number of atoms N in this mass of aluminum, we can set up a proportion using the fact that one mole of alu- 1.4 6.02 ⫻ 10 23 atoms N ⫽ 27 g 0.54 g N⫽ (0.54 g)(6.02 ⫻ 10 23 atoms) 27 g ⫽ 1.2 ⫻ 10 22 atoms DIMENSIONAL ANALYSIS The word dimension has a special meaning in physics. It usually denotes the physical nature of a quantity. Whether a distance is measured in the length unit feet or the length unit meters, it is still a distance. We say the dimension — the physical nature — of distance is length. The symbols we use in this book to specify length, mass, and time are L, M, and T, respectively. We shall often use brackets [ ] to denote the dimensions of a physical quantity. For example, the symbol we use for speed in this book is v, and in our notation the dimensions of speed are written [v] ⫽ L/T. As another example, the dimensions of area, for which we use the symbol A, are [A] ⫽ L2. The dimensions of area, volume, speed, and acceleration are listed in Table 1.6. In solving problems in physics, there is a useful and powerful procedure called dimensional analysis. This procedure, which should always be used, will help minimize the need for rote memorization of equations. Dimensional analysis makes use of the fact that dimensions can be treated as algebraic quantities. That is, quantities can be added or subtracted only if they have the same dimensions. Furthermore, the terms on both sides of an equation must have the same dimensions. 1.4 Dimensional Analysis TABLE 1.6 Dimensions and Common Units of Area, Volume, Speed, and Acceleration System SI British engineering Area (L2) Volume (L3) Speed (L/T) Acceleration (L/T 2) m2 ft2 m3 ft3 m/s ft/s m/s2 ft/s2 By following these simple rules, you can use dimensional analysis to help determine whether an expression has the correct form. The relationship can be correct only if the dimensions are the same on both sides of the equation. To illustrate this procedure, suppose you wish to derive a formula for the distance x traveled by a car in a time t if the car starts from rest and moves with constant acceleration a. In Chapter 2, we shall ﬁnd that the correct expression is x ⫽ 12at 2. Let us use dimensional analysis to check the validity of this expression. The quantity x on the left side has the dimension of length. For the equation to be dimensionally correct, the quantity on the right side must also have the dimension of length. We can perform a dimensional check by substituting the dimensions for acceleration, L/T 2, and time, T, into the equation. That is, the dimensional form of the equation x ⫽ 12at 2 is L ⭈T 2 ⫽ L T2 L⫽ The units of time squared cancel as shown, leaving the unit of length. A more general procedure using dimensional analysis is to set up an expression of the form x ⬀ a nt m where n and m are exponents that must be determined and the symbol ⬀ indicates a proportionality. This relationship is correct only if the dimensions of both sides are the same. Because the dimension of the left side is length, the dimension of the right side must also be length. That is, [a nt m] ⫽ L ⫽ LT 0 Because the dimensions of acceleration are L/T 2 and the dimension of time is T, we have 冢 TL 冣 T n 2 m ⫽ L1 Ln T m⫺2n ⫽ L1 Because the exponents of L and T must be the same on both sides, the dimensional equation is balanced under the conditions m ⫺ 2n ⫽ 0, n ⫽ 1, and m ⫽ 2. Returning to our original expression x ⬀ a nt mwe result , conclude that x ⬀ at 2This . differs by a factor of 2 from the correct expression, which is x ⫽ 12at 2. Because the factor 12 is dimensionless, there is no way of determining it using dimensional analysis. 11 12 CHAPTER 1 Physics and Measurements Quick Quiz 1.1 True or False: Dimensional analysis can give you the numerical value of constants of proportionality that may appear in an algebraic expression. EXAMPLE 1.2 Analysis of an Equation Show that the expression v ⫽ at is dimensionally correct, where v represents speed, a acceleration, and t a time interval. The same table gives us L/T 2 for the dimensions of acceleration, and so the dimensions of at are [at] ⫽ Solution For the speed term, we have from Table 1.6 Analysis of a Power Law Suppose we are told that the acceleration a of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say r n, and some power of v, say v m. How can we determine the values of n and m? Solution 2 Therefore, the expression is dimensionally correct. (If the expression were given as v ⫽ at 2, it would be dimensionally incorrect. Try it and see!) L [v] ⫽ T EXAMPLE 1.3 冢 TL 冣(T) ⫽ TL This dimensional equation is balanced under the conditions n⫹m⫽1 Therefore n ⫽ ⫺ 1, and we can write the acceleration expression as Let us take a to be a⫽ a ⫽ kr ⫺1v 2 ⫽ k kr nv m where k is a dimensionless constant of proportionality. Knowing the dimensions of a, r, and v, we see that the dimensional equation must be L/T 2 ⫽ Ln(L/T)m ⫽ Ln⫹m/T m m⫽2 and v2 r When we discuss uniform circular motion later, we shall see that k ⫽ 1 if a consistent set of units is used. The constant k would not equal 1 if, for example, v were in km/h and you wanted a in m/s2. QuickLab 1.5 Estimate the weight (in pounds) of two large bottles of soda pop. Note that 1 L of water has a mass of about 1 kg. Use the fact that an object weighing 2.2 lb has a mass of 1 kg. Find some bathroom scales and check your estimate. Sometimes it is necessary to convert units from one system to another. Conversion factors between the SI units and conventional units of length are as follows: CONVERSION OF UNITS 1 mi ⫽ 1 609 m ⫽ 1.609 km 1 ft ⫽ 0.304 8 m ⫽ 30.48 cm 1 m ⫽ 39.37 in. ⫽ 3.281 ft 1 in. ⬅ 0.025 4 m ⫽ 2.54 cm (exactly) A more complete list of conversion factors can be found in Appendix A. Units can be treated as algebraic quantities that can cancel each other. For example, suppose we wish to convert 15.0 in. to centimeters. Because 1 in. is deﬁned as exactly 2.54 cm, we ﬁnd that 15.0 in. ⫽ (15.0 in.)(2.54 cm/in.) ⫽ 38.1 cm cm This works because multiplying by (2.54 1 in. ) is the same as multiplying by 1, because the numerator and denominator describe identical things. 1.6 Estimates and Order-of-Magnitude Calculations (Left) This road sign near Raleigh, North Carolina, shows distances in miles and kilometers. How accurate are the conversions? (Billy E. Barnes/Stock Boston). (Right) This vehicle’s speedometer gives speed readings in miles per hour and in kilometers per hour. Try conﬁrming the conversion between the two sets of units for a few readings of the dial. (Paul Silverman/Fundamental Photographs) EXAMPLE 1.4 The Density of a Cube The mass of a solid cube is 856 g, and each edge has a length of 5.35 cm. Determine the density of the cube in basic SI units. Because 1 g ⫽ 10⫺3 kg and 1 cm ⫽ 10⫺2 m, the mass m and volume V in basic SI units are Solution m ⫽ 856 g ⫻ 10 ⫺3 kg/g ⫽ 0.856 kg 1.6 V ⫽ L3 ⫽ (5.35 cm ⫻ 10 ⫺2 m/cm)3 ⫽ (5.35)3 ⫻ 10 ⫺6 m3 ⫽ 1.53 ⫻ 10 ⫺4 m3 Therefore, ⫽ 0.856 kg m ⫽ ⫽ 5.59 ⫻ 10 3 kg/m3 V 1.53 ⫻ 10 ⫺4 m3 ESTIMATES AND ORDER-OFMAGNITUDE CALCULATIONS It is often useful to compute an approximate answer to a physical problem even where little information is available. Such an approximate answer can then be used to determine whether a more accurate calculation is necessary. Approximations are usually based on certain assumptions, which must be modiﬁed if greater accuracy is needed. Thus, we shall sometimes refer to the order of magnitude of a certain quantity as the power of ten of the number that describes that quantity. If, for example, we say that a quantity increases in value by three orders of magnitude, this means that its value is increased by a factor of 103 ⫽ 1000. Also, if a quantity is given as 3 ⫻ 103, we say that the order of magnitude of that quantity is 103 (or in symbolic form, 3 ⫻ 103 ⬃ 103). Likewise, the quantity 8 ⫻ 107 ⬃ 108. The spirit of order-of-magnitude calculations, sometimes referred to as “guesstimates” or “ball-park ﬁgures,” is given in the following quotation: “Make an estimate before every calculation, try a simple physical argument . . . before every derivation, guess the answer to every puzzle. Courage: no one else needs to 13 14 CHAPTER 1 Physics and Measurements know what the guess is.” 4 Inaccuracies caused by guessing too low for one number are often canceled out by other guesses that are too high. You will ﬁnd that with practice your guesstimates get better and better. Estimation problems can be fun to work as you freely drop digits, venture reasonable approximations for unknown numbers, make simplifying assumptions, and turn the question around into something you can answer in your head. EXAMPLE 1.5 Breaths in a Lifetime Estimate the number of breaths taken during an average life span. Solution We shall start by guessing that the typical life span is about 70 years. The only other estimate we must make in this example is the average number of breaths that a person takes in 1 min. This number varies, depending on whether the person is exercising, sleeping, angry, serene, and so forth. To the nearest order of magnitude, we shall choose 10 breaths per minute as our estimate of the average. (This is certainly closer to the true value than 1 breath per minute or 100 breaths per minute.) The number of minutes in a year is EXAMPLE 1.6 approximately 1 yr ⫻ 400 days h min ⫻ 25 ⫻ 60 ⫽ 6 ⫻ 105 min yr day h Notice how much simpler it is to multiply 400 ⫻ 25 than it is to work with the more accurate 365 ⫻ 24. These approximate values for the number of days in a year and the number of hours in a day are close enough for our purposes. Thus, in 70 years there will be (70 yr)(6 ⫻ 105 min/yr) ⫽ 4 ⫻ 107 min. At a rate of 10 breaths/min, an individual would take 4 ⫻ 10 8 breaths in a lifetime. It’s a Long Way to San Jose Estimate the number of steps a person would take walking from New York to Los Angeles. Now we switch to scientiﬁc notation so that we can do the calculation mentally: Solution Without looking up the distance between these two cities, you might remember from a geography class that they are about 3 000 mi apart. The next approximation we must make is the length of one step. Of course, this length depends on the person doing the walking, but we can estimate that each step covers about 2 ft. With our estimated step size, we can determine the number of steps in 1 mi. Because this is a rough calculation, we round 5 280 ft/mi to 5 000 ft/mi. (What percentage error does this introduce?) This conversion factor gives us (3 ⫻ 10 3 mi)(2.5 ⫻ 10 3 steps/mi) ⫽ 7.5 ⫻ 10 6 steps ⬃ 10 7 steps So if we intend to walk across the United States, it will take us on the order of ten million steps. This estimate is almost certainly too small because we have not accounted for curving roads and going up and down hills and mountains. Nonetheless, it is probably within an order of magnitude of the correct answer. 5 000 ft/mi ⫽ 2 500 steps/mi 2 ft/step EXAMPLE 1.7 How Much Gas Do We Use? Estimate the number of gallons of gasoline used each year by all the cars in the United States. Solution There are about 270 million people in the United States, and so we estimate that the number of cars in the country is 100 million (guessing that there are between two and three people per car). We also estimate that the aver- age distance each car travels per year is 10 000 mi. If we assume a gasoline consumption of 20 mi/gal or 0.05 gal/mi, then each car uses about 500 gal/yr. Multiplying this by the total number of cars in the United States gives an estimated total consumption of 5 ⫻ 1010 gal ⬃ 10 11 gal. 4 E. Taylor and J. A. Wheeler, Spacetime Physics, San Francisco, W. H. Freeman & Company, Publishers, 1966, p. 60. 1.7 1.7 15 Significant Figures SIGNIFICANT FIGURES When physical quantities are measured, the measured values are known only to within the limits of the experimental uncertainty. The value of this uncertainty can depend on various factors, such as the quality of the apparatus, the skill of the experimenter, and the number of measurements performed. Suppose that we are asked to measure the area of a computer disk label using a meter stick as a measuring instrument. Let us assume that the accuracy to which we can measure with this stick is ⫾ 0.1 cm. If the length of the label is measured to be 5.5 cm, we can claim only that its length lies somewhere between 5.4 cm and 5.6 cm. In this case, we say that the measured value has two signiﬁcant ﬁgures. Likewise, if the label’s width is measured to be 6.4 cm, the actual value lies between 6.3 cm and 6.5 cm. Note that the signiﬁcant ﬁgures include the ﬁrst estimated digit. Thus we could write the measured values as (5.5 ⫾ 0.1) cm and (6.4 ⫾ 0.1) cm. Now suppose we want to ﬁnd the area of the label by multiplying the two measured values. If we were to claim the area is (5.5 cm)(6.4 cm) ⫽ 35.2 cm2, our answer would be unjustiﬁable because it contains three signiﬁcant ﬁgures, which is greater than the number of signiﬁcant ﬁgures in either of the measured lengths. A good rule of thumb to use in determining the number of signiﬁcant ﬁgures that can be claimed is as follows: When multiplying several quantities, the number of signiﬁcant ﬁgures in the ﬁnal answer is the same as the number of signiﬁcant ﬁgures in the least accurate of the quantities being multiplied, where “least accurate” means “having the lowest number of signiﬁcant ﬁgures.” The same rule applies to division. Applying this rule to the multiplication example above, we see that the answer for the area can have only two signiﬁcant ﬁgures because our measured lengths have only two signiﬁcant ﬁgures. Thus, all we can claim is that the area is 35 cm2, realizing that the value can range between (5.4 cm)(6.3 cm) ⫽ 34 cm2 and (5.6 cm)(6.5 cm) ⫽ 36 cm2. Zeros may or may not be signiﬁcant ﬁgures. Those used to position the decimal point in such numbers as 0.03 and 0.007 5 are not signiﬁcant. Thus, there are one and two signiﬁcant ﬁgures, respectively, in these two values. When the zeros come after other digits, however, there is the possibility of misinterpretation. For example, suppose the mass of an object is given as 1 500 g. This value is ambiguous because we do not know whether the last two zeros are being used to locate the decimal point or whether they represent signiﬁcant ﬁgures in the measurement. To remove this ambiguity, it is common to use scientiﬁc notation to indicate the number of signiﬁcant ﬁgures. In this case, we would express the mass as 1.5 ⫻ 103 g if there are two signiﬁcant ﬁgures in the measured value, 1.50 ⫻ 103 g if there are three signiﬁcant ﬁgures, and 1.500 ⫻ 103 g if there are four. The same rule holds when the number is less than 1, so that 2.3 ⫻ 10⫺4 has two signiﬁcant ﬁgures (and so could be written 0.000 23) and 2.30 ⫻ 10⫺4 has three signiﬁcant ﬁgures (also written 0.000 230). In general, a signiﬁcant ﬁgure is a reliably known digit (other than a zero used to locate the decimal point). For addition and subtraction, you must consider the number of decimal places when you are determining how many signiﬁcant ﬁgures to report. QuickLab Determine the thickness of a page from this book. (Note that numbers that have no measurement errors — like the count of a number of pages — do not affect the signiﬁcant ﬁgures in a calculation.) In terms of signiﬁcant ﬁgures, why is it better to measure the thickness of as many pages as possible and then divide by the number of sheets? 16 CHAPTER 1 Physics and Measurements When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum. For example, if we wish to compute 123 ⫹ 5.35, the answer given to the correct number of signiﬁcant ﬁgures is 128 and not 128.35. If we compute the sum 1.000 1 ⫹ 0.000 3 ⫽ 1.000 4, the result has ﬁve signiﬁcant ﬁgures, even though one of the terms in the sum, 0.000 3, has only one signiﬁcant ﬁgure. Likewise, if we perform the subtraction 1.002 ⫺ 0.998 ⫽ 0.004, the result has only one signiﬁcant ﬁgure even though one term has four signiﬁcant ﬁgures and the other has three. In this book, most of the numerical examples and end-of-chapter problems will yield answers having three signiﬁcant ﬁgures. When carrying out estimates we shall typically work with a single signiﬁcant ﬁgure. Quick Quiz 1.2 Suppose you measure the position of a chair with a meter stick and record that the center of the seat is 1.043 860 564 2 m from a wall. What would a reader conclude from this recorded measurement? EXAMPLE 1.8 The Area of a Rectangle A rectangular plate has a length of (21.3 ⫾ 0.2) cm and a width of (9.80 ⫾ 0.1) cm. Find the area of the plate and the uncertainty in the calculated area. Solution Area ⫽ ᐉw ⫽ (21.3 ⫾ 0.2 cm) ⫻ (9.80 ⫾ 0.1 cm) EXAMPLE 1.9 ⬇ (21.3 ⫻ 9.80 ⫾ 21.3 ⫻ 0.1 ⫾ 0.2 ⫻ 9.80) cm2 ⬇ (209 ⫾ 4) cm2 Because the input data were given to only three signiﬁcant ﬁgures, we cannot claim any more in our result. Do you see why we did not need to multiply the uncertainties 0.2 cm and 0.1 cm? Installing a Carpet A carpet is to be installed in a room whose length is measured to be 12.71 m and whose width is measured to be 3.46 m. Find the area of the room. Solution If you multiply 12.71 m by 3.46 m on your calculator, you will get an answer of 43.976 6 m2. How many of these numbers should you claim? Our rule of thumb for multiplication tells us that you can claim only the number of signiﬁcant ﬁgures in the least accurate of the quantities being measured. In this example, we have only three signiﬁcant ﬁgures in our least accurate measurement, so we should express our ﬁnal answer as 44.0 m2. Note that in reducing 43.976 6 to three signiﬁcant ﬁgures for our answer, we used a general rule for rounding off numbers that states that the last digit retained (the 9 in this example) is increased by 1 if the ﬁrst digit dropped (here, the 7) is 5 or greater. (A technique for avoiding error accumulation is to delay rounding of numbers in a long calculation until you have the ﬁnal result. Wait until you are ready to copy the answer from your calculator before rounding to the correct number of signiﬁcant ﬁgures.) Problems 17 SUMMARY The three fundamental physical quantities of mechanics are length, mass, and time, which in the SI system have the units meters (m), kilograms (kg), and seconds (s), respectively. Preﬁxes indicating various powers of ten are used with these three basic units. The density of a substance is deﬁned as its mass per unit volume. Different substances have different densities mainly because of differences in their atomic masses and atomic arrangements. The number of particles in one mole of any element or compound, called Avogadro’s number, NA , is 6.02 ⫻ 1023. The method of dimensional analysis is very powerful in solving physics problems. Dimensions can be treated as algebraic quantities. By making estimates and making order-of-magnitude calculations, you should be able to approximate the answer to a problem when there is not enough information available to completely specify an exact solution. When you compute a result from several measured numbers, each of which has a certain accuracy, you should give the result with the correct number of significant ﬁgures. QUESTIONS 1. In this chapter we described how the Earth’s daily rotation on its axis was once used to deﬁne the standard unit of time. What other types of natural phenomena could serve as alternative time standards? 2. Suppose that the three fundamental standards of the metric system were length, density, and time rather than length, mass, and time. The standard of density in this system is to be deﬁned as that of water. What considerations about water would you need to address to make sure that the standard of density is as accurate as possible? 3. A hand is deﬁned as 4 in.; a foot is deﬁned as 12 in. Why should the hand be any less acceptable as a unit than the foot, which we use all the time? 4. Express the following quantities using the preﬁxes given in 5. 6. 7. 8. 9. Table 1.4: (a) 3 ⫻ 10⫺4 m (b) 5 ⫻ 10⫺5 s (c) 72 ⫻ 102 g. Suppose that two quantities A and B have different dimensions. Determine which of the following arithmetic operations could be physically meaningful: (a) A ⫹ B (b) A/B (c) B ⫺ A (d) AB. What level of accuracy is implied in an order-of-magnitude calculation? Do an order-of-magnitude calculation for an everyday situation you might encounter. For example, how far do you walk or drive each day? Estimate your age in seconds. Estimate the mass of this textbook in kilograms. If a scale is available, check your estimate. PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 1.3 Density 1. The standard kilogram is a platinum – iridium cylinder 39.0 mm in height and 39.0 mm in diameter. What is the density of the material? 2. The mass of the planet Saturn (Fig. P1.2) is 5.64 ⫻ 1026 kg, and its radius is 6.00 ⫻ 107 m. Calculate its density. 3. How many grams of copper are required to make a hollow spherical shell having an inner radius of 5.70 cm and an outer radius of 5.75 cm? The density of copper is 8.92 g/cm3. 4. What mass of a material with density is required to make a hollow spherical shell having inner radius r1 and outer radius r 2 ? 5. Iron has molar mass 55.8 g/mol. (a) Find the volume of 1 mol of iron. (b) Use the value found in (a) to determine the volume of one iron atom. (c) Calculate the cube root of the atomic volume, to have an estimate for the distance between atoms in the solid. (d) Repeat the calculations for uranium, finding its molar mass in the periodic table of the elements in Appendix C. 18 CHAPTER 1 Physics and Measurements (a) What is the mass of a section 1.50 m long? (b) How many atoms are there in this section? The density of steel is 7.56 ⫻ 103 kg/m3. 11. A child at the beach digs a hole in the sand and, using a pail, ﬁlls it with water having a mass of 1.20 kg. The molar mass of water is 18.0 g/mol. (a) Find the number of water molecules in this pail of water. (b) Suppose the quantity of water on the Earth is 1.32 ⫻ 1021 kg and remains constant. How many of the water molecules in this pail of water were likely to have been in an equal quantity of water that once ﬁlled a particular claw print left by a dinosaur? Section 1.4 Dimensional Analysis 12. The radius r of a circle inscribed in any triangle whose sides are a, b, and c is given by r ⫽ [(s ⫺ a)(s ⫺ b)(s ⫺ c)/s]1/2 Figure P1.2 where s is an abbreviation for (a ⫹ b ⫹ c)/2. Check this formula for dimensional consistency. 13. The displacement of a particle moving under uniform acceleration is some function of the elapsed time and the acceleration. Suppose we write this displacement s ⫽ ka mt n, where k is a dimensionless constant. Show by dimensional analysis that this expression is satisﬁed if m ⫽ 1 and n ⫽ 2. Can this analysis give the value of k? 14. The period T of a simple pendulum is measured in time units and is described by A view of Saturn from Voyager 2. (Courtesy of NASA) 6. Two spheres are cut from a certain uniform rock. One has radius 4.50 cm. The mass of the other is ﬁve times greater. Find its radius. WEB 7. Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in atomic mass units and in grams. The molar masses are 4.00, 55.9, and 207 g/mol, respectively, for the atoms given. 8. On your wedding day your lover gives you a gold ring of mass 3.80 g. Fifty years later its mass is 3.35 g. As an average, how many atoms were abraded from the ring during each second of your marriage? The molar mass of gold is 197 g/mol. 9. A small cube of iron is observed under a microscope. The edge of the cube is 5.00 ⫻ 10⫺6 cm long. Find (a) the mass of the cube and (b) the number of iron atoms in the cube. The molar mass of iron is 55.9 g/mol, and its density is 7.86 g/cm3. 10. A structural I-beam is made of steel. A view of its crosssection and its dimensions are shown in Figure P1.10. 15.0 cm T ⫽ 2 36.0 cm F⫽ Figure P1.10 GMm r2 Here F is the gravitational force, M and m are masses, and r is a length. Force has the SI units kg⭈ m/s2. What are the SI units of the proportionality constant G ? 17. The consumption of natural gas by a company satisﬁes the empirical equation V ⫽ 1.50t ⫹ 0.008 00t 2, where V is the volume in millions of cubic feet and t the time in months. Express this equation in units of cubic feet and seconds. Put the proper units on the coefﬁcients. Assume a month is 30.0 days. Section 1.5 1.00 cm ᐉ g where ᐉ is the length of the pendulum and g is the freefall acceleration in units of length divided by the square of time. Show that this equation is dimensionally correct. 15. Which of the equations below are dimensionally correct? (a) v ⫽ v 0 ⫹ ax (b) y ⫽ (2 m) cos(kx), where k ⫽ 2 m⫺1 16. Newton’s law of universal gravitation is represented by WEB 1.00 cm √ Conversion of Units 18. Suppose your hair grows at the rate 1/32 in. per day. Find the rate at which it grows in nanometers per second. Since the distance between atoms in a molecule is 19 Problems 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. on the order of 0.1 nm, your answer suggests how rapidly layers of atoms are assembled in this protein synthesis. A rectangular building lot is 100 ft by 150 ft. Determine the area of this lot in m2. An auditorium measures 40.0 m ⫻ 20.0 m ⫻ 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic feet and (b) the weight of air in the room in pounds? Assume that it takes 7.00 min to ﬁll a 30.0-gal gasoline tank. (a) Calculate the rate at which the tank is ﬁlled in gallons per second. (b) Calculate the rate at which the tank is ﬁlled in cubic meters per second. (c) Determine the time, in hours, required to ﬁll a 1-cubic-meter volume at the same rate. (1 U.S. gal ⫽ 231 in.3 ) A creature moves at a speed of 5.00 furlongs per fortnight (not a very common unit of speed). Given that 1 furlong ⫽ 220 yards and 1 fortnight ⫽ 14 days, determine the speed of the creature in meters per second. What kind of creature do you think it might be? A section of land has an area of 1 mi2 and contains 640 acres. Determine the number of square meters in 1 acre. A quart container of ice cream is to be made in the form of a cube. What should be the length of each edge in centimeters? (Use the conversion 1 gal ⫽ 3.786 L.) A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/m3 ). An astronomical unit (AU) is deﬁned as the average distance between the Earth and the Sun. (a) How many astronomical units are there in one lightyear? (b) Determine the distance from the Earth to the Andromeda galaxy in astronomical units. The mass of the Sun is 1.99 ⫻ 1030 kg, and the mass of an atom of hydrogen, of which the Sun is mostly composed, is 1.67 ⫻ 10⫺27 kg. How many atoms are there in the Sun? (a) Find a conversion factor to convert from miles per hour to kilometers per hour. (b) In the past, a federal law mandated that highway speed limits would be 55 mi/h. Use the conversion factor of part (a) to ﬁnd this speed in kilometers per hour. (c) The maximum highway speed is now 65 mi/h in some places. In kilometers per hour, how much of an increase is this over the 55-mi/h limit? At the time of this book’s printing, the U. S. national debt is about $6 trillion. (a) If payments were made at the rate of $1 000/s, how many years would it take to pay off a $6-trillion debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long. If six trillion dollar bills were laid end to end around the Earth’s equator, how many times would they encircle the Earth? Take the radius of the Earth at the equator to be 6 378 km. (Note: Before doing any of these calculations, try to guess at the answers. You may be very surprised.) WEB 30. (a) How many seconds are there in a year? (b) If one micrometeorite (a sphere with a diameter of 1.00 ⫻ 10⫺6 m) strikes each square meter of the Moon each second, how many years will it take to cover the Moon to a depth of 1.00 m? (Hint: Consider a cubic box on the Moon 1.00 m on a side, and ﬁnd how long it will take to ﬁll the box.) 31. One gallon of paint (volume ⫽ 3.78 ⫻ 10⫺3 m3 ) covers an area of 25.0 m2. What is the thickness of the paint on the wall? 32. A pyramid has a height of 481 ft, and its base covers an area of 13.0 acres (Fig. P1.32). If the volume of a pyramid is given by the expression V ⫽ 13Bh, where B is the area of the base and h is the height, ﬁnd the volume of this pyramid in cubic meters. (1 acre ⫽ 43 560 ft2 ) Figure P1.32 Problems 32 and 33. 33. The pyramid described in Problem 32 contains approximately two million stone blocks that average 2.50 tons each. Find the weight of this pyramid in pounds. 34. Assuming that 70% of the Earth’s surface is covered with water at an average depth of 2.3 mi, estimate the mass of the water on the Earth in kilograms. 35. The amount of water in reservoirs is often measured in acre-feet. One acre-foot is a volume that covers an area of 1 acre to a depth of 1 ft. An acre is an area of 43 560 ft2. Find the volume in SI units of a reservoir containing 25.0 acre-ft of water. 36. A hydrogen atom has a diameter of approximately 1.06 ⫻ 10⫺10 m, as deﬁned by the diameter of the spherical electron cloud around the nucleus. The hydrogen nucleus has a diameter of approximately 2.40 ⫻ 10⫺15 m. (a) For a scale model, represent the diameter of the hydrogen atom by the length of an American football ﬁeld (100 yards ⫽ 300 ft), and determine the diameter of the nucleus in millimeters. (b) The atom is how many times larger in volume than its nucleus? 37. The diameter of our disk-shaped galaxy, the Milky Way, is about 1.0 ⫻ 105 lightyears. The distance to Messier 31 — which is Andromeda, the spiral galaxy nearest to the Milky Way — is about 2.0 million lightyears. If a scale model represents the Milky Way and Andromeda galax- 20 CHAPTER 1 Physics and Measurements ies as dinner plates 25 cm in diameter, determine the distance between the two plates. 38. The mean radius of the Earth is 6.37 ⫻ 106 m, and that of the Moon is 1.74 ⫻ 108 cm. From these data calculate (a) the ratio of the Earth’s surface area to that of the Moon and (b) the ratio of the Earth’s volume to that of the Moon. Recall that the surface area of a sphere is 4r 2 and that the volume of a sphere is 43 r 3. WEB 39. One cubic meter (1.00 m3 ) of aluminum has a mass of 2.70 ⫻ 103 kg, and 1.00 m3 of iron has a mass of 7.86 ⫻ 103 kg. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius 2.00 cm on an equal-arm balance. 40. Let A1 represent the density of aluminum and Fe that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius r Fe on an equalarm balance. Estimates and Order-ofMagnitude Calculations Section 1.6 WEB 41. Estimate the number of Ping-Pong balls that would ﬁt into an average-size room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. 42. McDonald’s sells about 250 million packages of French fries per year. If these fries were placed end to end, estimate how far they would reach. 43. An automobile tire is rated to last for 50 000 miles. Estimate the number of revolutions the tire will make in its lifetime. 44. Approximately how many raindrops fall on a 1.0-acre lot during a 1.0-in. rainfall? 45. Grass grows densely everywhere on a quarter-acre plot of land. What is the order of magnitude of the number of blades of grass on this plot of land? Explain your reasoning. (1 acre ⫽ 43 560 ft2.) 46. Suppose that someone offers to give you $1 billion if you can ﬁnish counting it out using only one-dollar bills. Should you accept this offer? Assume you can count one bill every second, and be sure to note that you need about 8 hours a day for sleeping and eating and that right now you are probably at least 18 years old. 47. Compute the order of magnitude of the mass of a bathtub half full of water and of the mass of a bathtub half full of pennies. In your solution, list the quantities you take as data and the value you measure or estimate for each. 48. Soft drinks are commonly sold in aluminum containers. Estimate the number of such containers thrown away or recycled each year by U.S. consumers. Approximately how many tons of aluminum does this represent? 49. To an order of magnitude, how many piano tuners are there in New York City? The physicist Enrico Fermi was famous for asking questions like this on oral Ph.D. qual- ifying examinations and for his own facility in making order-of-magnitude calculations. Section 1.7 Signiﬁcant Figures 50. Determine the number of signiﬁcant ﬁgures in the following measured values: (a) 23 cm (b) 3.589 s (c) 4.67 ⫻ 103 m/s (d) 0.003 2 m. 51. The radius of a circle is measured to be 10.5 ⫾ 0.2 m. Calculate the (a) area and (b) circumference of the circle and give the uncertainty in each value. 52. Carry out the following arithmetic operations: (a) the sum of the measured values 756, 37.2, 0.83, and 2.5; (b) the product 0.003 2 ⫻ 356.3; (c) the product 5.620 ⫻ . 53. The radius of a solid sphere is measured to be (6.50 ⫾ 0.20) cm, and its mass is measured to be (1.85 ⫾ 0.02) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density. 54. How many signiﬁcant ﬁgures are in the following numbers: (a) 78.9 ⫾ 0.2, (b) 3.788 ⫻ 109, (c) 2.46 ⫻ 10⫺6, and (d) 0.005 3? 55. A farmer measures the distance around a rectangular ﬁeld. The length of the long sides of the rectangle is found to be 38.44 m, and the length of the short sides is found to be 19.5 m. What is the total distance around the ﬁeld? 56. A sidewalk is to be constructed around a swimming pool that measures (10.0 ⫾ 0.1) m by (17.0 ⫾ 0.1) m. If the sidewalk is to measure (1.00 ⫾ 0.01) m wide by (9.0 ⫾ 0.1) cm thick, what volume of concrete is needed, and what is the approximate uncertainty of this volume? ADDITIONAL PROBLEMS 57. In a situation where data are known to three signiﬁcant digits, we write 6.379 m ⫽ 6.38 m and 6.374 m ⫽ 6.37 m. When a number ends in 5, we arbitrarily choose to write 6.375 m ⫽ 6.38 m. We could equally well write 6.375 m ⫽ 6.37 m, “rounding down” instead of “rounding up,” since we would change the number 6.375 by equal increments in both cases. Now consider an orderof-magnitude estimate, in which we consider factors rather than increments. We write 500 m ⬃ 103 m because 500 differs from 100 by a factor of 5 whereas it differs from 1000 by only a factor of 2. We write 437 m ⬃ 103 m and 305 m ⬃ 102 m. What distance differs from 100 m and from 1000 m by equal factors, so that we could equally well choose to represent its order of magnitude either as ⬃ 102 m or as ⬃ 103 m? 58. When a droplet of oil spreads out on a smooth water surface, the resulting “oil slick” is approximately one molecule thick. An oil droplet of mass 9.00 ⫻ 10⫺7 kg and density 918 kg/m3 spreads out into a circle of radius 41.8 cm on the water surface. What is the diameter of an oil molecule? 21 Problems 59. The basic function of the carburetor of an automobile is to “atomize” the gasoline and mix it with air to promote rapid combustion. As an example, assume that 30.0 cm3 of gasoline is atomized into N spherical droplets, each with a radius of 2.00 ⫻ 10⫺5 m. What is the total surface area of these N spherical droplets? 60. In physics it is important to use mathematical approximations. Demonstrate for yourself that for small angles (⬍ 20°) tan ␣ ⬇ sin ␣ ⬇ ␣ ⫽ ␣ ⬘/180° where ␣ is in radians and ␣ ⬘ is in degrees. Use a calculator to ﬁnd the largest angle for which tan ␣ may be approximated by sin ␣ if the error is to be less than 10.0%. 61. A high fountain of water is located at the center of a circular pool as in Figure P1.61. Not wishing to get his feet wet, a student walks around the pool and measures its circumference to be 15.0 m. Next, the student stands at the edge of the pool and uses a protractor to gauge the angle of elevation of the top of the fountain to be 55.0°. How high is the fountain? 64. A crystalline solid consists of atoms stacked up in a repeating lattice structure. Consider a crystal as shown in Figure P1.64a. The atoms reside at the corners of cubes of side L ⫽ 0.200 nm. One piece of evidence for the regular arrangement of atoms comes from the ﬂat surfaces along which a crystal separates, or “cleaves,” when it is broken. Suppose this crystal cleaves along a face diagonal, as shown in Figure P1.64b. Calculate the spacing d between two adjacent atomic planes that separate when the crystal cleaves. L d (a) (b) Figure P1.64 55.0˚ Figure P1.61 62. Assume that an object covers an area A and has a uniform height h. If its cross-sectional area is uniform over its height, then its volume is given by V ⫽ Ah. (a) Show that V ⫽ Ah is dimensionally correct. (b) Show that the volumes of a cylinder and of a rectangular box can be written in the form V ⫽ Ah, identifying A in each case. (Note that A, sometimes called the “footprint” of the object, can have any shape and that the height can be replaced by average thickness in general.) 63. A useful fact is that there are about ⫻ 107 s in one year. Find the percentage error in this approximation, where “percentage error” is deﬁned as 兩 Assumed value ⫺ true value 兩 ⫻ 100% True value 65. A child loves to watch as you ﬁll a transparent plastic bottle with shampoo. Every horizontal cross-section of the bottle is a circle, but the diameters of the circles all have different values, so that the bottle is much wider in some places than in others. You pour in bright green shampoo with constant volume ﬂow rate 16.5 cm3/s. At what rate is its level in the bottle rising (a) at a point where the diameter of the bottle is 6.30 cm and (b) at a point where the diameter is 1.35 cm? 66. As a child, the educator and national leader Booker T. Washington was given a spoonful (about 12.0 cm3) of molasses as a treat. He pretended that the quantity increased when he spread it out to cover uniformly all of a tin plate (with a diameter of about 23.0 cm). How thick a layer did it make? 67. Assume there are 100 million passenger cars in the United States and that the average fuel consumption is 20 mi/gal of gasoline. If the average distance traveled by each car is 10 000 mi/yr, how much gasoline would be saved per year if average fuel consumption could be increased to 25 mi/gal? 68. One cubic centimeter of water has a mass of 1.00 ⫻ 10⫺3 kg. (a) Determine the mass of 1.00 m3 of water. (b) Assuming biological substances are 98% water, esti- 22 CHAPTER 1 Physics and Measurements mate the mass of a cell that has a diameter of 1.0 m, a human kidney, and a ﬂy. Assume that a kidney is roughly a sphere with a radius of 4.0 cm and that a ﬂy is roughly a cylinder 4.0 mm long and 2.0 mm in diameter. 69. The distance from the Sun to the nearest star is 4 ⫻ 1016 m. The Milky Way galaxy is roughly a disk of diameter ⬃ 1021 m and thickness ⬃ 1019 m. Find the order of magnitude of the number of stars in the Milky Way. Assume the 4 ⫻ 10 16-m distance between the Sun and the nearest star is typical. 70. The data in the following table represent measurements of the masses and dimensions of solid cylinders of alu- minum, copper, brass, tin, and iron. Use these data to calculate the densities of these substances. Compare your results for aluminum, copper, and iron with those given in Table 1.5. Substance Aluminum Copper Brass Tin Iron Mass (g) Diameter (cm) Length (cm) 51.5 56.3 94.4 69.1 216.1 2.52 1.23 1.54 1.75 1.89 3.75 5.06 5.69 3.74 9.77 ANSWERS TO QUICK QUIZZES 1.1 False. Dimensional analysis gives the units of the proportionality constant but provides no information about its numerical value. For example, experiments show that doubling the radius of a solid sphere increases its mass 8-fold, and tripling the radius increases the mass 27-fold. Therefore, its mass is proportional to the cube of its radius. Because m ⬀ r 3we , can write m ⫽ kr 3. Dimensional analysis shows that the proportionality constant k must have units kg/m3, but to determine its numerical value requires either experimental data or geometrical reasoning. 1.2 Reporting all these digits implies you have determined the location of the center of the chair’s seat to the nearest ⫾ 0.000 000 000 1 m. This roughly corresponds to being able to count the atoms in your meter stick because each of them is about that size! It would probably be better to record the measurement as 1.044 m: this indicates that you know the position to the nearest millimeter, assuming the meter stick has millimeter markings on its scale. P U Z Z L E R In a moment the arresting cable will be pulled taut, and the 140-mi/h landing of this F/A-18 Hornet on the aircraft carrier USS Nimitz will be brought to a sudden conclusion. The pilot cuts power to the engine, and the plane is stopped in less than 2 s. If the cable had not been successfully engaged, the pilot would have had to take off quickly before reaching the end of the ﬂight deck. Can the motion of the plane be described quantitatively in a way that is useful to ship and aircraft designers and to pilots learning to land on a “postage stamp?” (Courtesy of the USS Nimitz/U.S. Navy) c h a p t e r Motion in One Dimension Chapter Outline 2.1 2.2 2.3 2.4 2.5 Displacement, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams 2.6 Freely Falling Objects 2.7 (Optional) Kinematic Equations Derived from Calculus GOAL Problem-Solving Steps One-Dimensional Motion with Constant Acceleration 23 24 CHAPTER 2 Motion in One Dimension A s a ﬁrst step in studying classical mechanics, we describe motion in terms of space and time while ignoring the agents that caused that motion. This portion of classical mechanics is called kinematics. (The word kinematics has the same root as cinema. Can you see why?) In this chapter we consider only motion in one dimension. We ﬁrst deﬁne displacement, velocity, and acceleration. Then, using these concepts, we study the motion of objects traveling in one dimension with a constant acceleration. From everyday experience we recognize that motion represents a continuous change in the position of an object. In physics we are concerned with three types of motion: translational, rotational, and vibrational. A car moving down a highway is an example of translational motion, the Earth’s spin on its axis is an example of rotational motion, and the back-and-forth movement of a pendulum is an example of vibrational motion. In this and the next few chapters, we are concerned only with translational motion. (Later in the book we shall discuss rotational and vibrational motions.) In our study of translational motion, we describe the moving object as a particle regardless of its size. In general, a particle is a point-like mass having inﬁnitesimal size. For example, if we wish to describe the motion of the Earth around the Sun, we can treat the Earth as a particle and obtain reasonably accurate data about its orbit. This approximation is justiﬁed because the radius of the Earth’s orbit is large compared with the dimensions of the Earth and the Sun. As an example on a much smaller scale, it is possible to explain the pressure exerted by a gas on the walls of a container by treating the gas molecules as particles. 2.1 TABLE 2.1 Position of the Car at Various Times Position 훽 훾 훿 t(s) x(m) 0 10 20 30 40 50 30 52 38 0 ⫺ 37 ⫺ 53 DISPLACEMENT, VELOCITY, AND SPEED The motion of a particle is completely known if the particle’s position in space is known at all times. Consider a car moving back and forth along the x axis, as shown in Figure 2.1a. When we begin collecting position data, the car is 30 m to the right of a road sign. (Let us assume that all data in this example are known to two signiﬁcant ﬁgures. To convey this information, we should report the initial position as 3.0 ⫻ 101 m. We have written this value in this simpler form to make the discussion easier to follow.) We start our clock and once every 10 s note the car’s location relative to the sign. As you can see from Table 2.1, the car is moving to the right (which we have deﬁned as the positive direction) during the ﬁrst 10 s of motion, from position 훽 to position 훾. The position values now begin to decrease, however, because the car is backing up from position 훾 through position . In fact, at , 30 s after we start measuring, the car is alongside the sign we are using as our origin of coordinates. It continues moving to the left and is more than 50 m to the left of the sign when we stop recording information after our sixth data point. A graph of this information is presented in Figure 2.1b. Such a plot is called a position – time graph. If a particle is moving, we can easily determine its change in position. The displacement of a particle is deﬁned as its change in position. As it moves from an initial position x i to a ﬁnal position xf , its displacement is given by x f ⫺ x i . We use the Greek letter delta (⌬) to denote the change in a quantity. Therefore, we write the displacement, or change in position, of the particle as ⌬x ⬅ x f ⫺ x i (2.1) From this deﬁnition we see that ⌬x is positive if xf is greater than x i and negative if xf is less than x i . 2.1 –60 –50 –40 –30 –50 –20 훽 –10 0 훾 10 20 30 40 50 60 –40 –30 –20 –10 x(m) IT LIM /h 30 km 훿 0 10 20 30 40 50 60 x(m) (a) x(m) 60 훾 ∆x 40 훽 훿 ∆t 20 0 –20 –40 –60 t(s) 0 10 20 30 40 25 Figure 2.1 (a) A car moves back and forth along a straight line taken to be the x axis. Because we are interested only in the car’s translational motion, we can treat it as a particle. (b) Position – time graph for the motion of the “particle.” IT LIM /h 30 km –60 Displacement, Velocity, and Speed 50 (b) A very easy mistake to make is not to recognize the difference between displacement and distance traveled (Fig. 2.2). A baseball player hitting a home run travels a distance of 360 ft in the trip around the bases. However, the player’s displacement is zero because his ﬁnal and initial positions are identical. Displacement is an example of a vector quantity. Many other physical quantities, including velocity and acceleration, also are vectors. In general, a vector is a physical quantity that requires the speciﬁcation of both direction and magnitude. By contrast, a scalar is a quantity that has magnitude and no direction. In this chapter, we use plus and minus signs to indicate vector direction. We can do this because the chapter deals with one-dimensional motion only; this means that any object we study can be moving only along a straight line. For example, for horizontal motion, let us arbitrarily specify to the right as being the positive direction. It follows that any object always moving to the right undergoes a 26 CHAPTER 2 Motion in One Dimension Figure 2.2 Bird’s-eye view of a baseball diamond. A batter who hits a home run travels 360 ft as he rounds the bases, but his displacement for the round trip is zero. (Mark C. Burnett/Photo Researchers, Inc.) positive displacement ⫹⌬x, and any object moving to the left undergoes a negative displacement ⫺⌬x. We shall treat vectors in greater detail in Chapter 3. There is one very important point that has not yet been mentioned. Note that the graph in Figure 2.1b does not consist of just six data points but is actually a smooth curve. The graph contains information about the entire 50-s interval during which we watched the car move. It is much easier to see changes in position from the graph than from a verbal description or even a table of numbers. For example, it is clear that the car was covering more ground during the middle of the 50-s interval than at the end. Between positions 훿 and , the car traveled almost 40 m, but during the last 10 s, between positions and , it moved less than half that far. A common way of comparing these different motions is to divide the displacement ⌬x that occurs between two clock readings by the length of that particular time interval ⌬t. This turns out to be a very useful ratio, one that we shall use many times. For convenience, the ratio has been given a special name — average velocity. The average velocity vx of a particle is deﬁned as the particle’s displacement ⌬ x divided by the time interval ⌬t during which that displacement occurred: vx ⬅ Average velocity 3.2 ⌬x ⌬t (2.2) where the subscript x indicates motion along the x axis. From this deﬁnition we see that average velocity has dimensions of length divided by time (L/T) — meters per second in SI units. Although the distance traveled for any motion is always positive, the average velocity of a particle moving in one dimension can be positive or negative, depending on the sign of the displacement. (The time interval ⌬t is always positive.) If the coordinate of the particle increases in time (that is, if x f ⬎ x i), then ⌬x is positive and v x ⫽ ⌬x/⌬t is positive. This case corresponds to motion in the positive x direction. If the coordinate decreases in time (that is, if x f ⬍ x i), then ⌬x is negative and hence v x is negative. This case corresponds to motion in the negative x direction. 2.2 27 Instantaneous Velocity and Speed We can interpret average velocity geometrically by drawing a straight line between any two points on the position – time graph in Figure 2.1b. This line forms the hypotenuse of a right triangle of height ⌬x and base ⌬t. The slope of this line is the ratio ⌬x/⌬t. For example, the line between positions 훽 and 훾 has a slope equal to the average velocity of the car between those two times, (52 m ⫺ 30 m)/ (10 s ⫺ 0) ⫽ 2.2 m/s. In everyday usage, the terms speed and velocity are interchangeable. In physics, however, there is a clear distinction between these two quantities. Consider a marathon runner who runs more than 40 km, yet ends up at his starting point. His average velocity is zero! Nonetheless, we need to be able to quantify how fast he was running. A slightly different ratio accomplishes this for us. The average speed of a particle, a scalar quantity, is deﬁned as the total distance traveled divided by the total time it takes to travel that distance: Average speed ⫽ total distance total time Average speed The SI unit of average speed is the same as the unit of average velocity: meters per second. However, unlike average velocity, average speed has no direction and hence carries no algebraic sign. Knowledge of the average speed of a particle tells us nothing about the details of the trip. For example, suppose it takes you 8.0 h to travel 280 km in your car. The average speed for your trip is 35 km/h. However, you most likely traveled at various speeds during the trip, and the average speed of 35 km/h could result from an inﬁnite number of possible speed values. EXAMPLE 2.1 Calculating the Variables of Motion Find the displacement, average velocity, and average speed of the car in Figure 2.1a between positions 훽 and . Solution The units of displacement must be meters, and the numerical result should be of the same order of magnitude as the given position data (which means probably not 10 or 100 times bigger or smaller). From the position – time graph given in Figure 2.1b, note that x A ⫽ 30 m at t A ⫽ 0 s and that x F ⫽ ⫺53 m at t F ⫽ 50 s. Using these values along with the deﬁnition of displacement, Equation 2.1, we ﬁnd that ⌬x ⫽ x F ⫺ x A ⫽ ⫺53 m ⫺ 30 m ⫽ ⫺83 m magnitude as the supplied data. A quick look at Figure 2.1a indicates that this is the correct answer. It is difﬁcult to estimate the average velocity without completing the calculation, but we expect the units to be meters per second. Because the car ends up to the left of where we started taking data, we know the average velocity must be negative. From Equation 2.2, vx ⫽ ⫽ ⫺83 m ⫺53 m ⫺ 30 m ⫽ ⫽ ⫺1.7 m/s 50 s ⫺ 0 s 50 s We ﬁnd the car’s average speed for this trip by adding the distances traveled and dividing by the total time: This result means that the car ends up 83 m in the negative direction (to the left, in this case) from where it started. This number has the correct units and is of the same order of 2.2 xf ⫺ xi ⌬x x ⫺ xA ⫽ ⫽ F ⌬t tf ⫺ ti tF ⫺ tA Average speed ⫽ 22 m ⫹ 52 m ⫹ 53 m ⫽ 2.5 m/s 50 s INSTANTANEOUS VELOCITY AND SPEED Often we need to know the velocity of a particle at a particular instant in time, rather than over a ﬁnite time interval. For example, even though you might want to calculate your average velocity during a long automobile trip, you would be especially interested in knowing your velocity at the instant you noticed the police 28 60 CHAPTER 2 x(m) Motion in One Dimension 60 훾 훿 40 훽 훾 20 0 40 –20 –40 –60 훾훾 훾 0 10 20 30 (a) 40 50 t(s) 훽 (b) Figure 2.3 (a) Graph representing the motion of the car in Figure 2.1. (b) An enlargement of the upper left -hand corner of the graph shows how the blue line between positions 훽 and 훾 approaches the green tangent line as point 훾 gets closer to point 훽. car parked alongside the road in front of you. In other words, you would like to be able to specify your velocity just as precisely as you can specify your position by noting what is happening at a speciﬁc clock reading — that is, at some speciﬁc instant. It may not be immediately obvious how to do this. What does it mean to talk about how fast something is moving if we “freeze time” and talk only about an individual instant? This is a subtle point not thoroughly understood until the late 1600s. At that time, with the invention of calculus, scientists began to understand how to describe an object’s motion at any moment in time. To see how this is done, consider Figure 2.3a. We have already discussed the average velocity for the interval during which the car moved from position 훽 to position 훾 (given by the slope of the dark blue line) and for the interval during which it moved from 훽 to (represented by the slope of the light blue line). Which of these two lines do you think is a closer approximation of the initial velocity of the car? The car starts out by moving to the right, which we deﬁned to be the positive direction. Therefore, being positive, the value of the average velocity during the 훽 to 훾 interval is probably closer to the initial value than is the value of the average velocity during the 훽 to interval, which we determined to be negative in Example 2.1. Now imagine that we start with the dark blue line and slide point 훾 to the left along the curve, toward point 훽, as in Figure 2.3b. The line between the points becomes steeper and steeper, and as the two points get extremely close together, the line becomes a tangent line to the curve, indicated by the green line on the graph. The slope of this tangent line represents the velocity of the car at the moment we started taking data, at point 훽. What we have done is determine the instantaneous velocity at that moment. In other words, the instantaneous velocity vx equals the limiting value of the ratio ⌬x/⌬t as ⌬t approaches zero:1 Deﬁnition of instantaneous velocity v x ⬅ lim 3.3 ⌬t:0 ⌬x ⌬t (2.3) Note that the displacement ⌬x also approaches zero as ⌬t approaches zero. As ⌬x and ⌬t become smaller and smaller, the ratio ⌬x/⌬t approaches a value equal to the slope of the line tangent to the x-versus-t curve. 1 2.2 29 Instantaneous Velocity and Speed In calculus notation, this limit is called the derivative of x with respect to t, written dx/dt: v x ⬅ lim ⌬t:0 ⌬x dx ⫽ ⌬t dt (2.4) The instantaneous velocity can be positive, negative, or zero. When the slope of the position – time graph is positive, such as at any time during the ﬁrst 10 s in Figure 2.3, vx is positive. After point 훾, vx is negative because the slope is negative. At the peak, the slope and the instantaneous velocity are zero. From here on, we use the word velocity to designate instantaneous velocity. When it is average velocity we are interested in, we always use the adjective average. The instantaneous speed of a particle is deﬁned as the magnitude of its velocity. As with average speed, instantaneous speed has no direction associated with it and hence carries no algebraic sign. For example, if one particle has a velocity of ⫹ 25 m/s along a given line and another particle has a velocity of ⫺ 25 m/s along the same line, both have a speed2 of 25 m/s. EXAMPLE 2.2 Average and Instantaneous Velocity A particle moves along the x axis. Its x coordinate varies with time according to the expression x ⫽ ⫺4t ⫹ 2t 2, where x is in meters and t is in seconds.3 The position – time graph for this motion is shown in Figure 2.4. Note that the particle moves in the negative x direction for the ﬁrst second of motion, is at rest at the moment t ⫽ 1 s, and moves in the positive x direction for t ⬎ 1 s. (a) Determine the displacement of the particle in the time intervals t ⫽ 0 to t ⫽ 1 s and t ⫽ 1 s to t ⫽ 3 s. x(m) 10 8 6 Slope = –2 m/s 2 Solution During the ﬁrst time interval, we have a negative slope and hence a negative velocity. Thus, we know that the displacement between 훽 and 훾 must be a negative number having units of meters. Similarly, we expect the displacement between 훾 and to be positive. In the ﬁrst time interval, we set t i ⫽ t A ⫽ 0 and t f ⫽ t B ⫽ 1 s. Using Equation 2.1, with x ⫽ ⫺4t ⫹ 2t 2, we obtain for the ﬁrst displacement ⌬x A:B ⫽ x f ⫺ x i ⫽ x B ⫺ x A Slope = 4 m/s 4 0 –2 –4 훿 훽 t(s) 훾 0 1 2 3 4 Figure 2.4 Position – time graph for a particle having an x coordinate that varies in time according to the expression x ⫽ ⫺4t ⫹ 2t 2. ⫽ [⫺4(1) ⫹ 2(1)2] ⫺ [⫺4(0) ⫹ 2(0)2] ⫽ ⫺2 m To calculate the displacement during the second time interval, we set t i ⫽ t B ⫽ 1 s and t f ⫽ t D ⫽ 3 s: ⌬x B:D ⫽ x f ⫺ x i ⫽ x D ⫺ x B 2 ⫽ [⫺4(3) ⫹ 2(3)2] ⫺ [⫺4(1) ⫹ 2(1)2] ⫽ ⫹8 m These displacements can also be read directly from the position – time graph. As with velocity, we drop the adjective for instantaneous speed: “Speed” means instantaneous speed. Simply to make it easier to read, we write the empirical equation as x ⫽ ⫺4t ⫹ 2t 2 rather than as x ⫽ (⫺4.00 m/s)t ⫹ (2.00 m/s2)t 2.00. When an equation summarizes measurements, consider its coefﬁcients to have as many signiﬁcant digits as other data quoted in a problem. Consider its coefﬁcients to have the units required for dimensional consistency. When we start our clocks at t ⫽ 0 s, we usually do not mean to limit the precision to a single digit. Consider any zero value in this book to have as many signiﬁcant ﬁgures as you need. 3 30 CHAPTER 2 Motion in One Dimension These values agree with the slopes of the lines joining these points in Figure 2.4. (b) Calculate the average velocity during these two time intervals. In the ﬁrst time interval, ⌬t ⫽ t f ⫺ t i ⫽ t B ⫺ t A ⫽ 1 s. Therefore, using Equation 2.2 and the displacement calculated in (a), we ﬁnd that Solution v x(A:B) ⫽ ⫺2 m ⌬x A:B ⫽ ⫽ ⌬t 1s ⫺2 m/s In the second time interval, ⌬t ⫽ 2 s; therefore, v x(B:D) ⫽ ⌬x B:D 8m ⫽ ⫽ ⌬t 2s 2.3 (c) Find the instantaneous velocity of the particle at t ⫽ 2.5 s. Solution Certainly we can guess that this instantaneous velocity must be of the same order of magnitude as our previous results, that is, around 4 m/s. Examining the graph, we see that the slope of the tangent at position 훿 is greater than the slope of the blue line connecting points 훾 and . Thus, we expect the answer to be greater than 4 m/s. By measuring the slope of the position – time graph at t ⫽ 2.5 s, we ﬁnd that ⫹4 m/s vx ⫽ ⫹6 m/s ACCELERATION In the last example, we worked with a situation in which the velocity of a particle changed while the particle was moving. This is an extremely common occurrence. (How constant is your velocity as you ride a city bus?) It is easy to quantify changes in velocity as a function of time in exactly the same way we quantify changes in position as a function of time. When the velocity of a particle changes with time, the particle is said to be accelerating. For example, the velocity of a car increases when you step on the gas and decreases when you apply the brakes. However, we need a better deﬁnition of acceleration than this. Suppose a particle moving along the x axis has a velocity vxi at time ti and a velocity vxf at time tf , as in Figure 2.5a. The average acceleration of the particle is deﬁned as the change in velocity ⌬vx divided by the time interval ⌬t during which that change occurred: ax ⬅ Average acceleration v x f ⫺ v xi ⌬v x ⫽ ⌬t tf ⫺ ti (2.5) As with velocity, when the motion being analyzed is one-dimensional, we can use positive and negative signs to indicate the direction of the acceleration. Because the dimensions of velocity are L/T and the dimension of time is T, accelera- –a = ∆vx x ∆t vx Figure 2.5 (a) A “particle” moving along the x axis from 훽 to 훾 has velocity vxi at t ⫽ ti and velocity vx f at t ⫽ tf . (b) Velocity – time graph for the particle moving in a straight line. The slope of the blue straight line connecting 훽 and 훾 is the average acceleration in the time interval ⌬t ⫽ t f ⫺ t i . vxf vxi 훽 훾 ti v = vxi tf v = vxf (a) 훾 ∆vx 훽 ∆t x ti tf (b) t 2.3 31 Acceleration tion has dimensions of length divided by time squared, or L/T 2. The SI unit of acceleration is meters per second squared (m/s 2). It might be easier to interpret these units if you think of them as meters per second per second. For example, suppose an object has an acceleration of 2 m/s2. You should form a mental image of the object having a velocity that is along a straight line and is increasing by 2 m/s during every 1-s interval. If the object starts from rest, you should be able to picture it moving at a velocity of ⫹ 2 m/s after 1 s, at ⫹ 4 m/s after 2 s, and so on. In some situations, the value of the average acceleration may be different over different time intervals. It is therefore useful to deﬁne the instantaneous acceleration as the limit of the average acceleration as ⌬t approaches zero. This concept is analogous to the deﬁnition of instantaneous velocity discussed in the previous section. If we imagine that point 훾 is brought closer and closer to point 훽 in Figure 2.5a and take the limit of ⌬vx /⌬t as ⌬t approaches zero, we obtain the instantaneous acceleration: a x ⬅ lim ⌬t:0 dv x ⌬v x ⫽ ⌬t dt (2.6) Instantaneous acceleration That is, the instantaneous acceleration equals the derivative of the velocity with respect to time, which by deﬁnition is the slope of the velocity – time graph (Fig. 2.5b). Thus, we see that just as the velocity of a moving particle is the slope of the particle’s x-t graph, the acceleration of a particle is the slope of the particle’s vx -t graph. One can interpret the derivative of the velocity with respect to time as the time rate of change of velocity. If ax is positive, then the acceleration is in the positive x direction; if ax is negative, then the acceleration is in the negative x direction. From now on we shall use the term acceleration to mean instantaneous acceleration. When we mean average acceleration, we shall always use the adjective average. Because vx ⫽ dx/dt, the acceleration can also be written ax ⫽ dvx d ⫽ dt dt 冢 dxdt 冣 ⫽ ddt x 2 (2.7) 2 That is, in one-dimensional motion, the acceleration equals the second derivative of x with respect to time. Figure 2.6 illustrates how an acceleration – time graph is related to a velocity – time graph. The acceleration at any time is the slope of the velocity – time graph at that time. Positive values of acceleration correspond to those points in Figure 2.6a where the velocity is increasing in the positive x direction. The acceler- vx ax tA tB (a) tC t tC tA tB (b) t Figure 2.6 Instantaneous acceleration can be obtained from the vx -t graph. (a) The velocity – time graph for some motion. (b) The acceleration – time graph for the same motion. The acceleration given by the ax -t graph for any value of t equals the slope of the line tangent to the vx -t graph at the same value of t. 32 CHAPTER 2 Motion in One Dimension ation reaches a maximum at time t A , when the slope of the velocity – time graph is a maximum. The acceleration then goes to zero at time t B , when the velocity is a maximum (that is, when the slope of the vx -t graph is zero). The acceleration is negative when the velocity is decreasing in the positive x direction, and it reaches its most negative value at time t C . CONCEPTUAL EXAMPLE 2.3 Graphical Relationships Between x, vx , and ax The position of an object moving along the x axis varies with time as in Figure 2.7a. Graph the velocity versus time and the acceleration versus time for the object. Solution The velocity at any instant is the slope of the tangent to the x -t graph at that instant. Between t ⫽ 0 and t ⫽ t A , the slope of the x-t graph increases uniformly, and so the velocity increases linearly, as shown in Figure 2.7b. Between t A and t B , the slope of the x -t graph is constant, and so the velocity remains constant. At t D , the slope of the x-t graph is zero, so the velocity is zero at that instant. Between t D and t E , the slope of the x -t graph and thus the velocity are negative and decrease uniformly in this interval. In the interval t E to t F , the slope of the x -t graph is still negative, and at t F it goes to zero. Finally, after t F , the slope of the x -t graph is zero, meaning that the object is at rest for t ⬎ t F . The acceleration at any instant is the slope of the tangent to the vx -t graph at that instant. The graph of acceleration versus time for this object is shown in Figure 2.7c. The acceleration is constant and positive between 0 and t A, where the slope of the vx -t graph is positive. It is zero between t A and t B and for t ⬎ t F because the slope of the vx -t graph is zero at these times. It is negative between t B and t E because the slope of the vx -t graph is negative during this interval. Figure 2.7 (a) Position – time graph for an object moving along the x axis. (b) The velocity – time graph for the object is obtained by measuring the slope of the position – time graph at each instant. (c) The acceleration – time graph for the object is obtained by measuring the slope of the velocity – time graph at each instant. x (a) O tA tB tC tD tE tF tA tB tC tD tE tF t vx (b) O t ax (c) O tA tB tE tF t Quick Quiz 2.1 Make a velocity – time graph for the car in Figure 2.1a and use your graph to determine whether the car ever exceeds the speed limit posted on the road sign (30 km/h). EXAMPLE 2.4 Average and Instantaneous Acceleration The velocity of a particle moving along the x axis varies in time according to the expression vx ⫽ (40 ⫺ 5t 2) m/s, where t is in seconds. (a) Find the average acceleration in the time interval t ⫽ 0 to t ⫽ 2.0 s. Solution Figure 2.8 is a vx -t graph that was created from the velocity versus time expression given in the problem statement. Because the slope of the entire vx -t curve is negative, we expect the acceleration to be negative. 2.3 ax ⫽ vx(m/s) 40 훽 33 Acceleration v xf ⫺ v xi tf ⫺ ti ⫽ vxB ⫺ vxA tB ⫺ tA ⫽ (20 ⫺ 40) m/s (2.0 ⫺ 0) s ⫽ ⫺10 m/s2 30 Slope = –20 m/s2 20 The negative sign is consistent with our expectations — namely, that the average acceleration, which is represented by the slope of the line (not shown) joining the initial and ﬁnal points on the velocity – time graph, is negative. 훾 10 t(s) 0 (b) Determine the acceleration at t ⫽ 2.0 s. The velocity at any time t is vxi ⫽ (40 ⫺ 5t 2) m/s, and the velocity at any later time t ⫹ ⌬t is Solution –10 vxf ⫽ 40 ⫺ 5(t ⫹ ⌬t)2 ⫽ 40 ⫺ 5t 2 ⫺ 10t ⌬t ⫺ 5(⌬t)2 –20 Therefore, the change in velocity over the time interval ⌬t is –30 0 1 2 3 4 Figure 2.8 The velocity – time graph for a particle moving along the x axis according to the expression vx ⫽ (40 ⫺ 5t 2) m/s. The acceleration at t ⫽ 2 s is equal to the slope of the blue tangent line at that time. ⌬vx ⫽ vxf ⫺ vxi ⫽ [⫺10t ⌬t ⫺ 5(⌬t)2] m/s Dividing this expression by ⌬t and taking the limit of the result as ⌬t approaches zero gives the acceleration at any time t: a x ⫽ lim ⌬t:0 ⌬v x ⫽ lim (⫺10t ⫺ 5⌬t) ⫽ ⫺10t m/s2 ⌬t:0 ⌬t Therefore, at t ⫽ 2.0 s, a x ⫽ (⫺10)(2.0) m/s2 ⫽ ⫺20 m/s2 We ﬁnd the velocities at t i ⫽ t A ⫽ 0 and tf ⫽ t B ⫽ 2.0 s by substituting these values of t into the expression for the velocity: v x A ⫽ (40 ⫺ 5t A2) m/s ⫽ [40 ⫺ 5(0)2] m/s ⫽ ⫹40 m/s v x B ⫽ (40 ⫺ 5t B2) m/s ⫽ [40 ⫺ 5(2.0)2] m/s ⫽ ⫹20 m/s Therefore, the average acceleration in the speciﬁed time interval ⌬t ⫽ t B ⫺ t A ⫽ 2.0 s is What we have done by comparing the average acceleration during the interval between 훽 and 훾 (⫺10 m/s2) with the instantaneous value at 훾 (⫺20 m/s2) is compare the slope of the line (not shown) joining 훽 and 훾 with the slope of the tangent at 훾. Note that the acceleration is not constant in this example. Situations involving constant acceleration are treated in Section 2.5. So far we have evaluated the derivatives of a function by starting with the deﬁnition of the function and then taking the limit of a speciﬁc ratio. Those of you familiar with calculus should recognize that there are speciﬁc rules for taking derivatives. These rules, which are listed in Appendix B.6, enable us to evaluate derivatives quickly. For instance, one rule tells us that the derivative of any constant is zero. As another example, suppose x is proportional to some power of t, such as in the expression x ⫽ At n where A and n are constants. (This is a very common functional form.) The derivative of x with respect to t is dx ⫽ nAt n⫺1 dt Applying this rule to Example 2.4, in which vx ⫽ 40 ⫺ 5t 2, we ﬁnd that a x ⫽ dv x /dt ⫽ ⫺10t. 34 CHAPTER 2 2.4 Motion in One Dimension MOTION DIAGRAMS The concepts of velocity and acceleration are often confused with each other, but in fact they are quite different quantities. It is instructive to use motion diagrams to describe the velocity and acceleration while an object is in motion. In order not to confuse these two vector quantities, for which both magnitude and direction are important, we use red for velocity vectors and violet for acceleration vectors, as shown in Figure 2.9. The vectors are sketched at several instants during the motion of the object, and the time intervals between adjacent positions are assumed to be equal. This illustration represents three sets of strobe photographs of a car moving from left to right along a straight roadway. The time intervals between ﬂashes are equal in each diagram. In Figure 2.9a, the images of the car are equally spaced, showing us that the car moves the same distance in each time interval. Thus, the car moves with constant positive velocity and has zero acceleration. In Figure 2.9b, the images become farther apart as time progresses. In this case, the velocity vector increases in time because the car’s displacement between adjacent positions increases in time. The car is moving with a positive velocity and a positive acceleration. In Figure 2.9c, we can tell that the car slows as it moves to the right because its displacement between adjacent images decreases with time. In this case, the car moves to the right with a constant negative acceleration. The velocity vector decreases in time and eventually reaches zero. From this diagram we see that the acceleration and velocity vectors are not in the same direction. The car is moving with a positive velocity but with a negative acceleration. You should be able to construct motion diagrams for a car that moves initially to the left with a constant positive or negative acceleration. v (a) v (b) a v (c) a Figure 2.9 (a) Motion diagram for a car moving at constant velocity (zero acceleration). (b) Motion diagram for a car whose constant acceleration is in the direction of its velocity. The velocity vector at each instant is indicated by a red arrow, and the constant acceleration by a violet arrow. (c) Motion diagram for a car whose constant acceleration is in the direction opposite the velocity at each instant. 2.5 35 One-Dimensional Motion with Constant Acceleration Quick Quiz 2.2 (a) If a car is traveling eastward, can its acceleration be westward? (b) If a car is slowing down, can its acceleration be positive? 2.5 ONE-DIMENSIONAL MOTION WITH CONSTANT ACCELERATION If the acceleration of a particle varies in time, its motion can be complex and difﬁcult to analyze. However, a very common and simple type of one-dimensional motion is that in which the acceleration is constant. When this is the case, the average acceleration over any time interval equals the instantaneous acceleration at any instant within the interval, and the velocity changes at the same rate throughout the motion. If we replace a x by ax in Equation 2.5 and take t i ⫽ 0 and tf to be any later time t, we ﬁnd that ax ⫽ vx f ⫺ vxi t or vx f ⫽ vxi ⫹ axt (2.8) (for constant ax ) Velocity as a function of time This powerful expression enables us to determine an object’s velocity at any time t if we know the object’s initial velocity and its (constant) acceleration. A velocity – time graph for this constant-acceleration motion is shown in Figure 2.10a. The graph is a straight line, the (constant) slope of which is the acceleration ax ; this is consistent with the fact that ax ⫽ dvx /dt is a constant. Note that the slope is positive; this indicates a positive acceleration. If the acceleration were negative, then the slope of the line in Figure 2.10a would be negative. When the acceleration is constant, the graph of acceleration versus time (Fig. 2.10b) is a straight line having a slope of zero. Quick Quiz 2.3 Describe the meaning of each term in Equation 2.8. vx ax Slope = ax Slope = 0 a xt vxi x Slope = vxf vxf t 0 (a) xi ax vxi t t 0 (b) Figure 2.10 An object moving along the x axis with constant acceleration ax . (a) The velocity – time graph. (b) The acceleration – time graph. (c) The position – time graph. Slope = vxi t 0 (c) t 36 CHAPTER 2 Motion in One Dimension Because velocity at constant acceleration varies linearly in time according to Equation 2.8, we can express the average velocity in any time interval as the arithmetic mean of the initial velocity vxi and the ﬁnal velocity vx f : vx ⫽ v xi ⫹ v x f (for constant ax ) 2 (2.9) Note that this expression for average velocity applies only in situations in which the acceleration is constant. We can now use Equations 2.1, 2.2, and 2.9 to obtain the displacement of any object as a function of time. Recalling that ⌬x in Equation 2.2 represents xf ⫺ xi , and now using t in place of ⌬t (because we take ti ⫽ 0), we can say xf ⫺ xi ⫽ v x t ⫽ 12(vxi ⫹ vx f )t Displacement as a function of velocity and time (for constant ax ) (2.10) We can obtain another useful expression for displacement at constant acceleration by substituting Equation 2.8 into Equation 2.10: x f ⫺ x i ⫽ 12(v xi ⫹ v xi ⫹ a x t)t x f ⫺ x i ⫽ v xit ⫹ 12a x t 2 The position – time graph for motion at constant (positive) acceleration shown in Figure 2.10c is obtained from Equation 2.11. Note that the curve is a parabola. The slope of the tangent line to this curve at t ⫽ t i ⫽ 0 equals the initial velocity vxi , and the slope of the tangent line at any later time t equals the velocity at that time, vx f . We can check the validity of Equation 2.11 by moving the xi term to the righthand side of the equation and differentiating the equation with respect to time: ax vx vx f ⫽ t t (a) (d) vx ax t (e) vx ax dx f dt ⫽ d dt 冢x ⫹ v i xi t ⫹ 冣 1 a t 2 ⫽ v xi ⫹ a x t 2 x Finally, we can obtain an expression for the ﬁnal velocity that does not contain a time interval by substituting the value of t from Equation 2.8 into Equation 2.10: xf ⫺ xi ⫽ vx f ⫺ vxi 1 (vxi ⫹ vxf) 2 ax 冢 vx f 2 ⫽ vxi2 ⫹ 2ax(x f ⫺ x i) t (b) (2.11) 冣 ⫽ vx f 2 ⫺ vxi2 2ax (for constant ax ) (2.12) For motion at zero acceleration, we see from Equations 2.8 and 2.11 that vx f ⫽ vxi ⫽ vx x f ⫺ x i ⫽ vxt 冧 when ax ⫽ 0 That is, when acceleration is zero, velocity is constant and displacement changes linearly with time. t (c) t (f ) Figure 2.11 Parts (a), (b), and (c) are vx -t graphs of objects in one-dimensional motion. The possible accelerations of each object as a function of time are shown in scrambled order in (d), (e), and (f). Quick Quiz 2.4 In Figure 2.11, match each vx -t graph with the a x -t graph that best describes the motion. Equations 2.8 through 2.12 are kinematic expressions that may be used to solve any problem involving one-dimensional motion at constant accelera- 2.5 One-Dimensional Motion with Constant Acceleration 37 TABLE 2.2 Kinematic Equations for Motion in a Straight Line Under Constant Acceleration Equation Information Given by Equation vxf ⫽ vxi ⫹ a x t xf ⫺ x i ⫽ 12(vxi ⫹ vx f )t xf ⫺ x i ⫽ vxi t ⫹ 12a x t 2 vx f 2 ⫽ vxi 2 ⫹ 2a x (xf ⫺ x i) Velocity as a function of time Displacement as a function of velocity and time Displacement as a function of time Velocity as a function of displacement Note: Motion is along the x axis. tion. Keep in mind that these relationships were derived from the deﬁnitions of velocity and acceleration, together with some simple algebraic manipulations and the requirement that the acceleration be constant. The four kinematic equations used most often are listed in Table 2.2 for convenience. The choice of which equation you use in a given situation depends on what you know beforehand. Sometimes it is necessary to use two of these equations to solve for two unknowns. For example, suppose initial velocity vxi and acceleration ax are given. You can then ﬁnd (1) the velocity after an interval t has elapsed, using v x f ⫽ v xi ⫹ a x t, and (2) the displacement after an interval t has elapsed, using x f ⫺ x i ⫽ v xi t ⫹ 12a x t 2. You should recognize that the quantities that vary during the motion are velocity, displacement, and time. You will get a great deal of practice in the use of these equations by solving a number of exercises and problems. Many times you will discover that more than one method can be used to obtain a solution. Remember that these equations of kinematics cannot be used in a situation in which the acceleration varies with time. They can be used only when the acceleration is constant. CONCEPTUAL EXAMPLE 2.5 The Velocity of Different Objects ﬁned as ⌬x/⌬t.) There is one point at which the instantaneous velocity is zero — at the top of the motion. Consider the following one-dimensional motions: (a) A ball thrown directly upward rises to a highest point and falls back into the thrower’s hand. (b) A race car starts from rest and speeds up to 100 m/s. (c) A spacecraft drifts through space at constant velocity. Are there any points in the motion of these objects at which the instantaneous velocity is the same as the average velocity over the entire motion? If so, identify the point(s). (b) The car’s average velocity cannot be evaluated unambiguously with the information given, but it must be some value between 0 and 100 m/s. Because the car will have every instantaneous velocity between 0 and 100 m/s at some time during the interval, there must be some instant at which the instantaneous velocity is equal to the average velocity. Solution (a) The average velocity for the thrown ball is zero because the ball returns to the starting point; thus its displacement is zero. (Remember that average velocity is de- (c) Because the spacecraft’s instantaneous velocity is constant, its instantaneous velocity at any time and its average velocity over any time interval are the same. EXAMPLE 2.6 Entering the Trafﬁc Flow (a) Estimate your average acceleration as you drive up the entrance ramp to an interstate highway. Solution This problem involves more than our usual amount of estimating! We are trying to come up with a value of ax , but that value is hard to guess directly. The other three variables involved in kinematics are position, velocity, and time. Velocity is probably the easiest one to approximate. Let us assume a ﬁnal velocity of 100 km/h, so that you can merge with trafﬁc. W