Física para Ciencias e Ingenieria - Serway - 7ed SOL (1) PDF

1 Physics and Measurement CHAPTER OUTLINE ANSWERS TO QUESTIONS 1.1 Standards of Length, Mass, and Time...

1 Physics and Measurement CHAPTER OUTLINE ANSWERS TO QUESTIONS 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model-Building * An asterisk indicates an item new to this edition. 1.3 Dimensional Analysis 1.4 Conversion of Units 1.5 Estimates and Order-of- Q1.1 Density varies with temperature and pressure. It would Magnitude Calculations be necessary to measure both mass and volume very 1.6 Significant Figures accurately in order to use the density of water as a standard. Q1.2 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms *Q1.3 In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041 kg (e) 0.27 kg. Then the ranking is c=e>d>a>b Q1.4 No: A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimension- ally correct. Yes: If an equation is not dimensionally correct, it cannot be correct. *Q1.5 The answer is yes for (a), (c), and (f ). You cannot add or subtract a number of apples and a number of jokes. The answer is no for (b), (d), and (e). Consider the gauge of a sausage, 4 kgⲐ2 m, or the volume of a cube, (2 m)3. Thus we have (a) yes (b) no (c) yes (d) no (e) no (f ) yes *Q1.6 41 € ≈ 41 € (1 LⲐ1.3 €)(1 qtⲐ1 L)(1 galⲐ4 qt) ≈ (10Ⲑ1.3) gal ≈ 8 gallons, answer (c) *Q1.7 The meterstick measurement, (a), and (b) can all be 4.31 cm. The meterstick measurement and (c) can both be 4.24 cm. Only (d) does not overlap. Thus (a) (b) and (c) all agree with the meterstick measurement. *Q1.8 0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 107 kg. So (d) 3 digits are significant. SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time Modeling the Earth as a sphere, we find its volume as π r 3 = π ( 6.37 × 10 6 m ) = 1.08 × 10 21 m 3. 4 4 3 P1.1 3 3 m 5.98 × 10 24 kg Its density is then ρ = = = 5.52 × 10 3 kg m 3. This value is intermediate V 1.08 × 10 m 21 3 between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kgⲐm3. The average density of the Earth is significantly higher, so higher-density material must be down below the surface. 1 ISMV1_5103_01.indd 1 10/27/06 4:33:21 PM  Libros y Software Gratuitos www.youtube.com/MiLibroPDFyMas Mi Libro PDF y Más Libros y Software Gratuitos “PDF´s de calidad siempre” “Siempre con solucionario completo” “Programas más buscados” Siguenos: Contacto Correo: [email protected]  2 Chapter 1 m P1.2 With V = ( base area ) ( height ) V = (π r 2 ) h and ρ = , we have V m 1 kg ⎛ 10 9 mm 3 ⎞ ρ= = π r h π (19.5 mm ) ( 39.0 mm ) ⎜⎝ 1 m 3 ⎟⎠ 2 2 ρ = 2.15 × 10 4 kg m 3. m P1.3 Let V represent the volume of the model, the same in ρ = for both. Then ρiron = 9.35 kg V V mgold ρgold mgold ⎛ 19.3 × 10 3 kg/m 3 ⎞ and ρgold =. Next, = and mgold = 9.35 kg ⎜ = 23.0 kg. V ρiron 9.35 kg ⎝ 7.86 × 10 3 kg/m 3 ⎟⎠ *P1.4 ρ = m / V and V = (4 / 3)π r 3 = (4 / 3)π (d / 2)3 = π d 3 / 6 where d is the diameter. 6(1.67 × 10 −27 kg) Then ρ = 6 m / π d 3 = = 2.3 × 1017 kg/m 3 π (2.4 × 10 −15 m)3 2.3 × 1017 kg/m 3 /(11.3 × 10 3 kg/m 3 ) = it is 20 × 1012 times the density of lead. 4 3 4 P1.5 For either sphere the volume is V = π r and the mass is m = ρV = ρ π r 3. We divide 3 3 this equation for the larger sphere by the same equation for the smaller: mᐉ ρ 4π rᐉ3 3 rᐉ3 = = = 5. ms ρ 4π rs3 3 rs3 Then rᐉ = rs 3 5 = 4.50 cm (1.71) = 7.69 cm. Section 1.2 Matter and Model-Building P1.6 From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = L2 + L2. Thus, since the atoms are separated by a distance 1 2 L = 0.200 nm, the diagonal planes are separated by L + L2 = 0.141 nm. 2 Section 1.3 Dimensional Analysis P1.7 (a) This is incorrect since the units of [ ax ] are m 2 s 2 , while the units of [ v ] are m s. This is correct since the units of [ y ] are m, and cos ( kx ) is dimensionless if [ k ] is in m. −1 (b) P1.8 (a) Circumference has dimensions of L. (b) Volume has dimensions of L3. (c) Area has dimensions of L2. Expression (i) has dimension L ( L2 ) 1/ 2 = L2 , so this must be area (c). Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L ( L2 ) = L3, so it is (b). Thus, (a) = ii; (b) = iii;(c) = i. ISMV1_5103_01.indd 2 10/27/06 12:27:01 PM  Physics and Measurement 3 P1.9 Inserting the proper units for everything except G, ⎡ kg m ⎤ = G [ kg ]. 2 ⎢⎣ s 2 ⎥⎦ [ m ]2 Multiply both sides by [ m ]2 and divide by [ kg ] ; the units of G are 2 m3. kg ⋅ s 2 Section 1.4 Conversion of Units P1.10 Apply the following conversion factors: 1 in = 2.54 cm, 1 d = 86 400 s, 100 cm = 1 m, and 10 9 nm = 1 m ⎛ 1 in day⎞ ( 2.54 cm in ) (10 m cm ) (10 nm m ) = 9.19 nm s. −2 9 ⎝ 32 ⎠ 86 400 s day This means the proteins are assembled at a rate of many layers of atoms each second! P1.11 Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we ( )( ) should expect the area to be about A ≈ 30 m 50 m = 1500 m 2. Categorize: We model the lot as a perfect rectangle to use Area = Length × Width. Use the conversion:1 m = 3.281 ft. Analyze: A = LW = (100 ft ) ⎛ (150 ft ) ⎛ 1m ⎞ 1m ⎞ = 1 390 m 2 = 1.39 × 10 3 m 2. ⎝ 3.281 ft ⎠ ⎝ 3.281 ft ⎠ Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m 2. Unit conversion is a common technique that is applied to many problems. P1.12 (a) V = ( 40.0 m ) ( 20.0 m ) (12.0 m ) = 9.60 × 10 3 m 3 V = 9.60 × 10 3 m 3 ( 3.28 ft 1 m ) = 3.39 × 10 5 ft 3 3 (b) The mass of the air is m = ρairV = (1.20 kg m 3 ) ( 9.60 × 10 3 m 3 ) = 1.15 × 10 4 kg. The student must look up weight in the index to find Fg = mg = (1.15 × 10 4 kg ) ( 9.80 m s 2 ) = 1.13 × 10 5 N. Converting to pounds, Fg = (1.13 × 10 5 N ) (1 lb 4.45 N ) = 2.54 × 10 4 lb. *P1.13 The area of the four walls is (3.6 + 3.8 + 3.6 + 3.8)m (2.5 m) = 37 m2. Each sheet in the book has area (0.21 m) (0.28 m) = 0.059 m2. The number of sheets required for wallpaper is 37 m2Ⲑ0.059 m2 = 629 sheets = 629 sheets(2 pagesⲐ1 sheet) = 1260 pages. The pages from volume one are inadequate, but the full version has enough pages. ISMV1_5103_01.indd 3 10/27/06 4:18:09 PM  4 Chapter 1 P1.14 (a) Seven minutes is 420 seconds, so the rate is 30.0 gal r= = 7.14 × 10 −2 gal s. 420 s (b) Converting gallons first to liters, then to m 3, ⎛ 3.786 L ⎞ ⎛ 10 −3 m 3 ⎞ r = ( 7.14 × 10 −2 gal s ) ⎜ ⎝ 1 gal ⎟⎠ ⎜⎝ 1 L ⎟⎠ r = 2.70 × 10 −4 m 3 s. (c) At that rate, to fill a 1-m 3 tank would take ⎛ 1 m3 ⎞⎛ 1h ⎞ t=⎜ = 1.03 h. ⎝ 2.70 × 10 m s ⎟⎠ ⎜⎝ 3 600 ⎟⎠ −4 3 P1.15 From Table 14.1, the density of lead is 1.13 × 10 4 kg m 3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks. m Density is defined as mass per volume, in ρ =. We must convert to SI units in the calculation. V 3 23.94 g ⎛ 1 kg ⎞ ⎛ 100 cm ⎞ 23.94 g ⎛ 1 kg ⎞ ⎛ 1 000 000 cm 3 ⎞ ρ= 3⎜ ⎟ ⎜ ⎟ = 3⎜ ⎟ ⎜ ⎟ = 1.14 × 10 4 kg m 3 2.10 cm ⎝ 1000 g ⎠⎝ 1 m ⎠ 2.10 cm ⎝ 1000 g ⎠⎝ 1m 3 ⎠ At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g cm 3, and objects that float must be less dense than water. ton ⎛ 2 000 lb ⎞ ⎛ 1 h ⎞ ⎛ 1 min ⎞ P1.16 The weight flow rate is 1 200 = 667 lb s. h ⎝ ton ⎠ ⎝ 60 min ⎠ ⎝ 60 s ⎠ P1.17 (a) ⎛ 8 × 1012 $ ⎞ ⎛ 1 h ⎞ ⎛ 1 day ⎞ ⎛ 1 yr ⎞ ⎜⎝ 1 000 $ s ⎟⎠ ⎜⎝ 3 600 s ⎟⎠ ⎝ 24 h ⎠ ⎜⎝ 365 days ⎟⎠ = 250 years (b) The circumference of the Earth at the equator is 2π ( 6.378 × 10 3 m ) = 4.01 × 10 7 m. The length of one dollar bill is 0.155 m so that the length of 8 trillion bills is 1.24 × 1012 m. Thus, the 8 trillion dollars would encircle the Earth 1.24 × 1012 m = 3.09 × 10 4 times. 4.01 × 10 7 m ⎡(13.0 acres ) ( 43 560 ft 2 acre ) ⎤⎦ Bh = ⎣ 1 P1.18 V= ( 481 ft ) h 3 3 = 9.08 × 10 7 ft 3 , B or ⎛ 2.83 × 10 −2 m 3 ⎞ V = ( 9.08 × 10 7 ft 3 ) ⎜ ⎟⎠ FIG. P1.18 ⎝ 1 ft 3 = 2.57 × 10 6 m 3 ISMV1_5103_01.indd 4 10/27/06 12:27:03 PM  Physics and Measurement 5 P1.19 Fg = ( 2.50 tons block ) ( 2.00 × 10 6 blocks ) ( 2 000 lb ton ) = 1.00 × 1010 lbs ⎛d ⎞ dnucleus, scale = dnucleus, real ⎜ atom, scale ⎟ = ( 2.40 × 10 −15 m ) ⎛ ⎞ = 6.79 × 10 −3 ft , or 300 ft P1.20 (a) ⎝ datom, real ⎠ ⎝ 1.06 × 10 −10 m ⎠ dnucleus, scale = ( 6.79 × 10 −3 ft ) ( 304.8 mm 1 ft ) = 2.07 mm 3 3 3 (b) Vatom 4π ratom 3 / 3 ⎛ ratom ⎞ ⎛ datom ⎞ ⎛ 1.06 × 10 −10 m ⎞ = =⎜ = = ⎜⎝ 2.40 × 10 −15 m ⎟⎠ Vnucleus 4π rnucleus / 3 ⎝ rnucleus ⎟⎠ 3 ⎜⎝ d ⎟ nucleus ⎠ = 8.62 × 1013 times as large V 3.78 × 10 −3 m 3 P1.21 V = At so t = = = 1.51 × 10 −4 m ( or 151 µ m ) A 25.0 m 2 ⎛ ( 6.37 × 10 6 m ) (100 cm m ) ⎞ 2 2 P1.22 (a) AEarth 4π rEarth 2 ⎛r ⎞ = = ⎜ Earth ⎟ = ⎜ ⎟ = 13.4 AMoon 4π rMoon ⎝ rMoon ⎠ 2 ⎝ 1.74 × 108 cm ⎠ ⎛ ( 6.37 × 10 6 m ) (100 cm m ) ⎞ 3 3 (b) VEarth 4π rEarth 3 / 3 ⎛ rEarth ⎞ = =⎜ = ⎜ ⎟ = 49.1 VMoon 4π rMoon / 3 ⎝ rMoon ⎟⎠ 3 ⎝ 1.74 × 108 cm ⎠ P1.23 To balance, mFe = mAl or ρFeVFe = ρAlVAl ⎛ 4⎞ ρFe ⎜ ⎟ π rFe 3 = ρAl ⎛ ⎞ π rAl 3 4 ⎝ 3⎠ ⎝ 3⎠ 1/ 3 ⎛ρ ⎞ 1/ 3 = ( 2.00 cm ) ⎛ 7.86 ⎞ rAl = rFe ⎜ Fe ⎟ = 2.86 cm. ⎝ ρAl ⎠ ⎝ 2.70 ⎠ P1.24 The mass of each sphere is 4π ρAl rAl 3 mAl = ρAlVAl = 3 and 4π ρFe rFe 3 mFe = ρFeVFe =. 3 Setting these masses equal, 4π ρAl rAl 3 4π ρFe rFe 3 and r = r 3 ρFe. = 3 3 Al Fe ρAl The resulting expression shows that the radius of the aluminum sphere is directly proportional to the radius of the balancing iron sphere. The sphere of lower density has larger radius. The ρ fraction Fe is the factor of change between the densities, a number greater than 1. Its cube root ρAl is a number much closer to 1. The relatively small change in radius implies a change in volume sufficient to compensate for the change in density. ISMV1_5103_01.indd 5 10/27/06 12:27:04 PM  6 Chapter 1 Section 1.5 Estimates and Order-of-Magnitude Calculations P1.25 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 × 4 × 3 = 48 m 3, while the volume of one ball is 4π ⎛ 0.038 m ⎞ 3 = 2.87 × 10 −5 m 3. 3 ⎝ 2 ⎠ 48 Therefore, one can fit about ~ 10 6 ping-pong balls in the room. 2.87 × 10 −5 As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best packing fraction” is 1 π 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the 6 above estimate reduces to 1.67 × 10 6 × 0.740 ~ 10 6. P1.26 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. ( )( )( ) Thus, the tire would make 50 000 mi 5 280 ft mi 1 rev 8 ft = 3 × 10 7 rev ~ 10 7 rev. P1.27 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then V = ( 0.5) (1.3 m ) ( 0.5 m ) ( 0.3 m ) = 0.10 m 3. The mass of this volume of water is mwater = ρwaterV = (1 000 kg m 3 ) ( 0.10 m 3 ) = 100 kg ~ 10 2 kg. Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is mcopper = ρcopperV = (8 920 kg m 3 ) ( 0.10 m 3 ) = 892 kg ~ 10 3 kg. *P1.28 The time required for the task is ⎛ 1 s ⎞ 1 h ⎞ ⎛ 1 working day ⎞ ⎛ ⎞ 10 9 $ ⎜ ⎟ ⎛ 1 bad yr ⎜ = 58 yr ⎝ 1 $ ⎠ ⎝ 3600 s ⎠ ⎝ 16 h ⎠ ⎝ 300 working days ⎟⎠ Since you are already around 20 years old, you would have a miserable life and likely die before accomplishing the task. You have better things to do. Say no. P1.29 Assume: Total population = 10 7; one out of every 100 people has a piano; one tuner can serve about 1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, ⎛ 1 tuner ⎞ ⎛ 1 piano ⎞ # tuners ~ ⎜ (10 7 people ) = 100 tuners. ⎝ 1 000 pianos ⎟⎠ ⎜⎝ 100 people ⎟⎠ Section 1.6 Significant Figures P1.30 METHOD ONE We treat the best value with its uncertainty as a binomial ( 21.3 ± 0.2 ) cm ( 9.8 ± 0.1) cm, A = [ 21.3 ( 9.8 ) ± 21.3 ( 0.1) ± 0.2 ( 9.8 ) ± ( 0.2 ) ( 0.1)] cm 2. The first term gives the best value of the area. The cross terms add together to give the uncertainty and the fourth term is negligible. A = 209 cm 2 ± 4 cm 2. METHOD TWO We add the fractional uncertainties in the data. A = ( 21.3 cm ) ( 9.8 cm ) ± ⎛ 0.2 0.1 ⎞ + = 209 cm 2 ± 2% = 209 cm 2 ± 4 cm 2 ⎝ 21.3 9.8 ⎠ ISMV1_5103_01.indd 6 10/28/06 2:41:27 AM  Physics and Measurement 7 P1.31 (a) 3 (b) 4 (c) 3 (d) 2 P1.32 r = ( 6.50 ± 0.20 ) cm = ( 6.50 ± 0.20 ) × 10 −2 m m = (1.85 ± 0.02 ) kg m ρ= ( 4 3 )π r3 also, δ ρ = δ m + 3δ r. ρ m r In other words, the percentages of uncertainty are cumulative. Therefore, δ ρ 0.02 3 ( 0.20 ) = + = 0.103, ρ 1.85 6.50 1.85 ρ= = 1.61 × 10 3 kg m 3 ( ) π ( 6.5 × 10 m) −2 3 4 3 and ρ ± δ ρ = (1.61 ± 0.17 ) × 10 3 kg m 3 = (1.6 ± 0.2 ) × 10 3 kg m 3. P1.33 (a) 756.?? 37.2? 0.83 + 2.5? 796. / 5 / 3 = 797 (b) 0.003 2 ( 2 s.f.) × 356.3 ( 4 s.f.) = 1.140 16 = ( 2 s.f.) 1.1 (c) 5.620 ( 4 s.f.) × π ( >4 s.f.) = 17.656= ( 4 s.f.) 17.66 P1.34 We work to nine significant digits: ⎛ 365.242 199 d ⎞ ⎛ 24 h ⎞ ⎛ 60 min ⎞ ⎛ 60 s ⎞ 1 yr = 1 yr ⎜ ⎟⎠ ⎝ 1 d ⎠ ⎝ 1 h ⎠ ⎝ 1 min ⎠ = 31 556 926.0 s. ⎝ 1 yr *P1.35 The tax amount is $1.36 − $1.25 = $0.11. The tax rate is $0.11Ⲑ$1.25 = 0.088 0 = 8.80% *P1.36 (a) We read from the graph a vertical separation of 0.3 spaces = 0.015 g. (b) Horizontally, 0.6 spaces = 30 cm2. (c) Because the graph line goes through the origin, the same percentage describes the vertical and the horizontal scatter: 30 cm2Ⲑ380 cm2 = 8%. (d) Choose a grid point on the line far from the origin: slope = 0.31 g Ⲑ600 cm2 = 0.000 52 gⲐcm2 = (0.000 52 gⲐcm2)(10 000 cm2Ⲑ1 m2) = 5.2 g/m2. (e) For any and all shapes cut from this copy paper, the mass of the cutout is proportional to its area. The proportionality constant is 5.2 g/m2 ± 8%, where the uncertainty is estimated. (f ) This result should be expected if the paper has thickness and density that are uniform within the experimental uncertainty. The slope is the areal density of the paper, its mass per unit area. ISMV1_5103_01.indd 7 10/28/06 2:41:45 AM  8 Chapter 1 *P1.37 15 players = 15 players (1 shift
1.667 player) = 9 shifts *P1.38 Let o represent the number of ordinary cars and s the number of trucks. We have o = s + 0.947s = 1.947s, and o = s + 18. We eliminate o by substitution: s + 18 = 1.947s 0.947s = 18 and s = 18
0.947 = 19. *P1.39 Let s represent the number of sparrows and m the number of more interesting birds. We have s
m = 2.25 and s + m = 91. We eliminate m by substitution: m = s
2.25 s + s
2.25 = 91 1.444s = 91 s = 91
1.444 = 63. *P1.40 For those who are not familiar with solving equations numerically, we provide a detailed solution. It goes beyond proving that the suggested answer works. The equation 2 x 4 − 3 x 3 + 5 x − 70 = 0 is quartic, so we do not attempt to solve it with algebra. To find how many real solutions the equation has and to estimate them, we graph the expression: x −3 −2 −1 0 1 2 3 4 y = 2 x 4 − 3 x 3 + 5 x − 70 158 −24 −70 −70 −66 −52 26 270 We see that the equation y = 0 has two roots, one around x = −2.2 and the other near x = +2.7. To home in on the first of these solutions y we compute in sequence: When x = −2.2, y = −2.20. The root must be between x = −2.2 and x = −3. When x = −2.3, y = 11.0. The root is between x = −2.2 and x = −2.3. When x = −2.23, y = 1.58. The root x is between x = −2.20 and x = −2.23. When x = −2.22, y = 0.301. The root is between x = −2.20 and −2.22. When x = −2.215, y = −0.331. The root is between x = −2.215 and −2.22. We could next try FIG. P1.40 x = −2.218, but we already know to three-digit precision that the root is x = −2.22. *P1.41 We require sin θ = −3 cos θ , or sin θ = −3 , or tan θ = −3. tan θ cos θ For tan −1 ( −3) = arc tan ( −3) , your calculator may return −71.6°, but this angle is not between 0° and 360° as the problem θ requires. The tangent function is negative in the second quad- 0 360° rant (between 90° and 180°) and in the fourth quadrant (from 270° to 360°). The solutions to the equation are then 360° − 71.6° = 288° and 180° − 71.6° = 108°. FIG. P1.41 ISMV1_5103_01.indd 8 10/27/06 12:27:07 PM  Physics and Measurement 9 *P1.42 We draw the radius to the initial point and the radius to the final point. i The angle θ between these two radii has its sides perpendicular, right side to right side and left side to left side, to the 35° angle between the original 35.0° and final tangential directions of travel. A most useful theorem from R f geometry then identifies these angles as equal: θ = 35°. The whole N θ circumference of a 360° circle of the same radius is 2π R. By proportion, W E then 2π R = 840 m. S 360° 35° FIG. P1.42 360° 840 m 840 m R= = = 1.38 × 10 3 m 2π 35° 0.611 We could equally well say that the measure of the angle in radians is 2π radians ⎞ θ = 35° = 35° ⎛ 840 m = 0.611 rad =. ⎝ 360 ° ⎠ R Solving yields R = 1.38 km. *P1.43 Mass is proportional to cube of length: m = kᐉ3 mƒ Ⲑmi = (ᐉf Ⲑᐉi)3. Length changes by 15.8%: ᐉf = ᐉi + 0.158 ᐉi = 1.158 ᐉi. Mass increase: mf = mi + 17.3 kg. mf Eliminate by substitution: = 1.1583 = 1.553 m f − 17.3 kg mf = 1.553 mf − 26.9 kg 26.9 kg = 0.553 mf mf = 26.9 kg Ⲑ0.553 = 48. 6 kg. *P1.44 We use substitution, as the most generally applicable method for solving simultaneous equations. We substitute p = 3q into each of the other two equations to eliminate p: ⎧3qr = qs ⎪ ⎨1 1 2 1 2. ⎪⎩ 2 3qr + 2 qs = 2 qt 2 ⎧3r = s 3r 2 + ( 3r ) = t 2 2 These simplify to ⎨ 2 2 2. We substitute to eliminate s:. We solve for the ⎩3r + s = t 12r 2 = t 2 combination t : r t2 = 12. r2 t = either 3.46 or − 3.46 r *P1.45 Solve the given equation for ∆t: ∆t = 4QLⲐkπd 2(Th − Tc) = [4QLⲐkπ (Th − Tc)] [1Ⲑ d 2]. (a) Making d three times larger with d 2 in the bottom of the fraction makes ∆t nine times smaller. (b) ∆t is inversely proportional to the square of d. (c) Plot ∆t on the vertical axis and 1/d 2 on the horizontal axis. (d) From the last version of the equation, the slope is 4QL/kπ(Th − Tc). Note that this quantity is constant as both ∆t and d vary. ISMV1_5103_01.indd 9 10/28/06 2:42:03 AM  10 Chapter 1 Additional Problems P1.46 It is desired to find the distance x such that x 1 000 m = 100 m x (i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that x 2 = (100 m ) (1 000 m ) = 1.00 × 10 5 m 2 and therefore x = 1.00 × 10 5 m 2 = 316 m. *P1.47 (a) The mass is equal to the mass of a sphere of radius 2.6 cm and density 4.7 g
cm3, minus the mass of a sphere of radius a and density 4.7 g
cm3 plus the mass of a sphere of radius a and density 1.23 g
cm3. m = ρ14πr 3
3 − ρ14πa 3
3 + ρ24πa 3
3 = (4.7 g
cm3)4π(2.6 cm)3
3 − (4.7 g
cm3)4π(a)3
3 + (1.23 g
cm3)4π(a)3
3 m = 346 g − (14.5 g/cm3)a3 (b) For a = 0 the mass is a maximum, (c) 346 g. (d) Yes. This is the mass of the uniform sphere we considered in the first term of the calculation. (e) For a = 2.60 cm the mass is a minimum, (f ) 346 − 14.5(2.6)3 = 90.6 g. (g) Yes. This is the mass of a uniform sphere of density 1.23 g
cm3. (h) (346 g + 90.6 g)
2 = 218 g (i) No. The result of part (a) gives 346 g − (14.5 g
cm3) (1.3 cm)3 = 314 g, not the same as 218 g. ( j) We should expect agreement in parts b-c-d, because those parts are about a uniform sphere of density 4.7 g/cm3. We should expect agreement in parts e-f-g, because those parts are about a uniform liquid drop of density 1.23 g/cm3. The function m(a) is not a linear function, so a halfway between 0 and 2.6 cm does not give a value for m halfway between the minimum and maximum values. The graph of m versus a starts at a = 0 with a horizontal tangent. Then it curves down more and more steeply as a increases. The liquid drop of radius 1.30 cm has only one eighth the volume of the whole sphere, so its presence brings down the mass by only a small amount, from 346 g to 314 g. (k) No change, so long as the wall of the shell is unbroken. *P1.48 (a) We have B + C(0) = 2.70 g
cm3 and B + C(14 cm) = 19.3 g
cm3. We know B = 2.70 g/cm3 and we solve for C by subtracting: C(14 cm) = 16.6 g
cm3 so C = 1.19 g/cm4. 14 cm (b) m= ∫0 (2.70 g/cm 3 + 1.19 g/cm 4 x )(9 cm 2 )dx 14 cm 14 cm = 24.3 g/cm ∫ dx + 10.7 g/cm 2 ∫ xdx 0 0 = (24.3 g/cm)(14 cm – 0) + (10.7 g/cm 2 )[(14 cm)2 − 0] / 2 = 340 g + 1046 g = 1.39 kg ISMV1_5103_01.indd 10 10/27/06 12:27:09 PM  Physics and Measurement 11 P1.49 The scale factor used in the “dinner plate” model is 0.25 m S= = 2.5 × 10 −6 m lightyeears. 1.0 × 10 5 lightyears The distance to Andromeda in the scale model will be Dscale = Dactual S = ( 2.0 × 10 6 lightyears ) ( 2.5 × 10 −66 m lightyears ) = 5.0 m. *P1.50 The rate of volume increase is dV d 4 3 4 dr dr = π r = π 3r 2 = 4π r 2. dt dt 3 3 dt dt (a) dV
dt = 4 π(6.5 cm)2(0.9 cm
s) = 478 cm3/s dr dV / dt 478 cm 3 / s (b) = = = 0.225 cm 3 / s dt 4π r 2 4π (13 cm)2 (c) When the balloon radius is twice as large, its surface area is four times larger. The new volume added in one second in the inflation process is equal to this larger area times an extra radial thickness that is one-fourth as large as it was when the balloon was smaller. P1.51 One month is 1 mo = ( 30 day ) ( 24 h day ) ( 3 600 s h ) = 2.592 × 10 6 s. Applying units to the equation, V = (1.50 Mft 3 mo ) t + ( 0.008 00 Mft 3 mo2 ) t 2. Since 1 Mft 3 = 10 6 ft 3 , V = (1.50 × 10 6 ft 3 mo ) t + ( 0.008 00 × 10 6 ft 3 mo2 ) t 2. Converting months to seconds, 1.50 × 10 6 ft 3 mo 0.008 00 × 10 6 ft 3 mo2 2 V= t + t. 2.592 × 10 6 s mo ( 2.592 × 106 s mo)2 Thus, V [ft 3 ] = ( 0.579 ft 3 s ) t + (1.19 × 10 −9 ft 3 s 2 ) t 2. *P1.52 α ′(deg) α (rad) tan (α ) sin (α ) difference between α and tan α 15.0 0.262 0.268 0.259 2.30% 20.0 0.349 0.364 0.342 4.09% 30.0 0.524 0.577 0.500 9.32% 33.0 0.576 0.649 0.545 11.3% 31.0 0.541 0.601 0.515 9.95% 31.1 0.543 0.603 0.516 10.02% We see that  in radians, tan() and sin() start out together from zero and diverge only slightly in value for small angles. Thus 31.0 º is the largest angle for which tan α − α < 0.1. tan α ISMV1_5103_01.indd 11 10/27/06 12:27:09 PM  12 Chapter 1 P1.53 2π r = 15.0 m r = 2.39 m h = tan 55.0 ° r h h = ( 2.39 m ) tan (55.0° ) = 3.41 m P1.54 Let d represent the diameter of the coin and h its thickness. 55° The mass of the gold is r ⎛ 2π d 2 ⎞ m = ρV = ρ At = ρ ⎜ + π dh⎟ t ⎝ 4 ⎠ FIG. P1.53 where t is the thickness of the plating. ⎡ ( 2.41)2 ⎤ m = 19.3 ⎢ 2π + π ( 2.41) ( 0.178 ) ⎥ ( 0.18 × 10 −4 ) ⎣ 4 ⎦ = 0.003 64 grams cost = 0.003 64 grams × $10 grram = $0.036 4 = 3.64 cents This is negligible compared to $4.98. P1.55 The actual number of seconds in a year is (86 400 s day ) ( 365.25 day yr ) = 31 557 600 s yr. The percent error in the approximation is (π × 10 7 s yr ) − ( 31 557 600 s yr ) × 100% = 0.449%. 31 557 600 s yr ⎛ furlongs ⎞ ⎛ 220 yd ⎞ ⎛ 0.914 4 m ⎞ ⎛ 1 fortnight ⎞ ⎛ 1 day ⎞ ⎛ 1 hr ⎞ P1.56 v = ⎜ 5.00 = 8.32 × 10 −4 m s ⎝ fortnight ⎟⎠ ⎜⎝ 1 furlongg ⎟⎠ ⎜⎝ 1 yd ⎟⎠ ⎜⎝ 14 days ⎟⎠ ⎝ 24 hrs ⎠ ⎜⎝ 3 600 s ⎟⎠ This speed is almost 1 mm
s; so we might guess the creature was a snail, or perhaps a sloth. P1.57 (a) The speed of rise may be found from ( Vol rate of flow ) 16.5 cm 3 s v= = = 0.529 cm s. (Area: π D 2 / 4) π ( 6.30 cm )2 / 4 (b) Likewise, at a 1.35 cm diameter, 16.5 cm 3 s v= = 11.5 cm s. π (1.35 cm )2 / 4 ISMV1_5103_01.indd 12 10/27/06 12:27:10 PM  Physics and Measurement 13 m m 4m P1.58 The density of each material is ρ = = =. V π r 2h π D2h 4 ( 51.5 g ) The tabulated value ⎛ 2.70 g g ⎞ Al: ρ = = 2.75 3 is 2% smaller. π ( 2.52 cm ) ( 3.75 cm ) 2 cm ⎝ cm 3 ⎠ 4 ( 56.3 g ) The tabulated value ⎛ 8.92 3 ⎞ is 5% smaller. g g Cu: ρ = = 9.36 3 π (1.23 cm )2 ( 5.06 cm ) cm ⎝ cm ⎠ 4 ( 94.4 g ) g Brass: ρ = = 8.91 3 π (1.54 cm ) ( 5.69 cm ) 2 cm 4 ( 69.1 g ) g Sn: ρ = = 7.68 3 π (1.75 cm ) ( 3.74 cm ) 2 cm 4 ( 216.1 g ) The tabulated value ⎛ 7.86 3 ⎞ is 0.3% smaller. g g Fe: ρ = = 7.88 3 π (1.89 cm )2 ( 9.77 cm ) cm ⎝ cm ⎠ (108 cars)(10 4 mi yr ) P1.59 V20 mpg = = 5.0 × 1010 gal yr 20 mi gal (108 cars)(10 4 mi yr ) V25 mpg = = 4.0 × 1010 gal yr 25 mi gal Fuel saved = V25 mpg − V20 mpg = 1.0 × 1010 gal yr P1.60 The volume of the galaxy is π r 2t = π (10 21 m ) (1019 m ) ~ 10 61 m 3. 2 If the distance between stars is 4 × 1016 m, then there is one star in a volume on the order of ( 4 × 10 m ) ~ 10 50 m 3. 16 3 The number of stars is about 10 61 m 3 ~ 1011 stars. 10 50 m 3 star ANSWERS TO EVEN-NUMBERED PROBLEMS P1.2 2.15 × 104 kg
m3 P1.4 2.3 × 1017 kg
m3 is twenty trillion times larger than the density of lead. P1.6 0.141 nm P1.8 (a) ii (b) iii (c) i P1.10 9.19 nm
s P1.12 (a) 3.39 × 105 ft3 (b) 2.54 × 104 lb P1.14 (a) 0.071 4 gal
s (b) 2.70 × 10−4 m3
s (c) 1.03 h ISMV1_5103_01.indd 13 10/27/06 12:27:11 PM  14 Chapter 1 P1.16 667 lb
s P1.18 2.57 × 10 6 m 3 P1.20 (a) 2.07 mm (b) 8.57 × 1013 times as large P1.22 (a) 13.4; (b) 49.1 13 ⎛ρ ⎞ P1.24 rAl = rFe ⎜ Fe ⎟ ⎝ ρAl ⎠ P1.26 ~10 7 rev P1.28 No. There is a strong possibility that you would die before finishing the task, and you have much more productive things to do. P1.30 ( 209 ± 4 ) cm 2 P1.32 (1.61 ± 0.17) × 103 kg
m3 P1.34 31 556 926.0 s P1.36 (a) 0.015 g (b) 30 cm2 (c) 8% (d) 5.2 g
m2 (e) For any and all shapes cut from this copy paper, the mass of the cutout is proportional to its area. The proportionality constant is 5.2 g
m2 ± 8%, where the uncertainty is estimated. (f) This result is to be expected if the paper has thickness and density that are uniform within the experimental uncertainty. The slope is the areal density of the paper, its mass per unit area. P1.38 19 P1.40 see the solution P1.42 1.38 km P1.44 either 3.46 or −3.46 P1.46 316 m P1.48 (a) ρ = 2.70 g
cm3 + 1.19 g
cm4 x (b) 1.39 kg P1.50 (a) 478 cm3
s (b) 0.225 cm
s (c) When the balloon radius is twice as large, its surface area is four times larger. The new volume added in one increment of time in the inflation process is equal to this larger area times an extra radial thickness that is one-fourth as large as it was when the bal- loon was smaller. P1.52 0.542 rad P1.54 3.64 cents; no P1.56 8.32 × 10 −4 m s; a snail P1.58 see the solution P1.60 ~1011 stars ISMV1_5103_01.indd 14 10/27/06 12:27:11 PM  2 Motion in One Dimension CHAPTER OUTLINE ANSWERS TO QUESTIONS 2.1 Position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed * An asterisk indicates an item new to this edition. 2.3 Acceleration 2.4 Motion Diagrams 2.5 One-Dimensional Motion with *Q2.1 Count spaces (intervals), not dots. Count 5, not 6. The first Constant Acceleration drop falls at time zero and the last drop at 5 × 5 s = 25 s. 2.6 Freely Falling Objects 2.7 Kinematic Equations Derived The average speed is 600 m
25 s = 24 m
s, answer (b). from Calculus Q2.2 The net displacement must be zero. The object could have moved away from its starting point and back again, but it is at its initial position again at the end of the time interval. Q2.3 Yes. Yes. If the speed of the object varies at all over the interval, the instantaneous velocity will sometimes be greater than the average velocity and will sometimes be less. *Q2.4 (a) It speeds up and its acceleration is positive. (b) It slows down overall, since final speed 1 m
s is slower than 3 m
s. Its acceleration is positive, meaning to the right. (c) It slows down and its acceleration is negative. (d) It speeds up to final speed 7 m
s. Its acceleration is negative, mean- ing toward the left or towards increasing-magnitude negative numbers on the track. Q2.5 No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past to give car B greater acceleration just then. *Q2.6 (c) A graph of velocity versus time slopes down steadily from an original positive (northward) value. The graph cuts through zero and goes through increasing-magnitude negative values, all with the same constant acceleration. *Q2.7 (i) none. All of the disks are moving. (ii) (b) shows equal spacing, meaning constant nonzero velocity and constant zero acceleration. (iii) (b) This question has the same physical meaning as question (ii). (iv) (c) shows positive acceleration throughout. (v) (a) shows negative (leftward) acceleration in the last three images. *Q2.8 Tramping hard on the brake at zero speed on a level road, you do not feel pushed around inside the car. The forces of rolling resistance and air resistance have dropped to zero as the car coasted to a stop, so the car’s acceleration is zero at this moment and afterward. Tramping hard on the brake at zero speed on an uphill slope, you feel thrown backward against your seat. Before, during, and after the zero-speed moment, the car is moving with a downhill acceleration if you do not tramp on the brake. Brian Popp suggested the idea for this question. 15 ISMV1_5103_02.indd 15 10/27/06 2:45:19 PM  16 Chapter 2 *Q2.9 With original velocity zero, displacement is proportional to the square of time in (1
2)at 2. Making the time one-third as large makes the displacement one-ninth as large, answer (c). Q2.10 No. Constant acceleration only. Yes. Zero is a constant. Q2.11 They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity of magnitude vi. This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same. For the release from rest we have (4 m
s)2 = 02 + 2 gh. For case (i), we have vf = (3 m
s)2 + 2 gh = 2 *Q2.12 (3 m
s)2 + (4 m
s)2. Thus answer (d) is true. For case (ii) the same steps give the same answer (d). *Q2.13 (i) Its speed is zero at b and e. Its speed is equal at a and c, and somewhat larger at d. On the bounce it is moving somewhat slower at f than at d, and slower at g than at c. The assembled answer is d > f > a = c > g > b = e. (ii) The velocity is positive at a, f, and g, zero at b and e, and negative at c and d, with magnitudes as described in part (i). The assembled answer is f > a > g > b = e > c > d. (iii) The acceleration has a very large positive value at e. At all the other points it is −9.8 m
s 2. The answer is e > a = b = c = d = f = g. Q2.14 (b) Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point. So your ball must travel a smaller distance to the passing point that the ball your friend throws. SOLUTIONS TO PROBLEMS Section 2.1 Position, Velocity, and Speed ∆x 10 m P2.1 (a) vavg = = = 5 ms ∆t 2s 5m (b) vavg = = 1.2 m s 4s x2 − x1 5 m − 10 m (c) vavg = = = −2.5 m s t 2 − t1 4 s−2 s x2 − x1 −5 m − 5 m (d) vavg = = = −3.3 m s t 2 − t1 7 s−4 s x2 − x1 0 − 0 (e) vavg = = = 0 ms t 2 − t1 8 − 0 P2.2 (a) vavg = 2.30 m s ∆x 57.5 m − 9.20 m (b) v= = = 16.1 m s ∆t 3.00 s ∆x 57.5 m − 0 m (c) vavg = = = 11.5 m s ∆t 5.00 s ISMV1_5103_02.indd 16 10/27/06 2:45:20 PM  Motion in One Dimension 17 P2.3 (a) Let d represent the distance between A and B. Let t 1 be the time for which the walker has d the higher speed in 5.00 m s =. Let t 2 represent the longer time for the return trip in t1 d d d −3.00 m s = −. Then the times are t1 = and t 2 =. The average t2 ( 5. 00 m s ) ( 3. 00 m s) speed is: Total distance d+d 2d vavg = = = Total time d / ( 5. 00 m s ) + d / ( 3. 00 m s ) (8. 00 m s ) d / (15. 0 m 2 s 2 ) 2 (15. 0 m 2 s 2 ) vavg = = 3. 75 m s 8. 00 m s (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0. t ( s ) = 2.0 2.1 3.0 P2.4 x = 10t 2 : By substitution, for x ( m ) = 40 44.1 90 ∆x 50 m (a) vavg = = = 50.0 m s ∆t 1.0 s ∆x 4.1 m (b) vavg = = = 41.0 m s ∆t 0.1 s Section 2.2 Instantaneous Velocity and Speed P2.5 (a) at ti = 1.5 s, xi = 8.0 m (Point A) at t f = 4.0 s, x f = 2.0 m (Point B) x f − xi ( 2.0 − 8.0 ) m 6.0 m vavg = = =− = −2.4 m s t f − ti ( 4 − 1.5) s 2.5 s (b) The slope of the tangent line can be found from points C and D. ( tC = 1.0 s, xC = 9.5 m ) and ( t D = 3.5 s, x D = 0 ), v ≈ −3.8 m s. (c) The velocity is zero when x is a minimum. This is at t ≈ 4 s. FIG. P2.5 P2.6 (a) At any time, t, the position is given by x = ( 3.00 m s ) t. 2 2 Thus, at ti = 3.00 s: xi = ( 3.00 m s 2 ) ( 3.00 s ) = 27.0 m. 2 At t f = 3.00 s + ∆t : x f = ( 3.00 m s 2 ) ( 3.00 s + ∆t ) , or 2 (b) x f = 27.0 m + (18.0 m s ) ∆t + ( 3.00 m s 2 ) ( ∆t ). 2 (c) The instantaneous velocity at t = 3.00 s is: ⎛ x f − xi ⎞ v = lim ⎜ ∆t → 0 ⎝ ∆t ⎟⎠ ∆t →0 ( ) = lim 18.0 m s + ( 3.00 m s 2 ) ∆t = 18.0 m s. ISMV1_5103_02.indd 17 10/27/06 2:45:21 PM  18 Chapter 2 P2.7 (a) 58 m (b) At t = 5.0 s, the slope is v ≈ = 23 m s. 2.5 s 54 m At t = 4.0 s, the slope is v ≈ = 18 m s. 3s 49 m At t = 3.0 s, the slope is v ≈ = 14 m s. 3.4 s 36 m At t = 2.0 s, the slope is v ≈ = 9.0 m s. 4.0 s ∆v 23 m s (c) aavg = ≈ = 4.6 m s 2 ∆t 5.0 s (d) Initial velocity of the car was zero. P2.8 (a) v= (5 − 0 ) m = 5 m s (1 − 0 ) s (b) v= (5 − 10 ) m = −2.5 m s ( 4 − 2) s (c) v= (5 m − 5 m ) = 0 (5 s − 4 s) 0 − ( −5 m ) (d) v= = +5 m s (8 s − 7 s ) P2.9 Once it resumes the race, the hare will run for a time of FIG. P2.8 x f − xi 1 000 m − 800 m t= = = 25 s. vx 8 ms In this time, the tortoise can crawl a distance x f − xi = ( 0.2 m s ) ( 25 s ) = 5.00 m. Section 2.3 Acceleration P2.10 Choose the positive direction to be the outward direction, perpendicular to the wall. ∆v 22.0 m s − ( −25.0 m s ) v f = vi + at : a = = = 1.34 × 10 4 m s 2 ∆t 3.50 × 10 −3 s ISMV1_5103_02.indd 18 10/27/06 2:45:22 PM  Motion in One Dimension 19 P2.11 (a) Acceleration is constant over the first ten seconds, so at the end of this interval v f = vi + at = 0 + ( 2.00 m s 2 ) (10.0 s ) = 20.0 m s. Then a = 0 so v is constant from t =10.0 s to t =15.0 s. And over the last five seconds the velocity changes to v f = vi + at = 20.0 m s + ( −3.00 m s 2 ) ( 5.00 s ) = 5.00 m s. (b) In the first ten seconds, x f = xi + vi t + at 2 = 0 + 0 + ( 2.00 m s 2 ) (10.0 s ) = 100 m. 1 1 2 2 2 Over the next five seconds the position changes to 1 x f = xi + vi t + at 2 = 100 m + ( 20.0 m s ) ( 5.00 s ) + 0 = 200 m. 2 And at t = 20.0 s, x f = xi + vi t + at 2 = 200 m + ( 20.0 m s ) ( 5.00 s ) + ( −3.00 m s 2 ) ( 5.00 s ) = 262 m. 1 1 2 2 2 P2.12 (a) Acceleration is the slope of the graph of v versus t. a (m/s2) For 0 < t < 5.00 s, a = 0. 2.0 1.6 For 15.0 s < t < 20.0 s, a = 0. 1.0 v f − vi For 5.0 s < t < 15.0 s, a =. t f − ti 8.00 − ( −8.00 ) 0.0 t (s) a= = 1.60 m s 2 0 5 10 15 20 15.0 − 5.00 FIG. P2.12 We can plot a ( t ) as shown. v f − vi (b) a= t f − ti (i) For 5.00 s < t < 15.0 s, ti = 5.00 s, vi = −8.00 m s, t f = 15.0 s v f = 8.00 m s v f − vi 8.00 − ( −8.00 ) a= = = 1.60 m s 2. t f − ti 15.0 − 5.00 (ii) ti = 0, vi = −8.00 m s, t f = 20.0 s, v f = 8.00 m s v f − vi 8.00 − ( −8.00 ) a= = = 0.800 m s 2 t f − ti 20.0 − 0 ISMV1_5103_02.indd 19 10/27/06 2:45:23 PM  20 Chapter 2 dx dv P2.13 x = 2.00 + 3.00t − t 2, so v = = 3.00 − 2.00t , and a = = −2.00 dt dt At t = 3.00 s: (a) x = ( 2.00 + 9.00 − 9.00 ) m = 2.00 m (b) ( v = 3.00 − 6.00 m s = −3.00 m s ) (c) a = −2.00 m s 2 *P2.14 The acceleration is zero whenever the marble is on a horizontal section. The acceleration has a constant positive value when the marble is rolling on the 20-to-40-cm section and has a constant negative value when it is rolling on the second sloping section. The position graph is a straight sloping line whenever the speed is constant and a section of a parabola when the speed changes. Position as a function of time 100 Position along track, cm 80 60 40 20 0 time Velocity as a function of time x component of velocity, arbitrary units 0 time continued on next page ISMV1_5103_02.indd 20 11/2/06 1:29:11 PM  Motion in One Dimension 21 Acceleration as a function of time acceleration, arbitrary units  time At t = 2.00 s, x = ⎡⎣3.00 ( 2.00 ) − 2.00 ( 2.00 ) + 3.00 ⎤⎦ m = 11.0 m. 2 P2.15 (a) At t = 3.00 s, x = ⎡⎣3.00 ( 9.00 ) − 2.00 ( 3.00 ) + 3.00 ⎤⎦ m = 24.0 m 2 so ∆x 24.0 m − 11.0 m = vavg = = 13.0 m s. ∆t 3.00 s − 2.00 s (b) At all times the instantaneous velocity is v= d dt (3.00t 2 − 2.00t + 3.00 ) = (6.00t − 2.00 ) m s ( ) At t = 2.00 s, v = ⎡⎣6.00 2.00 − 2.00 ⎤⎦ m s = 10.0 m s. ( ) At t = 3.00 s, v = ⎡⎣6.00 3.00 − 2.00 ⎤⎦ m s = 16.0 m s. ∆v 16.0 m s − 10.0 m s (c) aavg = = = 6.00 m s 2 ∆t 3.00 s − 2.00 s d (d) At all times a = ( 6.00t − 2.00 ) = 6.00 m s 2. This includes both t = 2.00 s and t = 3.00 s. dt ∆v 8.00 m s P2.16 (a) a= = = 1.3 m s 2 ∆t 6.00 s (b) Maximum positive acceleration is at t = 3 s, and is the slope of the graph, approximately (6 − 2)
(4 − 2) = 2 m s 2. (c) a = 0 at t = 6 s , and also for t > 10 s. (d) Maximum negative acceleration is at t = 8 s, and is the slope of the graph, approximately −1.5 m s 2. ISMV1_5103_02.indd 21 10/27/06 2:45:25 PM  22 Chapter 2 Section 2.4 Motion Diagrams *P2.17 (a) The motion is slow at first, then fast, and then slow again. a 0 t 0 t x 0 t (b) The motion is constant in speed. a 0 t 0 t x 0 t (c) The motion is speeding up, and we suppose the acceleration is constant. a 0 t 0 t x 0 t P2.18 (a) (b) (c) (d) (e) (f ) One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction. ISMV1_5103_02.indd 22 10/27/06 4:53:04 PM  Motion in One Dimension 23 Section 2.5 One-Dimensional Motion with Constant Acceleration *P2.19 (a) vf = vi + at = 13 m
s − 4 m
s2 (1 s) = 9.00 m
s (b) vf = vi + at = 13 m
s − 4 m
s2 (2 s) = 5.00 m
s (c) vf = vi + at = 13 m
s − 4 m
s2 (2.5 s) = 3.00 m
s (d) vf = vi + at = 13 m
s − 4 m
s2 (4 s) = −3.00 m
s (e) vf = vi + at = 13 m
s − 4 m
s2 (−1 s) = 17.0 m
s (f ) The graph of velocity versus time is a slanting straight line, having the value 13 m
s at 10:05:00 a.m. on the certain date, and sloping down by 4 m
s for every second thereafter. (g) If we also know the velocity at any one instant, then knowing the value of the constant acceleration tells us the velocity at all other instants. P2.20 (a) x f − xi = 1 2 ( ) 1 vi + v f t becomes 40 m = ( vi + 2.80 m s ) (8.50 s ) which yields 2 vi = 6.61 m s. v f − vi 2.80 m s − 6.61 m s (b) a= = = −0.448 m s 2 t 8.50 s P2.21 Given vi = 12.0 cm s when xi = 3.00 cm ( t = 0 ) , and at t = 2.00 s, x f = −5.00 cm, 1 1 x f − xi = vi t + at 2 : −5.00 − 3.00 = 12.0 ( 2.00 ) + a ( 2.00 ) 2 2 2 32.0 −8.00 = 24.0 + 2a a=− = −16.0 cm s 2. 2 P2.22 (a) Total displacement = area under the ( v, t ) curve from t = 0 to 50 s. 1 ∆x = ( 50 m s ) (15 s ) + ( 50 m s ) ( 40 − 15) s 2 1 + ( 50 m s ) (10 s ) 2 ∆x = 1875 m = 1.88 km