2.1: a) During the later 4.75-s interval, the rocket moves a distance m63m1000.1 3  , and so the magnitude of the average velocity is .sm197 s754 m63m1000.1 3   . b) sm169 s5.90 m1000.1 3   2.2: a) The magnitude of the average velocity on the return flight is .sm42.4 )das400,86()da5.13( )m105150( 3   The direction has been defined to be the –x-direction ). ˆ ( i b) Because the bird ends up at the starting point, the average velocity for the round trip is 0. 2.3: Although the distance could be found, the intermediate calculation can be avoided by considering that the time will be inversely proportional to the speed, and the extra time will be min.701 hrkm70 hrkm105 min)140(           2.4: The eastward run takes )sm5.0m200( = 40.0 s and the westward run takes )sm4.0m280( = 70.0 s. a) (200 m + 280 m)/(40.0 s + 70.0 s) = sm4.4 to two significant figures. b) The net displacement is 80 m west, so the average velocity is )s110.0m80( = sm73.0 in the –x-direction ). ˆ ( i 2.5: In time t the fast runner has traveled 200 m farther than the slow runner: s286so,s)m(6.20m200s)m50.5(  ttt . Fast runner has run m.1770)sm20.6( t Slow runner has run m.1570)sm50.5( t 2.6: The s-waves travel slower, so they arrive 33 s after the p-waves. km250 s33 5.65.3 s33 s33 s km s km ps ps      d dd v d v d v d tvtd tt 2.7: a) The van will travel 480 m for the first 60 s and 1200 m for the next 60 s, for a total distance of 1680 m in 120 s and an average speed of .sm0.14 b) The first stage of the journey takes s30 sm8.0 m240  and the second stage of the journey takes ,s12s)m20m240(  so the time for the 480-m trip is 42 s, for an average speed of .sm11.4 c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same. 2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) sm80.2 s2.00 0m60.5   b) sm2.5 s4.00 0m8.20   c) sm6.7 s2.00 m5.60m8.20   2.9: a) At 0 ,0 11  xt , so Eq (2.2) gives .sm0.12 s)0.10( s)0.10)(sm120.0(s)0.10)(sm4.2( 3 3 22 2 2 av    t x v b) From Eq. (2.3), the instantaneous velocity as a function of time is ,)sm360.0()sm80.4(32 2 3 22 ttctbtv x  so i) ,0)0(  x v ii) ,sm0.15s)0.5)(sm360.0()s0.5)(sm80.4()s0.5( 2 32  x v and iii) .sm0.12s)0.10)(sm360.0()s0.10)(sm80.4()s0.10( 2 32  x v c) The car is at rest when 0 x v . Therefore 0)sm360.0()sm80.4( 2 32  tt . The only time after 0t when the car is at rest is s3.13 3 2 sm360.0 sm80.4 t 2.10: a) IV: The curve is horizontal; this corresponds to the time when she stops. b) I: This is the time when the curve is most nearly straight and tilted upward (indicating postive velocity). c) V: Here the curve is plainly straight, tilted downward (negative velocity). d) II: The curve has a postive slope that is increasing. e) III: The curve is still tilted upward (positive slope and positive velocity), but becoming less so. 2.11: Time (s) 0 2 4 6 8 10 12 14 16 Acceleration (m/s 2 ) 0 1 2 2 3 1.5 1.5 0 a) The acceleration is not constant, but is approximately constant between the times s4t and s.8t 2.12: The cruising speed of the car is 60 hrkm = 16.7 sm . a) 2 s10 sm7.16 sm7.1 (to two significant figures). b) 2 s10 sm7.160 sm7.1  c) No change in speed, so the acceleration is zero. d) The final speed is the same as the initial speed, so the average acceleration is zero. 2.13: a) The plot of the velocity seems to be the most curved upward near t = 5 s. b) The only negative acceleration (downward-sloping part of the plot) is between t = 30 s and t = 40 s. c) At t = 20 s, the plot is level, and in Exercise 2.12 the car is said to be cruising at constant speed, and so the acceleration is zero. d) The plot is very nearly a straight line, and the acceleration is that found in part (b) of Exercise 2.12, .sm7.1 2  e) 2.14: (a) The displacement vector is:   kjir ˆ )sm0.3()sm0.7( ˆ )sm0.10( ˆ )sm0.5()( 22 ttttt   The velocity vector is the time derivative of the displacement vector: kji r ˆ ))sm0.3(2sm0.7( ˆ )sm0.10( ˆ )sm0.5( )( 2 t dt td   and the acceleration vector is the time derivative of the velocity vector: k r ˆ sm0.6 )( 2 2 2  dt td  At t = 5.0 s:   kjir ˆ )s0.25)(sm0.3()s0.5)(sm0.7( ˆ )s0.5)(sm0.10( ˆ s)05)(sm0.5()( 2 2  .t  kji ˆ )m0.40( ˆ )m0.50( ˆ )m0.25(  kji kji r ˆ )sm0.23( ˆ )sm0.10( ˆ )sm0.5( ˆ ))s0.5)(sm0.6(sm0.7(( ˆ )sm0.10( ˆ )sm0.5( )( 2   dt td  k r ˆ sm0.6 )( 2 2 2  dt td  (b) The velocity in both the x- and the y-directions is constant and nonzero; thus the overall velocity can never be zero. (c) The object's acceleration is constant, since t does not appear in the acceleration vector. 2.15: t dt dx v x )scm125.0(scm00.2 2  2 scm125.0 dt dv a x x a) .scm125.0,scm00.2cm,0.50,0At 2  xx avxt b) s.0.16:for solveand0Set  ttv x c) Set x = 50.0 cm and solve for t. This gives s.0.32 and0  tt The turtle returns to the starting point after 32.0 s. d) Turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm. s.8.25ands20.6:for solveandcm0.60Set  tttx .scm23.1s,8.25At .scm23.1s,20.6At   x x vt vt Set cm0.40x and solve for s4.36: tt (other root to the quadratic equation is negative and hence nonphysical). .scm55.2 s,4.36At  x vt e) 2.16: Use of Eq. (2.5), with t = 10 s in all cases, a)        2 m/s0.1s10/m/s0.15m/s0.5  b)        2 m/s0.1s10/m/s0.5m/s0.15  c)        2 m/s0.3s10/m/s0.15m/s0.15  . In all cases, the negative acceleration indicates an acceleration to the left. 2.17: a) Assuming the car comes to rest from 65 mph (29 m/s) in 4 seconds, .sm25.7)s4()0sm29( 2  x a b) Since the car is coming to a stop, the acceleration is in the direction opposite to the velocity. If the velocity is in the positive direction, the acceleration is negative; if the velocity is in the negative direction, the acceleration is positive. 2.18: a) The velocity at t = 0 is (3.00 sm ) + (0.100 3 sm ) (0) = 3.00 sm , and the velocity at t = 5.00 s is (3.00 sm ) + (0.100 3 sm ) (5.00 s) 2 = 5.50 sm , so Eq. (2.4) gives the average acceleration as 2 sm50. )s00.5( )sm00.3()sm50.5(   . b) The instantaneous acceleration is obtained by using Eq. (2.5), .)sm2.0(2 3 tt dt dv a x   Then, i) at t = 0, a x = (0.2 3 sm ) (0) = 0, and ii) at t = 5.00 s, a x = (0.2 3 sm ) (5.00 s) = 1.0 2 sm . 2.19: a) b) 2.20: a) The bumper’s velocity and acceleration are given as functions of time by 5 62 )sm600.0()sm60.9( tt dt dx v x  .)sm000.3()sm60.9( 4 62 t dt dv a x  There are two times at which v = 0 (three if negative times are considered), given by t = 0 and t 4 = 16 s 4 . At t = 0, x = 2.17 m and a x = 9.60 sm 2 . When t 4 = 16 s 4 , x = (2.17 m) + (4.80 sm 2 ) )s16( 4 – (0.100) 6 sm )(16 s 4 ) 3/2 = 14.97 m, a x = (9.60 sm 2 ) – (3.000 6 sm )(16 s 4 ) = –38.4 sm 2 . b) 2.21: a) Equating Equations (2.9) and (2.10) and solving for v 0 , .sm00.5sm0.15 s007 m)70(2 )(2 0 0    . v t xx v xx b) The above result for v 0x may be used to find ,sm43.1 s00.7 sm00.5sm0.15 2 0      t vv a xx x or the intermediate calculation can be avoided by combining Eqs. (2.8) and (2.12) to eliminate v 0x and solving for a x , .sm43.1 s)007( m0.70 s00.7 sm0.15 22 2 22 0                   .t xx t v a x x 2.22: a) The acceleration is found from Eq. (2.13), which x v 0 = 0;       ,sm0.32 )ft307(2 )hrmi173( )(2 2 ft3.281 m1 2 hrmi1 sm4470.0 0 2    xx v a x x where the conversions are from Appendix E. b) The time can be found from the above acceleration,   .s42.2 sm0.32 )hrmi173( 2 hrmi1 sm4470.0  x x a v t The intermediate calculation may be avoided by using Eq. (2.14), again with v 0x = 0,      .s42.2 )hrmi173( ft307(2 )(2 hrmi1 sm4470.0 ft3.281 m1 0    x v xx t 2.23: From Eq. (2.13), with   ,0 Taking.,0 0max 2 0 2 0   xaav xx v xx m.70.1 )sm250(2 ))hrkms)(3.6mhr)(1km105(( 2 2 2 max 2 0  a v x x 2.24: In Eq. (2.14), with x – x 0 being the length of the runway, and v 0x = 0 (the plane starts from rest), .sm0.7022 s8 m280 0   t xx x v 2.25: a) From Eq. (2.13), with ,0 0  x v .sm67.1 m)120(2 )sm20( )(2 2 2 0 2    xx v a x x b) Using Eq. (2.14), s.12)sm20(m)120(2)(2 0  vxxt c) m.240)sm20)(s12(  2.26: a) x 0 < 0, v 0x < 0, a x < 0 b) x 0 > 0, v 0x < 0, a x > 0 c) x 0 > 0, v 0x > 0, a x < 0 [...]... purposes of graphing: 249 m s 2 2.47: a) (224 m s) (0.9 s)  249 m s 2 b) 9.80 m s 2  25.4 c) The most direct way to find the distance is vave t  ((224 m s) 2)(0.9 s)  101 m 2 2 d) (283 m s) (1.40 s)  202 m s but 40 g  392 m s , so the figures are not consistent 2 2.48: a) From Eq (2.8), solving for t gives (40.0 m s – 20.0 m s )/9.80 m s = 2.04 s b) Again from Eq (2.8), 40.0 m s  (20.0 m s) 9.80 . the length of the runway, and v 0x = 0 (the plane starts from rest), .sm0. 7022 s8 m280 0   t xx x v 2.25: a) From Eq. (2.13), with ,0 0  x v .sm67.1. acceleration and deceleration is 900 s, so the total time required for the trip is s 102. 22 4  , about 6.2 hr. 2.34: After the initial acceleration, the train has