1. 1. http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS.
2. 2. 1 Physics and Measurement CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model-Building 1.3 Dimensional Analysis 1.4 Conversion of Units 1.5 Estimates and Order-of- Magnitude Calculations 1.6 Signicant Figures ANSWERS TO QUESTIONS * An asterisk indicates an item new to this edition. Q1.1 Density varies with temperature and pressure. It would be necessary to measure both mass and volume very 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 kg2 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 L1.3 )(1 qt1 L)(1 gal4 qt) (101.3) 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 signicant. 1 SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time P1.1 Modeling the Earth as a sphere, we nd its volume as 4 3 4 3 6 37 10 1 08 103 6 3 21 3 r = ( ) = . .m m . Its density is then = = = m V 5 98 10 1 08 10 5 52 10 24 21 3 3 3. . . kg m kg m . This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kgm3 . The average density of the Earth is signicantly higher, so higher-density material must be down below the surface. ISMV1_5103_01.indd 1ISMV1_5103_01.indd 1 10/27/06 4:33:21 PM10/27/06 4:33:21 PM
3. 3. P1.2 With V = ( )( )base area height V r h= ( ) 2 and = m V , we have = = ( ) ( ) m r h2 2 9 1 19 5 39 0 10 1 kg mm mm mm m 3 3 . . = 2 15 104 3 . .kg m P1.3 Let V represent the volume of the model, the same in = m V for both. Then iron kg= 9 35. V and gold gold = m V . Next, gold iron gold kg = m 9 35. and mgold 3 3 3 kg 19.3 10 kg/m kg/m = 9 35 7 86 103 . . = 23 0. kg . *P1.4 = m V/ and V r d d= = =( / ) ( / ) ( / ) /4 3 4 3 2 63 3 3 where d is the diameter. Then = = = 6 6 1 67 10 2 4 10 2 33 27 15 3 m d/ ( . ) ( . ) . kg m 11017 3 kg/m 2.3 10 kg/m /(11.3 10 kg/m )=17 3 3 3 it is 20 1012 times the density of lead . P1.5 For either sphere the volume is V r= 4 3 3 and the mass is m V r= = 4 3 3 . We divide this equation for the larger sphere by the same equation for the smaller: m m r r r rs s s = = = 4 3 4 3 5 3 3 3 3 . Then r rs = = ( ) =5 4 50 1 71 7 693 . . .cm cm . Section 1.2 Matter and Model-Building P1.6 From the gure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a at plane. This diagonal distance may be obtained from the Pythagorean theorem, L L Ldiag = +2 2 . Thus, since the atoms are separated by a distance L = 0 200. nm, the diagonal planes are separated by 1 2 0 1412 2 L L+ = . nm . Section 1.3 Dimensional Analysis P1.7 (a) This is incorrect since the units of ax[ ] are m s2 2 , while the units of v[ ] are m s . (b) This is correct since the units of y[ ] are m, and cos kx( ) is dimensionless if k[ ] is in m1 . 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 L L2 1 2 2 ( ) = / , so this must be area (c). Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L L2 3 ( )= , so it is (b). Thus, (a) ii; (b) iii;(c) i= = = . 2 Chapter 1 ISMV1_5103_01.indd 2ISMV1_5103_01.indd 2 10/27/06 12:27:01 PM10/27/06 12:27:01 PM
4. 4. Physics and Measurement 3 P1.9 Inserting the proper units for everything except G, kg m s kg m2 = [ ] [ ] G 2 2 . Multiply both sides by m[ ]2 and divide by kg[ ]2 ; the units of G are m kg s 3 2 . Section 1.4 Conversion of Units P1.10 Apply the following conversion factors: 1 2 54in cm= . , 1 86 400d s= , 100 1cm m= , and 10 19 nm m= 1 32 2 54 10 102 9 in day cm in m cm nm m 8 ( )( )( ) . 66 400 s day nm s= 9 19. . 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 50 1500m m m2 . 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 1 3 281 150 1 3 28 ft m ft ft m . . 11 1 39 103 ft = 1 390 m m2 2 = . . Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m2 . 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 . m3 9 60 103 V = ( ) = 9 60 10 3 39 103 5 3 . m 3.28 ft 1 m ft3 3 . (b) The mass of the air is m V= = ( ) ( )= air 3 3 kg m 9.60 10 m . k1 20 1 15 103 4 . gg. The student must look up weight in the index to nd F mgg = = ( )( )= 1.15 10 kg 9.80 m s 1.13 10 N4 2 5 . Converting to pounds, Fg = ( )( ) = 1 13 10 2 54 105 4 . N 1 lb 4.45 N 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 m2 0.059 m2 = 629 sheets = 629 sheets(2 pages1 sheet) = 1260 pages. The pages from volume one are inadequate, but the full version has enough pages. ISMV1_5103_01.indd 3ISMV1_5103_01.indd 3 10/27/06 4:18:09 PM10/27/06 4:18:09 PM
5. 5. 4 Chapter 1 P1.14 (a) Seven minutes is 420 seconds, so the rate is r = = 30 0 420 7 14 10 2. . gal s gal s . (b) Converting gallons rst to liters, then to m3 , r = ( ) 7 14 10 3 786 102 3 . . gal s L 1 gal m3 11 L m s3 = r 2 70 10 4 . . (c) At that rate, to ll a 1-m3 tank would take t = = 1 2 70 10 1 1 034 m m s h 3 600 3 3 . . h . P1.15 From Table 14.1, the density of lead is 1 13 104 . kg m3 , 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. Density is dened as mass per volume, in = m V . We must convert to SI units in the calculation. = 23 94 2 10 1 1000 100 13 . . g cm kg g cm m = 3 3 23 94 2 10 1 1000 1 00. . g cm kg g 00 1 1 14 104000 cm m . kg m 3 3 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 signicant 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 cm3 , and objects that oat must be less dense than water. P1.16 The weight ow rate is 1 200 2 000 1 1ton h lb ton h 60 min min 60 s lb s = 667 . P1.17 (a) 8 10 1 112 $ 1 000 $ s h 3 600 s day 24 h yr 365 days years = 1 250 (b) The circumference of the Earth at the equator is 2 6 378 10 4 01 103 7 . .( )= m 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 10 1 3 09 10 12 4. . . = m 4.01 0 m times7 P1.18 V Bh= = ( )( ) 1 3 13 0 43 560 3 481 . acres ft acre2 fft ft3 ( ) = 9 08 107 . , or V = ( ) = 9 08 10 2 83 10 1 2 5 7 2 . . . ft m ft 3 3 3 77 106 m3 FIG. P1.18 B h B h ISMV1_5103_01.indd 4ISMV1_5103_01.indd 4 10/27/06 12:27:03 PM10/27/06 12:27:03 PM
6. 6. Physics and Measurement 5 P1.19 Fg = ( ) ( )2 50 2 00 10 2 0006 . .tons block blocks lb tton lbs( ) = 1 00 1010 . P1.20 (a) d d d nucleus, scale nucleus, real atom, scale = ddatom, real m ft 1.06 = ( ) 2 40 10 30015 . = 10 m ft10 3 6 79 10. , or dnucleus, scale ft mm 1 ft= ( )( 6 79 10 304 83 . . )) = 2 07. mm (b) V V r r ratom nucleus atom nucleus atom = = 4 3 4 3 3 3 / / rr d dnucleus atom nucleus = = 3 3 1 06. 110 2 40 10 8 62 10 10 15 3 13 = m m times . . as large P1.21 V At= so t V A = = = 3 78 10 25 0 1 51 10 151 3 4. . . m m m or 3 2 m( ) P1.22 (a) A A r r r r Earth Moon Earth Moon 2 Earth Moon = = 4 4 2 = ( )( ) 2 6 8 6 37 10 100 1 74 10 . . m cm m cm = 2 13 4. (b) V V r r r r Earth Moon Earth Moon Earth Mo = = 4 3 4 3 3 3 / / oon 3 m cm m cm = ( )( ) 6 37 10 100 1 74 10 6 8 . . = 3 49 1. P1.23 To balance, m mFe Al= or Fe Fe Al AlV V= Fe Fe Al Al Al Fe Fe A 4 3 4 3 3 3 = = r r r r ll cm c = ( ) = 1 3 1 3 2 00 7 86 2 70 2 86 / / . . . . mm . P1.24 The mass of each sphere is m V r Al Al Al Al Al = = 4 3 3 and m V r Fe Fe Fe Fe Fe = = 4 3 3 . Setting these masses equal, 4 3 4 3 3 3 Al Al Fe Fer r = and r rAl Fe Fe Al = 3 . 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 fraction Fe Al is the factor of change between the densities, a number greater than 1. Its cube root is a number much closer to 1. The relatively small change in radius implies a change in volume sufcient to compensate for the change in density. ISMV1_5103_01.indd 5ISMV1_5103_01.indd 5 10/27/06 12:27:04 PM10/27/06 12:27:04 PM
7. 7. 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 = m3 , while the volume of one ball is 4 3 0 038 2 87 10 3 5 . . m 2 m3 = . Therefore, one can t about 48 2 87 10 105 6 . ~ ping-pong balls in the room. 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 6 2 0 74 = . so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 1 67 10 0 740 106 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 5280 1 3 107 mi ft mi rev 8 ft rev ~( )( )( ) = 110 rev7 . 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 0 5 0 3 0 10. . . . .m m m m3 . The mass of this volume of water is m Vwater water 3 3 kg m m kg= = ( )( )= 1 000 0 10 100. ~ 1102 kg . Pennies are now mostly zinc, but consider copper pennies lling 50% of the volume of the tub. The mass of copper required is m Vcopper copper 3 3 kg m m k= = ( )( )= 8 920 0 10 892. gg kg~ 103 . *P1.28 The time required for the task is 10 1 1 1 1 9 $ s 1 $ h 3600 s working day 66 1 58 h bad yr 300 working days y = rr 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 = 107 ; 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, # tuners ~ 1 tuner 1 000 pianos 1 piano 10 00 people people tuners =( )10 1007 . Section 1.6 Signicant Figures P1.30 METHOD ONE We treat the best value with its uncertainty as a binomial 21 3 0 2 9 8 0 1. . . .( ) ( )cm cm, A = ( ) ( ) ( ) ( )( )[ ]21 3 9 8 21 3 0 1 0 2 9 8 0 2 0 1. . . . . . . . cm2 . The rst 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 42 2 cm cm . METHOD TWO We add the fractional uncertainties in the data. A = ( )( ) + =21 3 9 8 0 2 21 3 0 1 9 8 209. . . . . . cm cm ccm cm cm2 2 2 2 209 4 = % ISMV1_5103_01.indd 6ISMV1_5103_01.indd 6 10/28/06 2:41:27 AM10/28/06 2:41:27 AM
8. 8. Physics and Measurement 7 P1.31 (a) 3 (b) 4 (c) 3 (d) 2 P1.32 r m = ( ) = ( ) = 6 50 0 20 6 50 0 20 10 1 85 0 2 . . cm . . m . .. kg02 4 3 3 ( ) = ( ) m r also, = + m m r r 3 . In other words, the percentages of uncertainty are cumulative. Therefore, = + ( ) = 0 02 1 85 3 0 20 6 50 0 103 . . . . . , = ( ) ( ) = 1 85 6 5 10 1 61 10 4 3 2 3 3 3. . . m kg m and = ( ) = ( )1 61 0 17 10 1 6 0 2 103 3 . . . .kg m kg m3 33 . P1.33 (a) 756.?? 37.2? 0.83 + 2.5? 796./ /5 3 = 797 (b) 0 003 2 356 3 1 140 16. 2 s.f. . 4 s.f. . 2 s.f.( ) ( ) = = (( ) 1 1. (c) 5.620 4 s.f. >4 s.f. 17.656= 4 s.f. 17( ) ( ) = ( ) ..66 P1.34 We work to nine signicant digits: 1 1 365 242199 24 6 yr yr d 1 yr h 1 d = . 00 60 31556 926 0 min 1 h s 1 min s = . . *P1.35 The tax amount is $1.36 $1.25 = $0.11. The tax rate is $0.11$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 cm2 380 cm2 = 8% . (d) Choose a grid point on the line far from the origin: slope = 0.31 g600 cm2 = 0.000 52 gcm2 = (0.000 52 gcm2 )(10 000 cm2 1 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 7ISMV1_5103_01.indd 7 10/28/06 2:41:45 AM10/28/06 2:41:45 AM
9. 9. 8 Chapter 1 *P1.37 15 players = 15 players (1 shift1.667 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 = 180.947 = 19 . *P1.39 Let s represent the number of sparrows and m the number of more interesting birds. We have sm = 2.25 and s + m = 91. We eliminate m by substitution: m = s2.25 s + s2.25 = 91 1.444s = 91 s = 911.444 = 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 3 5 70 04 3 x x x + = is quartic, so we do not attempt to solve it with algebra. To nd 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 x x x= + 2 3 5 704 3 158 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 rst of these solutions 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 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 x = 2 218. , but we already know to three-digit precision that the root is x = 2 22. . *P1.41 We require sin cos = 3 , or sin cos = 3, or tan = 3. For tan a tan ( ) = ( )1 3 3rc , your calculator may return 71.6, but this angle is not between 0 and 360 as the problem requires. The tangent function is negative in the second quad- 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 =. . y x FIG. P1.40 3600 tan FIG. P1.41 ISMV1_5103_01.indd 8ISMV1_5103_01.indd 8 10/27/06 12:27:07 PM10/27/06 12:27:07 PM
10. 10. Physics and Measurement 9 *P1.42 We draw the radius to the initial point and the radius to the nal point. The angle 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 and nal tangential directions of travel. A most useful theorem from geometry then identies these angles as equal: = 35 . The whole circumference of a 360 circle of the same radius is 2 R. By proportion, then 2 360 840 R = m 35 . R = = = 360 2 840 840 1 38 103 m 35 m 0.611 m. We could equally well say that the measure of the angle in radians is = = = =35 35 2 0 611 840radians 360 rad m . RR . Solving yields R = 1 38. km. *P1.43 Mass is proportional to cube of length: m = k3 m mi = (f i )3 . Length changes by 15.8%: f = i + 0.158 i = 1.158 i . Mass increase: mf = mi + 17.3 kg. Eliminate by substitution: m m f f = = 17 3 1 158 1 5533 . . . kg mf = 1.553 mf 26.9 kg 26.9 kg = 0.553 mf mf = 26.9 kg0.553 = 48. 6 kg . *P1.44 We use substitution, as the most generally applicable method for solving simultaneous equations. We substitute p q= 3 into each of the other two equations to eliminate p: 3 1 2 3 1 2 1 2 2 2 2 qr qs qr qs qt = + = . These simplify to 3 3 2 2 2 r s r s t = + = . We substitute to eliminate s: 3 3 12 2 2 2 2 2 r r t r t +( ) = = . We solve for the combination t r : t r 2 2 12= . t r = either or3 46 3 46. . *P1.45 Solve the given equation for t: t = 4QLkd2 (Th Tc ) = [4QLk (Th Tc )] [1 d2 ]. (a) Making d three times larger with d2 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/d2 on the horizontal axis. (d) From the last version of the equation, the slope is 4QL/k(Th Tc ) . Note that this quantity is constant as both t and d vary. FIG. P1.42 R N i f S EW 35.0 ISMV1_5103_01.indd 9ISMV1_5103_01.indd 9 10/28/06 2:42:03 AM10/28/06 2:42:03 AM
11. 11. 10 Chapter 1 Additional Problems P1.46 It is desired to nd the distance x such that x x100 1 000 m m = (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 x2 5 100 1 000 1 00 10= ( )( ) = m m m2 . and therefore x = =1 00 10 3165 . m m2 . *P1.47 (a) The mass is equal to the mass of a sphere of radius 2.6 cm and density 4.7 gcm3 , minus the mass of a sphere of radius a and density 4.7 gcm3 plus the mass of a sphere of radius a and density 1.23 gcm3 . m = 1 4r3 3 1 4a3 3 + 2 4a3 3 = (4.7 gcm3 )4(2.6 cm)3 3 (4.7 gcm3 )4(a)3 3 + (1.23 gcm3 )4(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 rst 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 gcm3 . (h) (346 g + 90.6 g)2 = 218 g (i) No . The result of part (a) gives 346 g (14.5 gcm3 ) (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 gcm3 and B + C(14 cm) = 19.3 gcm3 . We know B = 2.70 g/cm3 and we solve for C by subtracting: C(14 cm) = 16.6 gcm3 so C = 1.19 g/cm4 . (b) m x dx= + (2.70 g/cm g/cm cm3 4 2 cm 1 19 9 0 14 . )( ) = 24.3 g/cm g/cm2 0 14 cm c dx xdx+ 10 7 0 14 . mm = (24.3 g/cm)(14 cm 0) + (10.7 g/c mm cm) = 340 g + 1046 g = 1. 2 2 )[( ] /14 0 2 339 kg ISMV1_5103_01.indd 10ISMV1_5103_01.indd 10 10/27/06 12:27:09 PM10/27/06 12:27:09 PM
12. 12. Physics and Measurement 11 P1.49 The scale factor used in the dinner plate model is S = = 0 25 1 0 10 2 105 6. . m lightyears .5 m lightyeears. The distance to Andromeda in the scale model will be D D Sscale actual 6 2.0 10 lightyears 2.5 10= = ( ) 66 m lightyears m( )= 5 0. . *P1.50 The rate of volume increase is dV dt d dt r r dr dt r dr dt = = = 4 3 4 3 3 43 2 2 . (a) dVdt = 4 (6.5 cm)2 (0.9 cms) = 478 cm3 /s (b) dr dt dV dt r = = = / / ( . 4 478 4 13 0 2252 cm s cm) cm 3 2 33 s/ (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 ination 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 30 24 3 600 2 592 106 mo day h day s h s= ( )( )( ) = . . Applying units to the equation, V t t= ( ) +( )1 50 0 008 00 2 . .Mft mo Mft mo3 3 2 . Since1 106 Mft ft3 3 = , V t t= ( ) + ( )1 50 10 0 008 00 106 6 2 . .ft mo ft mo3 3 2 . Converting months to seconds, V t= + 1 50 10 0 008 00 106 6 . .ft mo 2.592 10 s mo 3 6 fft mo 2.592 10 s mo 3 2 6 ( )2 2 t . Thus, V t tft ft s ft s3 3 3 2 [ ] . .= ( ) + ( ) 0 579 1 19 10 9 2 . *P1.52 (deg) (rad) tan ( ) sin ( ) difference between and tan 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. is the largest angle for which tan tan . < 0 1. ISMV1_5103_01.indd 11ISMV1_5103_01.indd 11 10/27/06 12:27:09 PM10/27/06 12:27:09 PM
13. 13. 12 Chapter 1 P1.53 2 15 0 2 39 2 39 r r h r h = = = = ( ) . m . m tan 55.0 . m tan(( . ) m55 0 3 41 = . P1.54 Let d represent the diameter of the coin and h its thickness. The mass of the gold is m V At d dh t= = = + 2 4 2 where t is the thickness of the plating. m = ( ) + ( )( ) 19 3 2 2 41 4 2 41 0 178 0 18 1 2 . . . . . 00 0 003 64 0 003 64 10 4 ( ) = = . . $ grams cost grams grram cents= =$ . .0 036 4 3 64 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 31557 600 s yr( )( ) = . The percent error in the approximation is ( )( ) 10 31557 600 31557 600 100 7 s yr s yr s yr % == 0 449. % . P1.56 v = 5 00 220 1 . furlongs fortnight yd furlongg m yd fortnight day 0 914 4 1 1 14 . ss day hrs hr s = 1 24 1 3 600 8 3. 22 10 4 m s This speed is almost 1 mms; so we might guess the creature was a snail, or perhaps a sloth. P1.57 (a) The speed of rise may be found from v = ( ) = Vol rate of flow (Area: cm3 D2 4 16 5 / ) . ss cm cm s 6 30 4 0 5292 . . ( ) = / . (b) Likewise, at a 1.35 cm diameter, v = ( ) = 16 5 1 35 4 11 52 . . . cm s cm cm s 3 / . FIG. P1.53 5555 h rr h ISMV1_5103_01.indd 12ISMV1_5103_01.indd 12 10/27/06 12:27:10 PM10/27/06 12:27:10 PM
14. 14. Physics and Measurement 13 P1.58 The density of each material is = = = m V m r h m D h2 2 4 . Al: g cm cm g cm3 = ( ) ( ) ( ) = 4 51 5 2 52 3 75 2 752 . . . . TThe tabulated value g cm is sma3 2 70 2. % lller. Cu: g .23 cm .06 cm g cm3 = ( ) ( ) ( ) = 4 56 3 1 5 9 362 . . TThe tabulated value g cm is sma3 8 92 5. % lller. Brass: 4.4 g .54 cm .69 cm g = ( ) ( ) ( ) = 4 9 1 5 8 912 . ccm3 Sn: g .75 cm .74 cm g cm3 = ( ) ( ) ( ) = 4 69 1 1 3 7 682 . . Fe: 16.1 g .89 cm .77 cm g cm = ( ) ( ) ( ) = 4 2 1 9 7 882 . 33 3 The tabulated value g cm is7 86 0 3. . % ssmaller. P1.59 V20 mpg cars mi yr mi gal 5.0 10= = ( )( )10 10 20 8 4 110 gal yr V25 mpg cars mi yr mi gal 4.0 10= = ( )( )10 10 25 8 4 110 gal yr Fuel saved gal yr25 mpg 20 mpg= = V V 1 0 1010 . P1.60 The volume of the galaxy is r t2 21 2 19 61 10 10 10= ( ) ( )m m m3 ~ . If the distance between stars is 4 1016 m, then there is one star in a volume on the order of 4 10 1016 3 50 ( )m m3 ~ . The number of stars is about 10 10 10 61 50 11m m star stars 3 3 ~ . ANSWERS TO EVEN-NUMBERED PROBLEMS P1.2 2.15 104 kgm3 P1.4 2.3 1017 kgm3 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 nms P1.12 (a) 3.39 105 ft3 (b) 2.54 104 lb P1.14 (a) 0.071 4 gals (b) 2.70 104 m3 s (c) 1.03 h ISMV1_5103_01.indd 13ISMV1_5103_01.indd 13 10/27/06 12:27:11 PM10/27/06 12:27:11 PM
15. 15. 14 Chapter 1 P1.16 667 lbs P1.18 2 57 106 3 . m P1.20 (a) 2.07 mm (b) 8.57 1013 times as large P1.22 (a) 13.4; (b) 49.1 P1.24 r rAl Fe Fe Al = 1 3 P1.26 ~10 rev7 P1.28 No. There is a strong possibility that you would die before nishing the task, and you have much more productive things to do. P1.30 209 4 2 ( ) cm P1.32 (1.61 0.17) 103 kgm3 P1.34 31 556 926.0 s P1.36 (a) 0.015 g (b) 30 cm2 (c) 8% (d) 5.2 gm2 (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 gm2 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) = 2.70 gcm3 + 1.19 gcm4 x (b) 1.39 kg P1.50 (a) 478 cm3 s (b) 0.225 cms (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 ination 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 14ISMV1_5103_01.indd 14 10/27/06 12:27:11 PM10/27/06 12:27:11 PM
16. 16. 2 Motion in One Dimension CHAPTER OUTLINE 2.1 Position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.4 Motion Diagrams 2.5 One-Dimensional Motion with Constant Acceleration 2.6 Freely Falling Objects 2.7 Kinematic Equations Derived from Calculus ANSWERS TO QUESTIONS * An asterisk indicates an item new to this edition. *Q2.1 Count spaces (intervals), not dots. Count 5, not 6. The rst drop falls at time zero and the last drop at 5 5 s = 25 s. The average speed is 600 m25 s = 24 ms, answer (b). 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 nal speed 1 ms is slower than 3 ms. Its acceleration is positive, meaning to the right. (c) It slows down and its acceleration is negative. (d) It speeds up to nal speed 7 ms. 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 cars 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 15ISMV1_5103_02.indd 15 10/27/06 2:45:19 PM10/27/06 2:45:19 PM
17. 17. *Q2.9 With original velocity zero, displacement is proportional to the square of time in (12)at2 . 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 rst 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. *Q2.12 For the release from rest we have (4 ms)2 = 02 + 2 gh. For case (i), we have vf 2 = (3 ms)2 + 2 gh = (3 ms)2 + (4 ms)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 ms2 . 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 ight to the passing point. So your ball must travel a smaller distance to the passing point that the ball your friend throws. 16 Chapter 2 SOLUTIONS TO PROBLEMS Section 2.1 Position, Velocity, and Speed P2.1 (a) vavg x t = = = 10 5 m 2 s m s (b) vavg = = 5 1 2 m 4 s m s. (c) vavg x x t t = = = 2 1 2 1 5 10 2 2 5 m m 4 s s m s. (d) vavg x x t t = = = 2 1 2 1 5 5 4 3 3 m m 7 s s m s. (e) vavg x x t t = = =2 1 2 1 0 0 8 0 0 m s P2.2 (a) vavg = 2 30. m s (b) v = = m m s = 16.1 m s x t 57 5 9 20 3 00 . . . (c) vavg x t = = = 57 5 0 11 5 . . m m 5.00 s m s ISMV1_5103_02.indd 16ISMV1_5103_02.indd 16 10/27/06 2:45:20 PM10/27/06 2:45:20 PM
18. 18. Motion in One Dimension 17 P2.3 (a) Let d represent the distance between A and B. Let t1 be the time for which the walker has the higher speed in 5 00 1 . m s = d t . Let t2 represent the longer time for the return trip in = 3 00 2 . m s d t . Then the times are t d 1 5 00 = ( ). m s and t d 2 3 00 = ( ). m s . The average speed is: vavg d d d = = + ( Total distance Total time m s/ .5 00 ))+ ( ) = ( ) ( )d d d avg / . . / .3 00 2 8 00 15 0m s m s m s2 2 v == ( ) = 2 15 0 8 00 3 75 . . . m s m s m s 2 2 (b) She starts and nishes at the same point A. With total displacement = 0, average velocity = 0 . P2.4 x t= 10 2 : By substitution, for t x s m ( ) = ( ) = 2 0 2 1 3 0 40 44 1 90 . . . . (a) vavg x t = = = 50 50 0 m 1.0 s m s. (b) vavg x t = = = 4 1 41 0 . . m 0.1 s m s Section 2.2 Instantaneous Velocity and Speed P2.5 (a) at ti = 1 5. s, xi = 8 0. m (Point A) at tf = 4 0. s, xf = 2 0. m (Point B) vavg f i f i x x t t = = ( ) ( ) = 2 0 8 0 4 1 5 6 0. . . .m s m 22.5 s m s= 2 4. (b) The slope of the tangent line can be found from points C and D. t xC C= =( )1 0 9 5. .s, m and t xD D= =( )3 5 0. s, , v 3 8. m s . (c) The velocity is zero when x is a minimum. This is at t 4 s . P2.6 (a) At any time, t, the position is given by x t= ( )3 00 2 . m s2 . Thus, at ti = 3 00. s: xi = ( )( ) =3 00 3 00 27 0 2 . . .m s s m2 . (b) At t tf = +3 00. s : x tf = ( ) +( )3 00 3 00 2 . .m s s2 , or x t tf = + ( ) + ( )( )27 0 18 0 3 00 2 . . .m m s m s2 . (c) The instantaneous velocity at t = 3 00. s is: v = = + lim lim . . t f i t x x t0 0 18 0 3 00m s mm s m s2 ( )( )=t 18 0. . FIG. P2.5 ISMV1_5103_02.indd 17ISMV1_5103_02.indd 17 10/27/06 2:45:21 PM10/27/06 2:45:21 PM
19. 19. 18 Chapter 2 P2.7 (a) (b) At t = 5 0. s, the slope is v = 58 23 m 2.5 s m s . At t = 4 0. s, the slope is v = 54 1 m 3 s 8 m s . At t = 3 0. s, the slope is v = 4 1 9 m 3.4 s 4 m s . At t = 2 0. s, the slope is v = 3 9 6 m 4.0 s .0 m s . (c) a t avg = = v 23 5 0 4 6 m s s m s2 . . (d) Initial velocity of the car was zero . P2.8 (a) v = ( ) ( ) = 5 0 1 0 5 m s m s (b) v = ( ) ( ) = 5 10 4 2 2 5 m s m s. (c) v = ( ) ( ) = 5 5 5 4 0 m m s s (d) v = ( ) ( ) = + 0 5 8 7 5 m s s m s P2.9 Once it resumes the race, the hare will run for a time of t x xf i x = = = v 1000 25 m 800 m 8 m s s. In this time, the tortoise can crawl a distance x xf i = ( )( ) =0 2 25 5 00. .m s s m . Section 2.3 Acceleration P2.10 Choose the positive direction to be the outward direction, perpendicular to the wall. v vf i at= + : a t = = ( ) = v 22 0 25 0 3 50 10 1 34 13 . . . . m s m s s 004 m s2 FIG. P2.8 ISMV1_5103_02.indd 18ISMV1_5103_02.indd 18 10/27/06 2:45:22 PM10/27/06 2:45:22 PM
20. 20. Motion in One Dimension 19 P2.11 (a) Acceleration is constant over the rst ten seconds, so at the end of this interval v vf i at= + = + ( )( ) =0 2 00 10 0 20 0. . .m s s m s2 . Then a = 0 so v is constant from t =10 0. s to t =15 0. s. And over the last ve seconds the velocity changes to v vf i at= + = + ( )( ) =20 0 3 00 5 00 5 00. . . .m s m s s m2 ss . (b) In the rst ten seconds, x x t atf i i= + + = + + ( )( ) =v 1 2 0 0 1 2 2 00 10 0 12 2 . .m s s2 000 m. Over the next ve seconds the position changes to x x t atf i i= + + = + ( )( )+ =v 1 2 100 20 0 5 00 0 22 m m s s. . 000 m. And at t = 20 0. s, x x t atf i i= + + = + ( )( )+ v 1 2 200 20 0 5 00 1 2 2 m m s s. . 33 00 5 00 262 2 . .m s s m2 ( )( ) = . P2.12 (a) Acceleration is the slope of the graph of v versus t. For 0 5 00<
21. 21. 20 Chapter 2 P2.13 x t t= + 2 00 3 00 2 . . , so v = = dx dt t3 00 2 00. . , and a d dt = = v 2 00. At t = 3 00. s: (a) x = + ( ) =2 00 9 00 9 00 2 00. . . .m m (b) v = ( ) = 3 00 6 00 3 00. . .m s m s (c) a = 2 00. m s2 *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 0 20 40 60 80 100 time Positionalongtrack,cm continued on next page Velocity as a function of time 0 time xcomponentofvelocity, arbitraryunits ISMV1_5103_02.indd 20ISMV1_5103_02.indd 20 11/2/06 1:29:11 PM11/2/06 1:29:11 PM
22. 22. Motion in One Dimension 21 Acceleration as a function of timetime acceleration,arbitraryunits P2.15 (a) At t = 2 00. s, x = ( ) ( )+ =3 00 2 00 2 00 2 00 3 00 11 0 2 . . . . . .m m. At t = 3 00. s, x = ( ) ( )+ =3 00 9 00 2 00 3 00 3 00 24 0 2 . . . . . .m m so vavg x t = = = 24 0 11 0 2 00 13 0 . . . . m m 3.00 s s m ss . (b) At all times the instantaneous velocity is v = +( )= ( ) d dt t t t3 00 2 00 3 00 6 00 2 002 . . . . . m s At t = 2 00. s, v = ( ) =6 00 2 00 2 00 10 0. . . .m s m s . At t = 3 00. s, v = ( ) =6 00 3 00 2 00 16 0. . . .m s m s . (c) a t avg = = = v 16 0 10 0 3 00 2 00 6 00 . . . . . m s m s s s m s2 (d) At all times a d dt t= ( ) =6 00 2 00 6 00. . . m s2 . This includes both t = 2 00. s and t = 3 00. s. P2.16 (a) a t = = = v 8 00 6 00 1 3 . . . m s s m s2 (b) Maximum positive acceleration is at t = 3 s, and is the slope of the graph, approximately (6 2)(4 2) = 2 m s2 . (c) a = 0 at t = 6 s , and also for t10 s . (d) Maximum negative acceleration is at t = 8 s, and is the slope of the graph, approximately 1 5. m s2 . ISMV1_5103_02.indd 21ISMV1_5103_02.indd 21 10/27/06 2:45:25 PM10/27/06 2:45:25 PM
23. 23. 22 Chapter 2 Section 2.4 Motion Diagrams *P2.17 (a) The motion is slow at rst, then fast, and then slow again. t t t0 a 0 0 x (b) The motion is constant in speed. t0 x t0 t0 a (c) The motion is speeding up, and we suppose the acceleration is constant. t0 x t0 t0 a 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 22ISMV1_5103_02.indd 22 10/27/06 4:53:04 PM10/27/06 4:53:04 PM
24. 24. Motion in One Dimension 23 Section 2.5 One-Dimensional Motion with Constant Acceleration *P2.19 (a) vf = vi + at = 13 ms 4 ms2 (1 s) = 9.00 ms (b) vf = vi + at = 13 ms 4 ms2 (2 s) = 5.00 ms (c) vf = vi + at = 13 ms 4 ms2 (2.5 s) = 3.00 ms (d) vf = vi + at = 13 ms 4 ms2 (4 s) = 3.00 ms (e) vf = vi + at = 13 ms 4 ms2 (1 s) = 17.0 ms (f) The graph of velocity versus time is a slanting straight line, having the value 13 ms at 10:05:00 a.m. on the certain date, and sloping down by 4 ms 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 x tf i i f = +( )1 2 v v becomes 40 1 2 2 80 8 50m m s s= +( )( )vi . . which yields vi = 6 61. m s . (b) a t f i = = = v v 2 80 6 61 8 50 0 448 . . . . m s m s s m s2 P2.21 Given vi = 12 0. cm s when x ti = =( )3 00 0. cm , and at t = 2 00. s, xf = 5 00. cm, x x t atf i i = +v 1 2 2 : = ( )+ ( )5 00 3 00 12 0 2 00 1 2 2 00 2 . . . . .a = +8 00 24 0 2. . a a = = 32 0 2 16 0 . . cm s2 . P2.22 (a) Total displacement = area under the v, t( ) curve from t = 0 to 50 s. x = ( )( )+ ( ) ( ) + 1 2 50 15 50 40 15 1 2 50 m s s m s s m s(( )( ) = = 10 1875 1 88 s m kmx . (b) From t = 10 s to t = 40 s, displacement is x = +( )( )+ ( )( ) = 1 2 50 33 5 50 25 1 46m s m s s m s s . kkm (c) 0 15 t s: a t 1 50 0 15 0 3 3= = ( ) = v m s s m s2 . 15 40s s t : a2 0= 40 50s s t : a t 3 0 50 50 40 5 0= = ( ) = v m s s s m s2 . FIG. P2.22 continued on next page ISMV1_5103_02.indd 23ISMV1_5103_02.indd 23 10/27/06 2:45:28 PM10/27/06 2:45:28 PM
25. 25. 24 Chapter 2 (d) (i) x a t t1 1 2 2 0 1 2 1 2 3 3= + = ( ). m s2 or x t1 2 1 67= ( ). m s2 (ii) x t2 1 2 15 50 0 50 15= ( ) [ ]+ ( ) ( )s m s m s s or x t2 50 375= ( ) m s m (iii) For 40 50s s t , x t t3 0 = = + area under vs from to 40 s v 11 2 40 50 403 2 a t t( ) + ( ) ( )s m s s or x t3 2 375 1250 1 2 5 0 40 50= + + ( ) ( ) +m m m s s m s2 . (( ) ( )t 40 s which reduces to x t t3 2 250 2 5 4 375= ( ) ( ) m s m s m2 . . (e) v = = total displacement total elapsed time 1875 m s m s 50 37 5= . P2.23 (a) vi = 100 m s, a = 5 00. m s2 , v vf i at= + so 0 100 5= t, v vf i f ia x x2 2 2= + ( ) so 0 100 2 5 00 0 2 = ( ) ( ) ( ). xf . Thus xf = 1000 m and t = 20 0. s . (b) 1 000 m is greater than 800 m. With this acceleration the plane would overshoot the runway: it cannot land . *P2.24 (a) For the rst car the speed as a function of time is v = vi + at = 3.5 cms + 2.4 cms2 t. For the second car, the speed is + 5.5 cms + 0. Setting the two expressions equal gives 3.5 cms + 2.4 cms2 t = 5.5 cms so t = (9 cms)(2.4 cms2 ) = 3.75 s . (b) The rst car then has speed 3.5 cms + (2.4 cms2 )(3.75 s) = 5.50 cms , and this is the constant speed of the second car also. (c) For the rst car the position as a function of time is xi + vi t + (12)at2 = 15 cm (3.5 cms)t + (0.5)(2.4 cms2 ) t2 . For the second car, the position is 10 cm + (5.5 cms)t + 0. At passing, the positions are equal: 15 cm (3.5 cms)t + (1.2 cms2 ) t2 = 10 cm + (5.5 cms)t (1.2 cms2 ) t2 (9 cms) t + 5 cm = 0. We solve with the quadratic formula: t = = + = 9 9 4 1 2 5 2 1 2 9 57 2 4 9 57 2 4 2 ( . )( ) ( . ) . . and 66 90. s and 0.604 s (d) At 0.604 s, the second and also the rst cars position is 10 cm + (5.5 cms)0.604 s = 13.3 cm . At 6.90 s, both are at position 10 cm + (5.5 cms)6.90 s = 47.9 cm . continued on next page ISMV1_5103_02.indd 24ISMV1_5103_02.indd 24 11/2/06 11:12:23 AM11/2/06 11:12:23 AM
26. 26. Motion in One Dimension 25 (e) The cars are initially moving toward each other, so they soon share the same position x when their speeds are quite different, giving one answer to (c) that is not an answer to (a). The rst car slows down in its motion to the left, turns around, and starts to move toward the right, slowly at rst and gaining speed steadily. At a particular moment its speed will be equal to the constant rightward speed of the second car. The distance between them will at that moment be staying constant at its maximum value. The dis- tance between the cars will be far from zero, as the accelerating car will be far to the left of the steadily moving car. Thus the answer to (a) is not an answer to (c). Eventually the accelerating car will catch up to the steadily-coasting car, whizzing past at higher speed than it has ever had before, and giving another answer to (c) that is not an answer to (a). A graph of x versus t for the two cars shows a parabola originally sloping down and then curving upward, intersecting twice with an upward-sloping straight line. The parabola and straight line are running parallel, with equal slopes, at just one point in between their intersections. P2.25 In the simultaneous equations: v v v v xf xi x f i xi xf a t x x t = + = +( ) 1 2 we have v v v v xf xi xi xf = ( )( ) = +( 5 60 4 20 62 4 1 2 . . . m s s m 2 ))( ) 4 20. s . So substituting for vxi gives 62 4 1 2 5 60 4 20 4 20. . . .m m s s2 = + ( )( )+ v vxf xf s( ) 14 9 1 2 5 60 4 20. . .m s m s s2 = + ( )( )vxf . Thus vxf = 3 10. m s . P2.26 Take any two of the standard four equations, such as v v v v xf xi x f i xi xf a t x x t = + = +( ) 1 2 . Solve one for vxi , and substitute into the other: v vxi xf xa t= x x a t tf i xf x xf = +( )1 2 v v . Thus x x t a tf i xf x = v 1 2 2 . We note that the equation is dimensionally correct. The units are units of length in each term. Like the standard equation x x t a tf i xi x = +v 1 2 2 , this equation represents that displacement is a quadratic function of time. Our newly derived equation gives us for the situation back in problem 25, 62 4 4 20 1 2 5 60 4 20 2 . . . .m s m s s2 = ( ) ( )( )vxf vxf = = 62 4 49 4 3 10 . . . m m 4.20 s m s . ISMV1_5103_02.indd 25ISMV1_5103_02.indd 25 10/27/06 2:45:30 PM10/27/06 2:45:30 PM
27. 27. 26 Chapter 2 P2.27 (a) a t f i = = ( ) = = v v 632 5280 3600 1 40 662 202 / . ft s2 mm s2 (b) x t atf i= + = ( ) ( )v 1 2 632 5 280 3 600 1 40 1 2 662 . 22 1 40 649 198 2 ( )( ) = =. ft m P2.28 (a) Compare the position equation x t t= + 2 00 3 00 4 00 2 . . . to the general form x x t atf i i= + +v 1 2 2 to recognize that xi = 2 00. m, vi = 3 00. m s, and a = 8 00. m s2 . The velocity equation, v vf i at= + , is then vf t= ( )3 00 8 00. .m s m s2 . The particle changes direction when vf = 0, which occurs at t = 3 8 s. The position at this time is x = + ( ) ( ) 2 00 3 00 3 8 4 00 3 8 . . .m m s s m s s2 = 2 2 56. m . (b) From x x t atf i i= + +v 1 2 2 , observe that when x xf i= , the time is given by t a i = 2v . Thus, when the particle returns to its initial position, the time is t = ( ) = 2 3 00 8 00 3 4 . . m s m s s2 and the velocity is vf = ( ) = 3 00 8 00 3 4 3 00. . .m s m s s m s2 . P2.29 We have vi = 2 00 104 . m s, vf = 6 00 106 . m s, x xf i = 1 50 10 2 . m. (a) x x tf i i f = +( )1 2 v v : t x xf i i f = ( ) + = ( ) + 2 2 1 50 10 2 00 10 6 2 4 v v . . m m s .. . 00 10 4 98 106 9 = m s s (b) v vf i x f ia x x2 2 2= + ( ): a x x x f i f i = = ( ) v v2 2 6 2 4 2 6 00 10 2 00 10 ( ) . .m s mm s m m s2( ) = 2 2 15 2 1 50 10 1 20 10 ( . ) . P2.30 (a) Along the time axis of the graph shown, let i = 0 and f tm= . Then v vxf xi xa t= + gives vc m ma t= +0 a t m c m = v . (b) The displacement between 0 and tm is x x t a t t t tf i xi x c m m c m = + = + =v v v 1 2 0 1 2 1 2 2 2 . The displacement between tm and t0 is x x t a t t tf i xi x c m = + = ( )+v v 1 2 02 0 . The total displacement is x t t t t tc m c c m c m= + = 1 2 1 2 0 0v v v v . continued on next page ISMV1_5103_02.indd 26ISMV1_5103_02.indd 26 10/27/06 2:45:31 PM10/27/06 2:45:31 PM
28. 28. Motion in One Dimension 27 (c) For constant vc and t0 , x is minimized by maximizing tm to t tm = 0. Then x t t t c c min = =v v 0 0 01 2 2 . (e) This is realized by having the servo motor on all the time. (d) We maximize x by letting tm approach zero. In the limit x t tc c= ( )=v v0 00 . (e) This cannot be attained because the acceleration must be nite. P2.31 Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit, x x t a t t a t t f i xi x i d d d d d = + + = + + = = v v v v 1 2 0 1 2 2 2 vvi da t+ 1 2 (a) The speed halfway through the photogate in space is given by v v v vhs i i d da a t2 2 2 2 2 = + = + . v v vhs i d da t= +2 and this is not equal to vd unless a = 0 . (b) The speed halfway through the photogate in time is given by v vht i d a t = + 2 and this is equal to vd as determined above. P2.32 Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sues car. For her we have xis = 0, vis = 30 0. m s, as = 2 00. m s2 so her position is given by x t x t a t ts is is s( ) = + + = ( ) + v 1 2 30 0 1 2 2 002 . .m s m s22 ( )t2 . For the van, xiv = 155 m, v vi = 5 00. m s, av = 0 and x t x t a t ti iv v v vv( ) = + + = + ( ) + 1 2 155 5 00 02 . m s . To test for a collision, we look for an instant tc when both are at the same place: 30 0 155 5 00 0 25 0 155 2 2 . . . . t t t t t c c c c c = + = + From the quadratic formula tc = ( ) ( ) = 25 0 25 0 4 155 2 13 6 2 . . . s or 11 4. s . The roots are real, not imaginary, so there is a collision . The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position 155 5 00 11 4 212m m s s m+ ( )( ) =. . . ISMV1_5103_02.indd 27ISMV1_5103_02.indd 27 10/27/06 2:45:33 PM10/27/06 2:45:33 PM
29. 29. 28 Chapter 2 *P2.33 (a) Starting from rest and accelerating at ab = 13 0. mi h s, the bicycle reaches its maximum speed of vb,max .= 20 0 mi h in a time t a b b b , ,max . . .1 0 20 0 13 0 1 54= = = v mi h mi h s s. Since the acceleration ac of the car is less than that of the bicycle, the car cannot catch the bicycle until some time t tb ,1 (that is, until the bicycle is at its maximum speed and coast- ing). The total displacement of the bicycle at time t is x a t t tb b b b b= + ( ) = 1 2 1 47 1 1 2 1, ,max , . v ft s mi h ( ) + ( ) 1 2 13 0 1 54 20 0 2 . . . mi h s s mi h t 11 54 29 4 22 6 . . . s ft s ft ( ) = ( ) t The total displacement of the car at this time is x a tc c= = 1 2 1 47 1 1 2 9 002 . . ft s mi h mi h s = ( )t t2 2 6 62. ft s At the time the car catches the bicycle x xc b= . This gives 6 62 29 4 22 62 . . .ft s ft s ft2 ( ) = ( ) t t or t t2 2 4 44 3 42 0 ( ) + =. .s s that has only one physically meaningful solution t tb ,1. This solution gives the total time the bicycle leads the car and is t = 3 45. s . (b) The lead the bicycle has over the car continues to increase as long as the bicycle is moving faster than the car. This means until the car attains a speed of v vc b= =,max .20 0 mi h. Thus, the elapsed time when the bicycles lead ceases to increase is t a b c = = = v ,max . . . 20 0 9 00 2 22 mi h mi h s s At this time, the lead is x x x xb c b c t ( ) = ( ) = ( )=max . . .2 22 29 4 2s ft s 222 22 6 6 62 2 22 2 s ft ft s s2 ( ) ( )( ) . . . or x xb c( ) =max .10 0 ft . P2.34 As in the algebraic solution to Example 2.9, we let t represent the time the trooper has been moving. We graph x tcar = +45 45 and x ttrooper = 1 5 2 . . They intersect at t = 31 s . FIG. P2.34 x (km) t (s) 10 20 30 40 0.5 1 1.5 car police officer ISMV1_5103_02.indd 28ISMV1_5103_02.indd 28 10/27/06 4:54:11 PM10/27/06 4:54:11 PM
30. 30. Motion in One Dimension 29 *P2.35 (a) Let a stopwatch start from t = 0 as the front end of the glider passes point A. The average speed of the glider over the interval between t = 0 and t = 0.628 s is 12.4 cm(0.628 s) = 19.7 cms , and this is the instantaneous speed halfway through the time interval, at t = 0.314 s. (b) The average speed of the glider over the time interval between 0.628 + 1.39 = 2.02 s and 0.628 + 1.39 + 0.431 = 2.45 s is 12.4 cm(0.431 s) = 28.8 cms and this is the instantaneous speed at the instant t = (2.02 + 2.45)2 = 2.23 s. Now we know the velocities at two instants, so the acceleration is found from [(28.8 19.7) cms][(2.23 0.314) s] = 9.031.92 cms2 = 4.70 cms2 . (c) The time required to pass between A and B is sufcient to nd the acceleration, more directly than we could nd it from the distance between the points. Section 2.6 Freely Falling Objects *P2.36 Choose the origin y t= =( )0 0, at the starting point of the cat and take upward as positive. Then yi = 0, vi = 0, and a g= = 9 80. m s2 . The position and the velocity at time t become: y y t atf i i = +v 1 2 2 : y gt tf = = ( )1 2 1 2 9 802 2 . m s2 and v vf i at= + : vf gt t= = ( )9 80. m s2 . (a) at t = 0 1. s: yf = ( )( ) = 1 2 9 80 0 0 049 0 2 . .m s .1 s m2 at t = 0 2. s: yf = ( )( ) = 1 2 9 80 0 0 196 2 . .m s .2 s m2 at t = 0 3. s: yf = ( )( ) = 1 2 9 80 0 0 441 2 . .m s .3 s m2 (b) at t = 0 1. s: vf = ( )( ) = 9 80 0 0 980. .m s .1 s m s2 at t = 0 2. s: vf = ( )( ) = 9 80 0 1 96. .m s .2 s m s2 at t = 0 3. s: vf = ( )( ) = 9 80 0 2 94. .m s .3 s m s2 ISMV1_5103_02.indd 29ISMV1_5103_02.indd 29 10/27/06 2:45:36 PM10/27/06 2:45:36 PM
31. 31. 30 Chapter 2 P2.37 Assume that air resistance may be neglected. Then, the acceleration at all times during the ight is that due to gravity, a g= = 9 80. m s2 . During the ight, Goff went 1 mile (1 609 m) up and then 1 mile back down. Determine his speed just after launch by considering his upward ight: v v vf i f i ia y y2 2 2 2 0 2 9 80 1609= + ( ) = ( )( ): . m s m2 vvi = 178 m s. His time in the air may be found by considering his motion from just after launch to just before impact: y y t atf i i = +v 1 2 2 : 0 178 1 2 9 80 2 = ( ) ( )m s m s2 t t. . The root t = 0 describes launch; the other root, t = 36 2. s, describes his ight time. His rate of pay is then pay rate = = ( )( )= $ . . . $ . 1 00 36 2 0 027 6 3 600 99 3 s $ s s h h . We have assumed that the workmans ight time, a mile, and a dollar, were measured to three-digit precision. We have interpreted up in the sky as referring to the free fall time, not to the launch and landing times. Both the takeoff and landing times must be several seconds away from the job, in order for Goff to survive to resume work. P2.38 We have y gt t yf i i = + + 1 2 2 v 0 4 90 8 00 30 02 = ( ) ( ) +. . .m s m s m2 t t . Solving for t, t = + 8 00 64 0 588 9 80 . . . . Using only the positive value for t, we nd that t = 1 79. s . P2.39 (a) y y t atf i i = +v 1 2 2 : 4 00 1 50 4 90 1 50 2 . . . .= ( ) ( )( )vi and vi = 10 0. m s upward . (b) v vf i at= + = ( )( ) = 10 0 9 80 1 50 4 68. . . . m s vf = 4 68. m s downward P2.40 The bill starts from rest vi = 0 and falls with a downward acceleration of 9 80. m s2 (due to gravity). Thus, in 0.20 s it will fall a distance of y t gti= = ( )( ) = v 1 2 0 4 90 0 20 0 202 2 . . .m s s m2 . This distance is about twice the distance between the center of the bill and its top edge ( )8 cm . Thus, David will be unsuccessful. ISMV1_5103_02.indd 30ISMV1_5103_02.indd 30 10/28/06 2:35:43 AM10/28/06 2:35:43 AM
32. 32. Motion in One Dimension 31 P2.41 (a) v vf i gt= : vf = 0 when t = 3 00. s, and g = 9 80. m s2 . Therefore, vi gt= = ( )( ) =9 80 3 00 29 4. . .m s s m s2 . (b) y y tf i f i = +( )1 2 v v y yf i = ( )( )= 1 2 29 4 3 00 44 1. . .m s s m *P2.42 We can solve (a) and (b) at the same time by assuming the rock passes the top of the wall and nding its speed there. If the speed comes out imaginary, the rock will not reach this elevation. vf 2 = vi 2 + 2a(xf xi ) = (7.4 ms)2 2(9.8 ms2 )(3.65 m 1.55 m) = 13.6 m2 s2 so the rock does reach the top of the wall with vf = 3.69 ms . (c) We nd the nal speed, just before impact, of the rock thrown down: vf 2 = vi 2 + 2a(xf xi ) = (7.4 ms)2 2(9.8 ms2 )(1.55 m 3.65 m) = 95.9 m2 s2 vf = 9.79 ms. The change in speed of the rock thrown down is 9.79 7.4 = 2.39 ms (d) The magnitude of the speed change of the rock thrown up is 7.4 3.69 = 3.71 ms. This does not agree with 2.39 ms. The upward-moving rock spends more time in ight, so the planet has more time to change its speed. P2.43 Time to fall 3.00 m is found from the equation describing position as a function of time, with vi = 0, thus: 3 00 1 2 9 80 2 . .m m s2 = ( )t , giving t = 0 782. s. (a) With the horse galloping at 10 0. m s, the horizontal distance is vt = 7 82. m . (b) from above t = 0 782. s P2.44 y t= 3 00 3 . : At t = 2 00. s, y = ( ) =3 00 2 00 24 0 3 . . . m and vy dy dt t= = = 9 00 36 02 . . m s . If the helicopter releases a small mailbag at this time, the mailbag starts its free fall with velocity 36 ms upward. The equation of motion of the mailbag is y y t gt t tb bi i= + = + ( )v 1 2 24 0 36 0 1 2 9 802 2 . . . . Setting yb = 0, 0 24 0 36 0 4 90 2 = + . . .t t . Solving for t, (only positive values of t count), t = 7 96. s . ISMV1_5103_02.indd 31ISMV1_5103_02.indd 31 10/28/06 2:35:45 AM10/28/06 2:35:45 AM
33. 33. 32 Chapter 2 P2.45 We assume the object starts from rest. Consider the last 30 m of its fall. We nd its speed 30 m above the ground: y y t a tf i yi y yi = + + = + ( )+ v v 1 2 0 30 1 5 1 2 9 8 2 m s m. . ss s m m 1.5 s m s 2 ( )( ) = + = 1 5 30 11 0 12 6 2 . . .vyi .. Now consider the portion of its fall above the 30 m point: v vyf yi y f ia y y2 2 2 2 12 6 0 2 9 8 = + ( ) ( ) = + . .m s m s2 (( ) = = y y 160 19 6 8 16 m s m s m 2 2 2 . . . Its original height was then 30 8 16 38 2m m m+ =. . . Section 2.7 Kinematic Equations Derived from Calculus P2.46 (a) See the graphs at the right. Choose x = 0 at t = 0. At t = 3 s, x = ( )( )= 1 2 8 3 12m s s m. At t = 5 s, x = + ( )( )=12 8 2 28m m s s m. At t = 7 s, x = + ( )( )=28 1 2 8 2 36m m s s m. (b) For 0 3 t s, a = = 8 3 2 67 m s s m s2 . . For 3 5 t s, a = 0 . (c) For 5 9s s t , a = = 16 4 4 m s s m s2 . (d) At t = 6 s, x = + ( )( )=28 6 1 34m m s s m . (e) At t = 9 s, x = + ( )( )=36 1 2 8 2 28m m s s m . FIG. P2.46 ISMV1_5103_02.indd 32ISMV1_5103_02.indd 32 10/27/06 2:45:39 PM10/27/06 2:45:39 PM
34. 34. Motion in One Dimension 33 P2.47 (a) J da dt = = constant da Jdt= a J dt Jt c= = + 1 but a ai= when t = 0 so c ai1 = . Therefore, a Jt ai= + a d dt d adt adt Jt a dt Jt a t ci i = = = = +( ) = + + v v v 1 2 2 2 but v v= i when t = 0, so c i2 = v and v v= + + 1 2 2 Jt a ti i v v v v = = = = + + = dx dt dx dt x dt Jt a t dt x J i i 1 2 1 6 2 tt a t t c x x i i i 3 2 3 1 2 + + + = v when t = 0, so c xi3 = . Therefore, x Jt a t t xi i i= + + + 1 6 1 2 3 2 v . (b) a Jt a J t a Ja ti i i 2 2 2 2 2 2= +( ) = + + a a J t Ja ti i 2 2 2 2 2= + +( ) a a J Jt a ti i 2 2 2 2 1 2 = + + Recall the expression for v: v v= + + 1 2 2 Jt a ti i . So v v( )= +i iJt a t 1 2 2 . Therefore, a a Ji i 2 2 2= + ( )v v . ISMV1_5103_02.indd 33ISMV1_5103_02.indd 33 10/27/06 2:45:40 PM10/27/06 2:45:40 PM
35. 35. 34 Chapter 2 P2.48 (a) a d dt d dt t t= = + v 5 00 10 3 00 107 2 5 . . a t= ( ) + 10 0 10 3 00 107 5 . .m s m s3 2 Take xi = 0 at t = 0. Then v = dx dt x dt t t dt x t t = = + ( ) = 0 5 00 10 3 00 10 5 0 7 2 5 0 v . . .. . . 00 10 3 3 00 10 2 1 67 10 7 3 5 2 7 3 + = ( ) + t t x tm s3 11 50 105 2 . .( )m s2 t (b) The bullet escapes when a = 0, at ( ) + =10 0 10 3 00 10 07 5 . .m s m s3 2 t t = = 3 00 10 3 00 10 5 3. . s 10.0 10 s7 . (c) New v = ( ) ( ) + ( ) 5 00 10 3 00 10 3 00 10 3 00 107 3 2 5 . . . . ( )3 v = + =450 900 450m s m s m s . (d) x = ( ) ( ) + ( ) 1 67 10 3 00 10 1 50 10 3 00 107 3 3 5 . . . . ( )3 2 x = + =0 450 1 35 0 900. . .m m m Additional Problems *P2.49 (a) The velocity is constant between ti = 0 and t = 4 s. Its acceleration is 0 . (b) a = (v9 v4 )(9 s 4 s) = (18 [12]) (ms)5 s = 6.0 ms2 . (c) a = (v18 v13 )(18 s 13 s) = (0 18) (ms)5 s = 3.6 ms2 . (d) We read from the graph that the speed is zero at t = 6 s and at 18 s . (e) and (f) The object moves away from x = 0 into negative coordinates from t = 0 to t = 6 s, but then comes back again, crosses the origin and moves farther into positive coordinates until t = 18 s , then attaining its maximum distance, which is the cumulative distance under the graph line: (12 ms)(4 s) + (6 ms)(2 s) + (9 ms)(3 s) + (18 ms)(4 s) + (9 ms)(5 s) = 60 m + 144 m = 84 m . (g) To gauge the wear on the tires, we consider the total distance rather than the resultant dis- placement, by counting the contributions computed in part (f) as all positive: + 60 m + 144 m = 204 m . ISMV1_5103_02.indd 34ISMV1_5103_02.indd 34 10/27/06 2:45:41 PM10/27/06 2:45:41 PM
36. 36. Motion in One Dimension 35 P2.50 (a) As we see from the graph, from about 50 s to 50 s Acela is cruising at a constant positive velocity in the +x direction. From 50 s to 200 s, Acela accelerates in the +x direction reaching a top speed of about 170 mih. Around 200 s, the engineer applies the brakes, and the train, still traveling in the +x direction, slows down and then stops at 350 s. Just after 350 s, Acela reverses direction (v becomes negative) and steadily gains speed in the x direction. t (s) 100 200 300 100 100 200 400 v t 50 0 0 FIG. P2.50(a) (b) The peak acceleration between 45 and 170 mih is given by the slope of the steepest tangent to the v versus t curve in this interval. From the tangent line shown, we nd a t = = = ( ) ( ) =slope mi h s mi h v 155 45 100 50 2 2. (( ) =s m s2 0 98. . (c) Let us use the fact that the area under the v versus t curve equals the displacement. The trains displacement between 0 and 200 s is equal to the area of the gray shaded region, which we have approximated with a series of triangles and rectangles. x0 200 50 = + + + + s 1 2 3 4 5area area area area area mmi h s mi h s 60 mi h s ( )( )+ ( )( ) + ( )( 50 50 50 1 100 )) + ( )( ) + ( ) 1 2 100 1 2 100 170 160 50 s mi h s mi h mii h mi h s ( ) = ( )( )24 000 Now, at the end of our calculation, we can nd the displacement in miles by converting hours to seconds. As 1 3 600h s= , x0 200 24 000 6 7 ( )=s mi 3 600 s s mi. . t (s) 100 200 300 100 200 400 1 2 4 3 5 0 0 FIG. P2.50(c) ISMV1_5103_02.indd 35ISMV1_5103_02.indd 35 10/27/06 2:45:42 PM10/27/06 2:45:42 PM
37. 37. 36 Chapter 2 P2.51 Let point 0 be at ground level and point 1 be at the end of the engine burn. Let point 2 be the highest point the rocket reaches and point 3 be just before impact. The data in the table are found for each phase of the rockets motion. (0 to 1) vf 2 2 80 0 2 4 00 1000 ( ) = ( )( ). . so vf = 120 m s 120 80 0 4 00= + ( ). . t giving t = 10 0. s (1 to 2) 0 120 2 9 80 2 ( ) = ( ) ( ). x xf i giving x xf i = 735 m 0 120 9 80 = . t giving t = 12 2. s This is the time of maximum height of the rocket. (2 to 3) vf 2 0 2 9 80 1735 = ( ) ( ). vf t= = ( )184 9 80. giving t = 18 8. s FIG. P2.51 (a) ttotal s= + + =10 12 2 18 8 41 0. . . (b) x xf i( ) =total km1 73. (c) vfinal m s= 184 t x v a 0 Launch 0.0 0 80 +4.00 #1 End Thrust 10.0 1 000 120 +4.00 #2 Rise Upwards 22.2 1 735 0 9.80 #3 Fall to Earth 41.0 0 184 9.80 P2.52 Area A1 is a rectangle. Thus, A hw txi1 = = v . Area A2 is triangular. Therefore A bh t x xi2 1 2 1 2 = = ( )v v . The total area under the curve is A A A t t xi x xi = + = + ( ) 1 2 2 v v v and since v vx xi xa t = A t a txi x= +v 1 2 2 . The displacement given by the equation is: x t a txi x= +v 1 2 2 , the same result as above for the total area. FIG. P2.52 vx vx vxi 0 t t A2 A1 ISMV1_5103_02.indd 36ISMV1_5103_02.indd 36 10/27/06 2:45:43 PM10/27/06 2:45:43 PM
38. 38. Motion in One Dimension 37 P2.53 (a) Let x be the distance traveled at acceleration a until maximum speed v is reached. If this is achieved in time t1 we can use the following three equations: x ti= +( )1 2 1v v , 100 10 2 1 = ( )x tv . , and v v= +i at1. The rst two give 100 10 2 1 2 10 2 1 2 200 20 1 1 1= = = . . . t t at a v 44 1 1( )t t . For Maggie: m s For J 2 a = ( )( ) = 200 18 4 2 00 5 43 . . . uudy: m s2 a = ( )( ) = 200 17 4 3 00 3 83 . . . (b) v = at1 Maggie: m s Judy: v v = ( )( ) = = ( 5 43 2 00 10 9 3 83 . . . . ))( ) =3 00 11 5. . m s (c) At the six-second mark x at t= + ( )1 2 6 001 2 1v . Maggie: x = ( )( ) + ( )( ) = 1 2 5 43 2 00 10 9 4 00 54 3 2 . . . . . m Judy: x = ( )( ) + ( )( ) = 1 2 3 83 3 00 11 5 3 00 51 2 . . . . .77 m Maggie is ahead by 54.3 m 51.7 m = 2 62. m . Note that your students may need a reminder that to get the answer in the back of the book they must use calculator memory or a piece of paper to save intermediate results without rounding off until the very end. *P2.54 (a) We rst nd the distance sstop over which you can stop. The car travels this distance during your reaction time: x1 = v0 (0.6 s). As you brake to a stop, the average speed of the car is v0 2, the interval of time is (vf vi )a = v0 (2.40 ms2 ) = v0 s2 2.40 m, and the braking distance is x2 = vavg t = (v0 s2 2.40 m)(v0 2) = v0 2 s2 4.80 m. The total stopping distance is then sstop = x1 + x2 = v0 (0.6 s) + v0 2 s2 4.80 m. If the car is at this distance from the intersection, it can barely brake to a stop, so it should also be able to get through the intersection at constant speed while the light is yellow, moving a total distance sstop + 22 m = v0 (0.6 s) + v0 2 s2 4.80 m + 22 m. This constant- speed motion requires time ty = (sstop + 22 m)v0 = (v0 (0.6 s) + v0 2 s2 4.80 m + 22 m)v0 = 0.6 s + v0 s2 4.80 m + 22 mv0 . continued on next page ISMV1_5103_02.indd 37ISMV1_5103_02.indd 37 10/27/06 4:55:57 PM10/27/06 4:55:57 PM
39. 39. 38 Chapter 2 (b) Substituting, ty = 0.6 s + (8 ms) s2 4.80 m + 22 m(8 ms) = 0.6 s + 1.67 s + 2.75 s = 5.02 s . (c) We are asked about higher and higher speeds. For 11 ms instead of 8 ms, the time is 0.6 s + (11 ms) s2 4.80 m + 22 m(11 ms) = 4.89 s less than we had at the lower speed. (d) Now the time 0.6 s + (18 ms) s2 4.80 m + 22 m(18 ms) = 5.57 s begins to increase (e) 0.6 s + (25 ms) s2 4.80 m + 22 m(25 ms) = 6.69 s (f) As v0 goes to zero, the 22 mv0 term in the expression for ty becomes large, approaching innity . (g) As v0 grows without limit, the v0 s2 4.80 m term in the expression for ty becomes large, approaching innity . (h) ty decreases steeply from an innite value at v0 = 0, goes through a rather at mini- mum, and then diverges to innity as v0 increases without bound. For a very slowly moving car entering the intersection and not allowed to speed up, a very long time is required to get across the intersection. A very fast-moving car requires a very long time to slow down at the constant acceleration we have assumed. (i) To nd the minimum, we set the derivative of ty with respect to v0 equal to zero: d dv v v 0 0 0 1 0 6 4 8 22 0. . s + s m m s2 2 + = + 44 8 22 00 2 . m m = v 22 m v0 2 = s2 4.8 m v0 = (22m [4.8 ms2 ])12 = 10.3 ms (j) Evaluating again, ty = 0.6 s + (10.3 ms) s2 4.80 m + 22 m(10.3 ms) = 4.88 s , just a little less than the answer to part (c). For some students an interesting project might be to measure the yellow-times of trafc lights on local roadways with various speed limits and compare with the minimum treaction + (width2 abraking )12 implied by the analysis here. But do not let the students string a tape measure across the intersection. P2.55 a1 0 100= . m s2 a2 0 500= . m s2 x a t t a t= = + +1000 1 2 1 2 1 1 2 1 2 2 2 2 m v t t t= +1 2 and v1 1 1 2 2= = a t a t 1000 1 2 1 2 1 1 2 1 1 1 1 2 2 1 1 2 = + + a t a t a t a a a t a 2 1000 1 2 11 1 2 1 2 = a a a t 1 000 = 0.5(0.1)[1 (0.10.5)]t1 2 20 000 = 1.20 t1 2 t1 20 000 1 20 129= = . s t a t a 2 1 1 2 12 9 0 500 26= = . . s Total time = =t 155 s ISMV1_5103_02.indd 38ISMV1_5103_02.indd 38 10/27/06 4:58:47 PM10/27/06 4:58:47 PM
40. 40. Motion in One Dimension 39 *P2.56 (a) From the information in the problem, we model the Ferrari as a particle under constant acceleration. The important particle for this part of the problem is the nose of the car. We use the position equation from the particle under constant acceleration model to nd the velocity v0 of the particle as it enters the intersection: x x t at= + + = + ( )+0 0 2 0 1 2 1 2 28 0 0 3 10 2 10v v. . .m s m/s22 2 0 3 10 12 3 ( )( ) = . . s m/sv Now we use the velocity-position equation in the particle under constant acceleration model to nd the displacement of the particle from the rst edge of the intersection when the Ferrari stops: v v v v 0 2 0 2 2 0 0 2 2 2 0 12 3 = + ( ) = = = ( a x x x x x a . m /s)) ( ) = 2 2 2 2 10 35 9 . . m /s m (b) The time interval during which any part of the Ferrari is in the intersection is that time interval between the instant at which the nose enters the intersection and the instant when the tail leaves the intersection. Thus, the change in position of the nose of the Ferrari is 4.52 m + 28.0 m = 32.52 m. We nd the time at which the car is at position x = 32.52 m if it is at x = 0 and moving at 12.3 ms at t = 0: x x t at t= + + = + ( ) +0 0 21 2 1 2 32 52 0 12 3 2 10v . . .m m/s mm/s2 2 2 1 05 12 3 32 52 0 ( ) + = t t t . . . The solutions to this quadratic equation are t = 4.04 s and 7.66 s. Our desired solution is the lower of these, so t = 4.04 s . (The later time corresponds to the Ferrari stopping and revers- ing, which it must do if the acceleration truly remains constant, and arriving again at the posi- tion x = 32.52 m.) (c) We again dene t = 0 as the time at which the nose of the Ferrari enters the intersection. Then at time t = 4.04 s, the tail of the Ferrari leaves the intersection. Therefore, to nd the minimum distance from the intersection for the Corvette, its nose must enter the intersec- tion at t = 4.04 s. We calculate this distance from the position equation: x x t at= + + = + + ( )( ) =0 0 2 2 21 2 0 0 1 2 5 60 4 04 45v . . .m /s s 88 m (d) We use the velocity equation: v v= + = + ( )( ) =0 2 0 5 60 4 04 22 6at . . .m /s s m/s ISMV1_5103_02.indd 39ISMV1_5103_02.indd 39 10/28/06 2:36:24 AM10/28/06 2:36:24 AM
41. 41. 40 Chapter 2 P2.57 (a) y t at t tf i= + = = + ( )v 1 2 21 2 50 0 2 00 1 2 9 80. . . , 4 90 2 00 50 0 02 . . .t t+ = t = + ( ) ( ) ( ) 2 00 2 00 4 4 90 50 0 2 4 90 2 . . . . . Only the positive root is physically meaningful: t = 3 00. s after the rst stone is thrown. (b) y t atf i= +v 2 21 2 and t = =3 00 1 00 2 00. . . s substitute 50 0 2 00 1 2 9 80 2 002 2 . . . .= ( )+ ( )( )vi : vi2 15 3= . m s downward (c) v v1 1 2 00 9 80 3 00 31 4f i at= + = + ( )( ) =. . . . m s downward v v2 2 15 3 9 80 2 00 34 8f i at= + = + ( )( ) =. . . . m s downward P2.58 Let the ball fall freely for 1.50 m after starting from rest. It strikes at speed given by v vxf xi f ia x x2 2 2= + ( ): vxf 2 0 2 9 80 1 50= + ( ) ( ). .m s m2 vxf = 5 42. m s If its acceleration were constant its stopping would be described by v vxf xi x f i x a x x a 2 2 2 2 2 0 5 42 2 10 = + ( ) = ( ) + . m s mm m s m m s 2 2 2 ( ) = = + ax 29 4 2 00 10 1 47 102 3. . . .. Upward acceleration of this same order of magnitude will continue for some additional time after the dent is at its maximum depth, to give the ball the speed with which it rebounds from the pave- ment. The balls maximum acceleration will be larger than the average acceleration we estimate by imagining constant acceleration, but will still be of order of magnitude ~103 m s2 . P2.59 (a) We require x xs k= when t ts k= +1 00. x t ts k k= ( ) +( ) = ( )( ) 1 2 3 50 1 00 1 2 4 90 2 2 . . .m s m s2 2 == + = = x t t t k k k k 1 00 1 183 5 46 . . . .s (b) xk = ( )( ) = 1 2 4 90 5 46 73 0 2 . . .m s s m2 (c) vk = ( )( ) =4 90 5 46 26 7. . .m s s m s2 vs = ( )( ) =3 50 6 46 22 6. . .m s s m s2 ISMV1_5103_02.indd 40ISMV1_5103_02.indd 40 10/28/06 2:36:25 AM10/28/06 2:36:25 AM
42. 42. Motion in One Dimension 41 P2.60 (a) d t= ( ) 1 2 9 80 1 2 . d t= 336 2 t t1 2 2 40+ = . 336 4 90 2 402 2 2 t t= ( ). . 4 90 359 5 28 22 02 2 2. . .t t + = t2 2 359 5 359 5 4 4 90 28 22 9 80 = ( )( ). . . . . t2 359 5 358 75 9 80 0 076 5= = . . . . s so d t= =336 26 42 . m (b) Ignoring the sound travel time, d = ( )( ) = 1 2 9 80 2 40 28 2 2 . . . m , an error of 6 82. % . *P2.61 (a) and (b) We divide each given thinking distance by the corresponding speed to test for the constancy of a proportionality constant. First, 27 ft25(5280 ft3600 s) = 0.736 s. Similarly, constant speed, mih 25 35 45 55 65 constant speed, fts 36.7 51.3 66 80.7 95.3 thinking distance, ft 27 38 49 60 71 thinking time, s 0.736 0.740 0.742 0.744 0.745 The times can be summarized as 0.742 s 0.7%. Their near constancy means that the car can be modeled as traveling at constant speed, and that the reaction time is 0.742 s . The proportionality could also be displayed on a graph of thinking distance versus speed, to which a straight line through the origin could be tted, passing very close to all the points. The line slope is the reaction time. (c) and (d) According to vf 2 = vi 2 + 2a(xf xi ) 0 = vi 2 2|a| (braking distance) |a| = vi 2 2 (braking distance), we test for proportionality of speed squared to distance by dividing each squared speed by the given braking distance. The rst is (36.7 fts)2 34 ft = 39.5 fts2 = 2a. Similarly, initial speed, fts 36.7 51.3 66 80.7 95.3 braking distance, ft 34 67 110 165 231 proportionality constant, fts2 39.5 39.3 39.6 39.4 39.3 The constancy within experimental uncertainty of the last line indicates that the square of initial speed is indeed proportional to the braking distance, so that the braking acceleration is constant . This could be displayed also by graphing initial speed squared versus braking distance. A straight line ts the points convincingly and its slope is 2a = 39.5 fts2 , indicating that the braking acceleration is 19.7 fts2 = 6.01 ms2 . ISMV1_5103_02.indd 41ISMV1_5103_02.indd 41 11/2/06 11:16:34 AM11/2/06 11:16:34 AM
43. 43. 42 Chapter 2 P2.62 Time Height h t v midpt time t (s) h (m) (m) (s) (ms) t (s) 0.00 5.00 0.75 0.25 3.00 0.13 0.25 5.75 0.65 0.25 2.60 0.38 0.50 6.40 0.54 0.25 2.16 0.63 0.75 6.94 0.44 0.25 1.76 0.88 1.00 7.38 0.34 0.25 1.36 1.13 1.25 7.72 0.24 0.25 0.96 1.38 1.50 7.96 0.14 0.25 0.56 1.63 1.75 8.10 0.03 0.25 0.12 1.88 2.00 8.13 0.06 0.25 0.24 2.13 2.25 8.07 0.17 0.25 0.68 2.38 2.50 7.90 0.28 0.25 1.12 2.63 2.75 7.62 0.37 0.25 1.48 2.88 3.00 7.25 0.48 0.25 1.92 3.13 3.25 6.77 0.57 0.25 2.28 3.38 3.50 6.20 0.68 0.25 2.72 3.63 3.75 5.52 0.79 0.25 3.16 3.88 4.00 4.73 0.88 0.25 3.52 4.13 4.25 3.85 0.99 0.25 3.96 4.38 4.50 2.86 1.09 0.25 4.36 4.63 4.75 1.77 1.19 0.25 4.76 4.88 5.00 0.58 TABLE P2.62 The very convincing t of a single straight line to the points in the graph of velocity versus time indicates that the rock does fall with constant acceleration. The acceleration is the slope of line: aa gv = =1 63 1 63. .m s m s downward2 2 P2.63 The distance x and y are always related by x y L2 2 2 + = . Differentiating through this equation with respect to time, we have 2 2 0x dx dt y dy dt + = Now dy dt is vB , the unknown velocity of B; and dx dt = v. From the equation resulting from differentiation, we have dy dt x y dx dt x y = = ( )v . But y x = tan so v vB = 1 tan . When = 60 0. , v v v vB = = = tan . . 60 0 3 3 0 577 . FIG. P2.62 FIG. P2.63 B O y x L v x y A ISMV1_5103_02.indd 42ISMV1_5103_02.indd 42 10/27/06 2:45:49 PM10/27/06 2:45:49 PM
44. 44. Motion in One Dimension 43 ANSWERS TO EVEN PROBLEMS P2.2 (a) 2.30 ms (b) 16.1 ms (c) 11.5 ms P2.4 (a) 50 0. m s (b) 41 0. m s P2.6 (a) 27 0. m (b) 27 0 18 0 3 00 2 . . .m m s m s2 + ( ) + ( )( ) t t (c) 18 0. m s P2.8 (a) 5.0 ms (b) 2.5 ms (c) 0 (d) 5.0 ms P2.10 1.34 104 ms2 P2.12 (a) see the solution (b) 1 60. m s2 ; 0 800. m s2 P2.14 see the solution P2.16 (a) 1.3 ms2 (b) 2.0 ms2 at 3 s (c) at t = 6 s and for t10 s (d) 1.5 m s2 at 8 s P2.18 see the solution P2.20 (a) 6 61. m s (b) 0 448. m s2 P2.22 (a) 1.88 km; (b) 1.46 km (c) see the solution (d) (i) x t1 2 1 67= ( ). m s2 (ii) x t2 50 375= ( ) m s m (iii) x t t3 2 250 2 5 4 375= ( ) ( ) m s m s m2 . (e) 37 5. m s P2.24 (a) 3.75 s after release (b) 5.50 cms (c) 0.604 s and 6.90 s (d) 13.3 cm and 47.9 cm (e) The cars are initially moving toward each other, so they soon share the same position x when their speeds are quite different, giving one answer to (c) that is not an answer to (a). The rst car slows down in its motion to the left, turns around, and starts to move toward the right, slowly at rst and gaining speed steadily. At a particular moment its speed will be equal to the constant rightward speed of the second car. The distance between them will at that moment be at its maximum value. The distance between the cars will be far from zero, as the accelerating car will be far to the left of the steadily moving car. Thus the answer to (a) is not an answer to (c). Eventually the accelerating car will catch up to the steadily- coasting car, whizzing past at higher speed than it has ever had before, and giving another answer to (c) that is not an answer to (a). A graph of x versus t for the two cars shows a parabola originally sloping down and then curving upward, intersecting twice with an upward-sloping straight line. The parabola and straight line are running parallel, with equal slopes, at just one point in between their intersections. P2.26 x x t a tf i xf x = v 1 2 2 ; 3 10. m s P2.28 (a) 2.56 m (b) 3.00 ms P2.30 (a) vc tm (c) vc to 2 (d) vc to (e) The minimum displacement can be attained by having the servo motor on all the time. The maximum displacement cannot be attained because the acceleration must be nite. P2.32 Yes; 212 m; 11.4 s P2.34 31.0 s ISMV1_5103_02.indd 43ISMV1_5103_02.indd 43 10/27/06 2:45:51 PM10/27/06 2:45:51 PM
45. 45. 44 Chapter 2 P2.36 (a) 0.049 0 m, 0.196 m, 0.441 m, with the negative signs all indicating downward (b) 0.980 ms, 1.96 m/s, 2.94 m s P2.38 1.79 s P2.40 No; see the solution P2.42 (a) Yes (b) 3.69 m s (c) 2.39 ms (d) +2.39 ms does not agree with the magnitude of 3.71 ms. The upward-moving rock spends more time in ight, so its speed change is larger. P2.44 7.96 s P2.46 (a) and (b) see the solution (c) 4 m s2 (d) 34 m (e) 28 m P2.48 (a) a = (10.0 107 m s3 )t + 3.00 105 m s2 ; x = (1.67 107 m/s3 )t3 + (1.50 105 ms2 )t2 (b) 3.00 103 s (c) 450 ms (d) 0.900 m P2.50 (a) Acela steadily cruises out of the city center at 45 mi/h. In less than a minute it smoothly speeds up to 150 mih; then its speed is nudged up to 170 mih. Next it smoothly slows to a very low speed, which it maintains as it rolls into a railroad yard. When it stops, it immediately begins backing up and smoothly speeds up to 50 mih in reverse, all in less than seven minutes after it started. (b) 2.2 mihs = 0.98 ms2 (c) 6.7 mi. P2.52 vxi xt a t+ 1 2 2 ; displacements agree P2.54 (a) ty = 0.6 s + v0 s2 4.8 m + 22 mv0 (b) 5.02 s (c) 4.89 s (d) 5.57 s (e) 6.69 s (f) ty (g) ty (h) ty decreases steeply from an innite value at v0 = 0, goes through a rather at minimum, and then diverges to innity as v0 increases without bound. For a very slowly moving car entering the intersection and not allowed to speed up, a very long time is required to get across the intersection. A very fast-moving car requires a very long time to slow down at the constant accelera- tion we have assumed. (i) at v0 = 10.3 ms, (j) ty = 4.88 s. P2.56 (a) 35.9 m (b) 4.04 s (c) 45.8 m (d) 22.6 ms P2.58 ~103 ms2 P2.60 (a) 26.4 m (b) 6.82% P2.62 see the solution; ax = 1 63. m s2 ISMV1_5103_02.indd 44ISMV1_5103_02.indd 44 10/27/06 2:45:52 PM10/27/06 2:45:52 PM
46. 46. 3 Vectors CHAPTER OUTLINE 3.1 Coordinate Systems 3.2 Vector and Scalar Quantities 3.3 Some Properties of Vectors 3.4 Components of a Vector and Unit Vectors ANSWERS TO QUESTIONS Q3.1 Only force and velocity are vectors. None of the other quantities requires a direction to be described. The answers are (a) yes (b) no (c) no (d) no (e) no (f) yes (g) no. Q3.2 The books displacement is zero, as it ends up at the point from which it started. The distance traveled is 6.0 meters. *Q3.3 The vector 2D1 will be twice as long as D1 and in the opposite direction, namely northeast. Adding D2, which is about equally long and southwest, we get a sum that is still longer and due east, choice (a). *Q3.4 The magnitudes of the vectors being added are constant, and we are considering the magnitude onlynot the directionof the resultant. So we need look only at the angle between the vectors being added in each case. The smaller this angle, the larger the resultant magnitude. Thus the ranking is c = eadb. *Q3.5 (a) leftward: negative. (b) upward: positive (c) rightward: positive (d) downward: negative (e) Depending on the signs and angles of A and B, the sum could be in any quadrant. (f) Now A will be in the fourth quadrant, so +A B will be in the fourth quadrant. *Q3.6 (i) The magnitude is 10 102 2 + m s , answer (f). (ii) Having no y component means answer (a). *Q3.7 The vertical component is opposite the 30 angle, so sin 30 = (vertical component)/50 m and the answer is (h). *Q3.8 Take the difference of the coordinates of the ends of the vector. Final rst means head end rst. (i) 4 2 = 6 cm, answer (j) (ii) 1 (2) = 3 cm, answer (c) Q3.9 (i) If the direction-angle of A is between 180 degrees and 270 degrees, its components are both negative: answer (c). If a vector is in the second quadrant or the fourth quadrant, its components have opposite signs: answer (b) or (d). Q3.10 Vectors A and B are perpendicular to each other. Q3.11 No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude. Q3.12 Addition of a vector to a scalar is not dened. Think of numbers of apples and of clouds. 45 13794_03_ch03_p045-064.indd 4513794_03_ch03_p045-064.indd 45 11/28/06 4:40:06 PM11/28/06 4:40:06 PM
47. 47. SOLUTIONS TO PROBLEMS Section 3.1 Coordinate Systems P3.1 x r= = ( ) = ( ) ( ) = cos cos . 5 50 240 5 50 0 5 2 7. m . m . 55 m y r= = ( ) = ( ) ( ) = sin sin 5 50 240 5 50 0 8 4. m . m . 66 ..76 m P3.2 (a) x r= cos and y r= sin, therefore x1 2 50 30 0= ( ). m .cos , y1 2 50 30 0= ( ). m .sin , and x y1 1 2 17 1 25, . , . m( ) = ( ) x2 3 80 120= ( ) . cosm , y2 3 80 120= ( ) . sinm , and x y2 2 1 90 3 29, . , . m( ) = ( ) (b) d x y= + = + =( ) ( ) . . . 2 2 2 2 4 07 2 04 4 55m m P3.3 The x distance out to the y is 2.00 m and the y distance up to the y is 1.00 m. (a) We can use the Pythagorean theorem to nd the distance from the origin to the y. distance m m m2 = + = ( ) + ( ) = =x y2 2 2 2 2 00 1 00 5 00. . . 22 24. m (b) = = tan .1 1 2 26 6 ; r = 2 24 26 6. , .m P3.4 We have 2 00 30 0. .= r cos r = = 2 00 30 0 2 31 . cos . . and y r= = =sin sin .30 0 2 31 30 0 1 15. . . . P3.5 We have r x y= +2 2 and = tan 1 y x . (a) The radius for this new point is ( ) + = + =x y x y r 2 2 2 2 and its angle is tan =1 180 y x . (b) ( ) ( ) + =2 2 22 2 x y r This point is in the third quadrant if x y,( ) is in the rst quadrant or in the fourth quadrant if x y,( ) is in the second quadrant. It is at an angle of 180 + . (c) ( ) ( )3 3 32 2 x y r+ = This point is in the fourth quadrant if x y,( ) is in the rst quadrant or in the third quadrant if x y,( ) is in the second quadrant. It is at an angle of . 46 Chapter 3 13794_03_ch03_p045-064.indd 4613794_03_ch03_p045-064.indd 46 11/28/06 4:40:07 PM11/28/06 4:40:07 PM
48. 48. Vectors 47 Section 3.2 Vector and Scalar Quantities Section 3.3 Some Properties of Vectors P3.6 =R 310 km at 57 S of W (Scale: 1 unit = 20 km) P3.7 tan . tan . . 35 0 100 100 35 0 70 0 = = ( ) = x x m m m P3.8 Find the resultant F F1 2+ graphically by placing the tail of F2 at the head of F1. The resultant force vector F F1 2+ is of magnitude 9 5. N and at an angle of 57 abovethe axisx . FIG. P3.6 FIG. P3.7 0 1 2 3 N x y 2 1 F1 2+F F F FIG. P3.8 13794_03_ch03_p045-064.indd 4713794_03_ch03_p045-064.indd 47 11/28/06 4:40:08 PM11/28/06 4:40:08 PM
49. 49. 48 Chapter 3 P3.9 (a) d i= =10 0 10 0. . m since the displacement is in a straight line from point A to point B. (b) The actual distance skated is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle (ACB). s r= ( ) = = 1 2 2 5 15 7 . m (c) If the circle is complete, d begins and ends at point A. Hence, d = 0 . P3.10 (a) The large majority of people are standing or sitting at this hour. Their instantaneous foot- to-head vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~ 105 m upward . (b) Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, and west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if one-tenth of one percent of the population are on-duty nurses or police ofcers, we estimate the total vector height as ~ . m m10 0 03 10 15 2 ( )+ ( ) ~ 103 m upward . P3.11 To nd these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 0 5unit m= . ) (a) A + B = 5.2 m at 60 (b) A B = 3.0 m at 330 (c) B A = 3.0 m at 150 (d) A 2B = 5.2 m at 300 C B A 5.00 m d FIG. P3.9 FIG. P3.11 13794_03_ch03_p045-064.indd 4813794_03_ch03_p045-064.indd 48 11/28/06 4:40:09 PM11/28/06 4:40:09 PM
50. 50. Vectors 49 P3.12 The three diagrams shown below represent the graphical solutions for the three vector sums: R A B C1 = + + , R B C A2 = + + , and R C B A3 = + + . We observe that R R R1 2 3= = , illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added . 2 B A R1 C C C R 3R A A B B 100 m FIG. P3.12 P3.13 The scale drawing for the graphical solution should be similar to the gure to the right. The magnitude and direction of the nal displacement from the starting point are obtained by measuring d and on the drawing and applying the scale factor used in making the drawing. The results should be d = = 420 3ft and . Section 3.4 Components of a Vector and Unit Vectors *P3.14 We assume the oor is level. Take the x axis in the direction of the rst displacement. If both of the 90 turns are to the right or both to the left , the displacements add like 40 0 15 0 20 0 20 0 15 0. . . . .m m mi i i+ = +j j( )m to give (a) displacement magnitude (202 + 152 )12 m = 25.0 m at (b) tan1 (1520) = 36.9 . If one turn is right and the other is left , the displacements add like 40 0 15 0 20 0 60 0 15 0. . . . .m m mi i+ + = +j i j( )m to give (a) displacement magnitude (602 + 152 )12 m = 61.8 m at (b) tan1 (1560) = 14.0. Just two answers are possible. FIG. P3.13 13794_03_ch03_p045-064.indd 4913794_03_ch03_p045-064.indd 49 11/28/06 4:40:10 PM11/28/06 4:40:10 PM
51. 51. 50 Chapter 3 P3.15 A A A A A x y x y = = = + = ( ) + ( ) = 25 0 40 0 25 0 40 0 42 2 2 2 . . . . 77 2. units We observe that tan = A A y x . So = = = ( ) = tan tan . . tan .1 140 0 25 0 1 60 5 A A y x 88 0. . The diagram shows that the angle from the +x axis can be found by subtracting from 180: = =180 58 122 . P3.16 The person would have to walk 3 10 1 31. 25.0 km northsin .( ) = , and 3 10 25 0 2 81. . km eastcos .( ) = . *P3.17 Let v represent the speed of the camper. The northward component of its velocity is v cos 8.5. To avoid crowding the minivan we require v cos 8.5 28 ms. We can satisfy this requirement simply by taking v (28 ms)cos 8.5 = 28.3 ms. P3.18 (a) Her net x (east-west) displacement is + + = +3 00 0 6 00 3 00. . . blocks, while her net y (north-south) displacement is 0 4 00 0 4 00+ + = +. . blocks. The magnitude of the resultant displacement is R x y= ( ) + ( ) = ( ) + ( ) =net net 2 2 2 2 3 00 4 00 5 00. . . blocks and the angle the resultant makes with the x axis (eastward direction) is = = ( ) = tan . . tan . .1 14 00 3 00 1 33 53 1 . The resultant displacement is then 5 00 53 1. .blocks at N of E . (b) The total distance traveled is 3 00 4 00 6 00 13 0. . . .+ + = blocks . P3.19 x r= cos and y r= sin, therefore: (a) x = 12 8 150. cos , y = 12 8 150. sin , and x y, . . m( ) = +( )111 6 40 i j (b) x = 3 30 60 0. cos . , y = 3 30 60 0. sin . , and x y, cm( ) = +( )1 65 2 86. . i j (c) x = 22 0 215. cos , y = 22 0 215. sin , and x y, in( ) = ( )18 0 12 6. . i j P3.20 x d= = ( ) ( ) = cos cos 50 0 120 25 0. m . m y d= = ( ) ( ) = = ( ) sin sin . . 50 0 120 43 3 25 0 . m m md . i j+ ( )43 3 m FIG. P3.15 13794_03_ch03_p045-064.indd 5013794_03_ch03_p045-064.indd 50 11/28/06 4:40:10 PM11/28/06 4:40:10 PM
52. 52. Vectors 51 P3.21 Let +x be East and +y be North. x y = = = = 250 125 30 358 75 125 30 150 + + cos sin m 112 5 358 12 5 358 2 2 2 2 . m md x y= ( ) + ( ) = ( ) + ( ) = . taan . = ( ) ( ) = = = = y x 12 5 358 0 0349 2 00 35 . . d 88 2 00m at S of E. P3.22 The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: d d dDC east DA east AC east .= =+ 730 5 00 560cos s iin 21 0 527. miles. DC north DA north AC = =d d d+ nnorth . . miles= =730 5 00 560 21 0 586sin cos + By the Pythagorean theorem, d d d= + =( ) ( )DC east DC north mi2 2 788 . Then tan = = d d DC north DC east .111 and = 48 0. . Thus, Chicago is 788 miles at 48.0 northeast of Dallas . P3.23 We have B R A= : A A x y = = = = 150 75 0 150 120 130 cos120 cm cm . sin RR R x y = = = = 140 35 0 115 140 35 0 80 3 cos . sin . . cm cmm Therefore, B i j i j= ( )[ ] + [ ] = (115 75 80 3 130 190 49 7 . . )) = + = = cm cmB 190 49 7 196 49 7 190 2 2 1 . tan . = 14 7. . P3.24 (a) See gure to the right. (b) C A B i j i j i= + = + + =2 00 6 00 3 00 2 00 5 00. . . . . ++ 4 00. j C = + = 25 0 16 0 6 401 . . tan .at 4 5 at 38.7 D A B i j i j= = + + = 2 00 6 00 3 00 2 00 1 00. . . . . ii j+ 8 00. D = ( ) + ( ) 1 00 8 00 8 00 1 00 2 2 1 . . tan . . at D = ( ) =8 06 180 82 9 8 06 97 2. . . .at at x FIG. P3.23 FIG. P3.24 13794_03_ch03_p045-064.indd 5113794_03_ch03_p045-064.indd 51 11/28/06 4:40:11 PM11/28/06 4:40:11 PM
53. 53. 52 Chapter 3 P3.25 (a) A B i j i j i j+( )= ( )+ ( )= 3 2 4 2 6 (b) A B i j i j i j( )= ( ) ( )= +3 2 4 4 2 (c) A B+ = + =2 6 6 322 2 . (d) A B = + =4 2 4 472 2 . (e) A B+ = = =tan .1 6 2 71 6 288 A B = =tan .1 2 4 26 6 *P3.26 We take the x axis along the slope uphill. Students, get used to this choice! The y axis is perpen- dicular to the slope, at 35 to the vertical. Then the displacement of the snow makes an angle of 90 35 20 = 35 with the x axis. (a) Its component parallel to the surface is 5 m cos 35 = 4.10 m toward the top of the hill . (b) Its component perpendicular to the surface is 5 m sin 35 = 2.87 m . P3.27 d j1 3 50= ( ). m d i j i2 8 20 45 0 8 20 45 0 5 80 5= + = +. cos . . sin . . . 880j( ) m d i3 15 0= ( ). m R d d d i j= + + = +( ) + ( )1 2 3 15 0 5 80 5 80 3 50. . . . == +( )9 20 2 30. . i j m (or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is R = + = ( ) + ( ) =R Rx y 2 2 2 2 9 20 2 30 9 48. . . m . The direction is = =arctan . . 2 30 9 20 166 . P3.28 Refer to the sketch R A C= + + = + = B i j i i 10 0 15 0 50 0 40 0 15 . . . . .. . . . / 0 40 0 15 0 42 7 2 2 1 2 j R = ( ) + ( ) = yards P3.29 East North x y 0 m 4.00 m 1.41 1.41 0.500 0.866 +0.914 4.55 R = + =x y y x 2 2 4 64at tan m at 78.6 N of1 ( ) . E A = 0.10 B = 15 0. C = 50 0. R FIG. P3.28 13794_03_ch03_p045-064.indd 5213794_03_ch03_p045-064.indd 52 11/28/06 4:40:12 PM11/28/06 4:40:12 PM
54. 54. Vectors 53 P3.30 A i j= +8 70 15 0. . and B i j= 13 2 6 60. . A B C + =3 0: 3 21 9 21 6 7 30 7 20 C B A i j C i j = = = . . . . or Cx = 7 30. cm ; Cy = 7 20. cm P3.31 (a) F F F F i = + = ( ) + ( ) 1 2 120 60 0 1