Physics exercises_solution:Chapter 02

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Physics exercises_solution:Chapter 02

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 02

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  1. 2.1: a) During the later 4.75-s interval, the rocket moves a distance 1.00  10 3 m  63 m , and so the magnitude of the average velocity is 1.00  10 3 m  63 m  197 m s . 4.75 s b) 1.00 10 3 m 5.90 s  169 m s 2.2: a) The magnitude of the average velocity on the return flight is (5150  10 3 m)  4.42 m s . (13.5 da ) (86, 400 s da ) ˆ The direction has been defined to be the –x-direction ( 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  105 km hr  (140 min)   70 km hr  1  70 min.    2.4: The eastward run takes (200 m 5.0 m s) = 40.0 s and the westward run takes (280 m 4.0 m s) = 70.0 s. a) (200 m + 280 m)/(40.0 s + 70.0 s) = 4.4 m s to two significant figures. b) The net displacement is 80 m west, so the average velocity is ˆ (80 m 110.0 s) = 0.73 m s in the –x-direction ( i ). 2.5: In time t the fast runner has traveled 200 m farther than the slow runner: (5.50 m s)t  200 m  (6.20 m s)t , so t  286 s . Fast runner has run (6.20 m s)t  1770 m. Slow runner has run (5.50 m s)t  1570 m.
  2. 2.6: The s-waves travel slower, so they arrive 33 s after the p-waves. t s  t p  33 s d d  vt  t  v d d   33 s vs v p d d   33 s 3.5 km 6.5 km s s d  250 km 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 14.0 m s . b) The first stage of the journey takes 8.0 mms  30 s and the second stage of the journey takes 240 (240 m 20 m s)  12 s , so the time for the 480-m trip is 42 s, for an average speed of 11.4 m s . 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) 5.60 m  0 20.8 m  5.60 m 2.00 s  2.80 m s b) 4.00 s  5.2 m s c) 20.8 m 0 2.00 s  7.6 m s 2.9: a) At t1  0, x1  0 , so Eq (2.2) gives 3 x 2 (2.4 m s 2 )(10.0 s) 2  (0.120 m s )(10.0 s) 3 vav    12.0 m s . t2 (10.0 s) b) From Eq. (2.3), the instantaneous velocity as a function of time is 3 v x  2bt  3ct 2  (4.80 m s 2 )t  (0.360 m s )t 2 , so i) v x (0)  0, 2 3 ii) v x (5.0 s)  (4.80 m s )(5.0 s)  (0.360 m s )(5.0 s) 2  15.0 m s , 2 3 and iii) v x (10.0 s)  (4.80 m s )(10.0 s)  (0.360 m s )(10.0 s) 2  12.0 m s . 2 3 c) The car is at rest when v x  0 . Therefore (4.80 m s )t  (0.360 m s )t 2  0 . The 4.80 m s 2 only time after t  0 when the car is at rest is t  0.360 m s 3  13.3 s
  3. 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/s2) 0 1 2 2 3 1.5 1.5 0 a) The acceleration is not constant, but is approximately constant between the times t  4 s and t  8 s. 2 2.12: The cruising speed of the car is 60 km hr = 16.7 m s . a) 16.7 m s 10 s  1.7 m s (to two significant figures). b) 0 16 .7sm s  1.7 m s 2 c) No change in speed, so the acceleration 10 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,  1.7 m s 2 . e)
  4. 2.14: (a) The displacement vector is:  ˆ j   r (t )  (5.0 m s)ti  (10.0 m s)tˆ  (7.0 m s)t  (3.0 m s 2 )t 2 k ˆ The velocity vector is the time derivative of the displacement vector:  d r (t ) ˆ  (5.0 m s)i  (10.0 m s) ˆ  (7.0 m s  2(3.0 m s 2 )t ) k j ˆ dt and the acceleration vector is the time derivative of the velocity vector:  d 2 r (t ) ˆ  6.0 m s 2 k dt 2 At t = 5.0 s:  ˆ j  2  r (t )  (5.0 m s)(5.0 s)i  (10.0 m s)(5.0 s) ˆ  (7.0 m s)(5.0 s)  (3.0 m s )(25.0 s 2 ) k ˆ  (25.0 m)i  (50.0 m) ˆ  (40.0 m)k ˆ j ˆ  d r (t ) ˆ  (5.0 m s)i  (10.0 m s) ˆ  ((7.0 m s  (6.0 m s )(5.0 s)) kˆ 2 j dt ˆ  (5.0 m s)i  (10.0 m s) ˆ  (23.0 m s) k j ˆ  d 2 r (t ) 2 ˆ 2  6.0 m s k dt (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.
  5. dx 2.15: vx   2.00 cm s  (0.125 cm s 2 )t dt dv x ax   0.125 cm s 2 dt a) At t  0, x  50.0 cm, v x  2.00 cm s , a x  0.125 cm s 2 . b) Set v x  0 and solve for t : t  16.0 s. c) Set x = 50.0 cm and solve for t. This gives t  0 and t  32.0 s. 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. Set x  60.0 cm and solve for t : t  6.20 s and t  25.8 s. At t  6.20 s, v x  1.23 cm s . At t  25.8 s, v x  1.23 cm s . Set x  40.0 cm and solve for t : t  36.4 s (other root to the quadratic equation is negative and hence nonphysical). At t  36.4 s, v x  2.55 cm s . e) 2.16: Use of Eq. (2.5), with t = 10 s in all cases, a) 5.0 m/s  15.0 m/s / 10 s   1.0 m/s 2 b)  15.0 m/s   5.0 m/s / 10 s   1.0 m/s 2 c)  15.0 m/s   15.0 m/s / 10 s   3.0 m/s 2 . 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, a x  (29 m s  0) (4 s)  7.25 m s 2 . 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.
  6. 2.18: a) The velocity at t = 0 is (3.00 m s ) + (0.100 m s 3 ) (0) = 3.00 m s , and the velocity at t = 5.00 s is (3.00 m s ) + (0.100 m s 3 ) (5.00 s)2 = 5.50 m s , so Eq. (2.4) gives the average acceleration as (5.50 m s)  (3.00 m s) 2  .50 m s . (5.00 s) b) The instantaneous acceleration is obtained by using Eq. (2.5), dv  2 t  (0.2 m s )t. 3 ax  dt 3 Then, i) at t = 0, ax = (0.2 m s ) (0) = 0, and 3 2 ii) at t = 5.00 s, ax = (0.2 m s ) (5.00 s) = 1.0 m s .
  7. 2.19: a) b)
  8. 2.20: a) The bumper’s velocity and acceleration are given as functions of time by dx 2 6 vx   (9.60 m s )t  (0.600 m s )t 5 dt dv 2 6 ax   (9.60 m s )  (3.000 m s )t 4 . dt There are two times at which v = 0 (three if negative times are considered), given by t = 0 and t4 = 16 s4. At t = 0, x = 2.17 m and ax = 9.60 m s 2. When t4 = 16 s4, x = (2.17 m) + (4.80 m s 2) (16 s 4 ) – (0.100) m s )(16 s4)3/2 = 14.97 m, 6 ax = (9.60 m s 2) – (3.000 m s )(16 s4) = –38.4 m s 2. 6 b) 2.21: a) Equating Equations (2.9) and (2.10) and solving for v0, 2( x  x0 ) 2(70 m) v0 x   vx   15.0 m s  5.00 m s . t 7.00 s b) The above result for v0x may be used to find v v 15.0 m s  5.00 m s 2 ax  x 0x   1.43 m s , t 7.00 s or the intermediate calculation can be avoided by combining Eqs. (2.8) and (2.12) to eliminate v0x and solving for ax, v x  x   15.0 m s 70.0 m  a x  2 x  2 0   2  7.00 s  (7.00 s) 2   1.43 m s . 2   t t   
  9. 2.22: a) The acceleration is found from Eq. (2.13), which v0 x = 0; ax  2 vx    (173 mi hr ) 0.1 mi hr s 4470 m  2 2  32.0 m s , 2( x  x0 ) 2(307 ft )  3.281 ft  1m where the conversions are from Appendix E. b) The time can be found from the above acceleration, t  0.4470 m s v x (173 mi hr ) 1 mi hr    2.42 s. 2 ax 32.0 m s The intermediate calculation may be avoided by using Eq. (2.14), again with v0x = 0, 2( x  x0 ) 2(307 ft  3.281 ft  1m t   2.42 s. vx  (173 mi hr ) 0.1 mi hr s 4470 m  2 v0 2.23: From Eq. (2.13), with v x  0, a x  2  x  x0   amax . Taking x0  0, 2 v0 x ((105 km hr)(1 m s)(3.6 km hr )) 2 x   1.70 m. 2a max 2(250 m s 2 ) 2.24: In Eq. (2.14), with x – x0 being the length of the runway, and v0x = 0 (the plane starts from rest), v x  2 x tx0  2 280sm  70.0 m s . 8 2.25: a) From Eq. (2.13), with v0 x  0, 2 vx (20 m s) 2 ax    1.67 m s 2 . 2( x  x0 ) 2(120 m) b) Using Eq. (2.14), t  2( x  x0 ) v  2(120 m) (20 m s)  12 s. c) (12 s)(20 m s)  240 m.
  10. 2.26: a) x0 < 0, v0x < 0, ax < 0 b) x0 > 0, v0x < 0, ax > 0 c) x0 > 0, v0x > 0, ax < 0
  11. 2.27: a) speeding up: x  x0  1320 ft, v0 x  0, t  19.9 s, a x  ? x  x0  v0 x t  1 a x t 2 gives a x  6.67 ft s 2 2 slowing down: x  x0  146 ft, v0 x  88.0 ft s , v x  0, a x  ? v x  v0 x  2a x ( x  x0 ) gives a x  26.5 ft s 2 . 2 2 2 b) x  x0  1320 ft, v0 x  0, a x  6.67 ft s , v x  ? v x  v0 x  2a x ( x  x0 ) gives v x  133 ft s  90.5 mph. 2 2 a x must not be constant. c) v0 x  88.0 ft s, a x  26.5 ft s 2 , v x  0, t  ? v x  v0 x  a x t gives t  3.32 s.
  12. 2.28: a) Interpolating from the graph: At 4.0 s, v  2.7 cm s (to the right) At 7.0 s, v  1.3 cm s (to the left) b) a  slope of v - t graph   8.06.cms / s  1.3 cm s 2 which is constant 0 c) x  area under v-t graph First 4.5 s: x  ARectangle  ATriangle  cm  1  cm   4.5 s  2   4.5 s  6   22.5 cm  s  2  s  From 0 to 7.5 s: The distance is the sum of the magnitudes of the areas. 1  cm  1  cm  d  6 s  8   1.5 s  2   25.5 cm 2  s  2  s  d)
  13. 2.29: a) b) 2.30: a)
  14. 2.31: a) At t = 3 s the graph is horizontal and the acceleration is 0. From t = 5 s to t = 9 s, the acceleration is constant (from the graph) and equal to 45 m s4s20m s  6.3 m s . From 2 t = 9 s to t = 13 s the acceleration is constant and equal to 0 45 m s 2 4s  11.2 m s . b) In the first five seconds, the area under the graph is the area of the rectangle, (20 m)(5 s) = 100 m. Between t = 5 s and t = 9 s, the area under the trapezoid is (1/2)(45 m/s + 20 m/s)(4 s) = 130 m (compare to Eq. (2.14)), and so the total distance in the first 9 s is 230 m. Between t = 9 s and t = 13 s, the area under the triangle is (1 2)(45 m s)(4 s)  90 m , and so the total distance in the first 13 s is 320 m. 2.32: 2.33: a) The maximum speed will be that after the first 10 min (or 600 s), at which time the speed will be 2 (20.0 m s )(900 s)  1.8  10 4 m s  18 km s . b) During the first 15 minutes (and also during the last 15 minutes), the ship will travel (1 2)(18 km s)(900 s)  8100 km , so the distance traveled at non-constant speed is 16,200 km and the fraction of the distance traveled at constant speed is 16,200 km 1  0.958, 384,000 km keeping an extra significant figure. km  c) The time spent at constant speed is 384, 00018 km16, 200 km  2.04  104 s and the time spent s during both the period of acceleration and deceleration is 900 s, so the total time required for the trip is 2.22  10 4 s , about 6.2 hr.
  15. 2.34: After the initial acceleration, the train has traveled 1 2 (1.60 m s )(14.0 s) 2  156.8 m 2 (from Eq. (2.12), with x0 = 0, v0x = 0), and has attained a speed of 2 (1.60 m s )(14.0 s)  22.4 m s . During the 70-second period when the train moves with constant speed, the train travels 22.4 m s 70 s   1568 m. The distance traveled during deceleration is given by Eq. (2.13), with vx  0, v0 x  22.4 m s and a x  3.50 m s 2 , so the train moves a distance x  x 0   (3.50mm/s ) )  71.68 m. The total distance covered in then 156.8 m + 1568 m + 71.7 m 2 22.4 / s 2 2( = 1.8 km. In terms of the initial acceleration a1, the initial acceleration time t1, the time t2 during which the train moves at constant speed and the magnitude a2 of the final acceleration, the total distance xT is given by 1 2 1 (a1t1 ) 2  a1t1  at  xT  a1t1  (a1t1 )t 2    t1  2t 2  1 1 ,  2 2 | a2 |  2  | a2 |   which yields the same result.
  16. 2.35: a) b) From the graph (Fig. (2.35)), the curves for A and B intersect at t = 1 s and t = 3 s. c) d) From Fig. (2.35), the graphs have the same slope at t = 2 s . e) Car A passes car B when they have the same position and the slope of curve A is greater than that of curve B in Fig. (2.30); this is at t = 3 s. f) Car B passes car A when they have the same position and the slope of curve B is greater than that of curve A; this is at t = 1 s.
  17. 2.36: a) The truck’s position as a function of time is given by xT = vTt, with vT being the truck’s constant speed, and the car’s position is given by xC = (1/2) aCt2. Equating the two expressions and dividing by a factor of t (this reflects the fact that the car and the truck are at the same place at t = 0) and solving for t yields 2v 2(20.0 m s) t T  2  12.5 s aC 3.20 m s and at this time xT = xC = 250 m. b) aCt = (3.20 m/s2)(12.5 s) = 40.0 m/s (See Exercise 2.37 for a discussion of why the car’s speed at this time is twice the truck’s speed.) c) d)
  18. 2.37: a) The car and the motorcycle have gone the same distance during the same time, so their average speeds are the same. The car's average speed is its constant speed vC, and for constant acceleration from rest, the motorcycle's speed is always twice its average, or 2vC. b) From the above, the motorcyle's speed will be vC after half the time needed to catch the car. For motion from rest with constant acceleration, the distance traveled is proportional to the square of the time, so for half the time one-fourth of the total distance has been covered, or d 4 . 2.38: a) An initial height of 200 m gives a speed of 60 m s when rounded to one significant figure. This is approximately 200 km/hr or approximately 150 mi hr . (Different values of the approximate height will give different answers; the above may be interpreted as slightly better than order of magnitude answers.) b) Personal experience will vary, but speeds on the order of one or two meters per second are reasonable. c) Air resistance may certainly not be neglected. 2.39: a) From Eq. (2.13), with v y  0 and a y   g , 2 v0 y  2 g ( y  y0 )  2(9.80 m s )(0.440 m)  2.94 m s , which is probably too precise for the speed of a flea; rounding down, the speed is about 2.9 m s . b) The time the flea is rising is the above speed divided by g, and the total time is twice this; symbolically, 2g ( y  y0 ) 2( y  y 0 ) 2(0.440 m) t2 2 2  0.599 s, g g (9.80 m/s 2 ) or about 0.60 s. 2.40: Using Eq. (2.13), with downward velocities and accelerations being positive, v 2 = y 2 (0.8 m s )2 + 2(1.6 m s )(5.0 m) = 16.64 m 2 s 2 (keeping extra significant figures), so vy = 4.1 m s .
  19. 2.41: a) If the meter stick is in free fall, the distance d is related to the reaction time t by d  (1 2) gt 2 , so t  2d g . If d is measured in centimeters, the reaction time is 2 2 t d  2 d  (4.52 10 2 s) d (1 cm). g 980 cm s b) Using the above result, (4.52  10 2 s) 17.6  0.190 s. 2.42: a) (1 2) gt 2  (1 2)(9.80 m s 2 )(2.5 s) 2  30.6 m. b) gt  (9.80 m s 2 )(2.5 s)  24.5. m s . c)
  20. 2.43: a) Using the method of Example 2.8, the time the ring is in the air is v 0 y  v0 y  2 g ( y  y 0 ) 2 t g 2 (5.00 m s)  (5.00 m s)  2(9.80 m s 2 )(12.0 m)  (9.80 m s 2 )  2.156 s, keeping an extra significant figure. The average velocity is then 12.0 ms  5.57 m s , down. 2.156 As an alternative to using the quadratic formula, the speed of the ring when it hits the ground may be obtained from v y  v 0 y  2 g ( y  y 0 ) , and the average velocity found 2 2 v v from y 2 0 y ; this is algebraically identical to the result obtained by the quadratic formula. b) While the ring is in free fall, the average acceleration is the constant acceleration due to gravity, 9.80 m / s 2 down. 1 c) y  y0  v0 y t  gt 2 2 1 0  12.0 m  (5.00 m s)t  (9.8 m s 2 )t 2 2 Solve this quadratic as in part a) to obtain t = 2.156 s. d) v y  v0 y  2 g ( y  y 0 )  (5.00 m s) 2  2(9.8 m s 2 )(12.0 m) 2 2 v y  16.1 m s e)
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