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Problems & Exercises

1.

(a) 7 m

(b) 7 m

(c) +7 m+7 m

3.

a. 8 m+2 m+3 m= 13 m ¯ 8 m+2 m+3 m= 13 m ¯

b. 9 m ¯ 9 m ¯

c. Δx=11 m2 m= 9 m ¯ Δx=11 m2 m= 9 m ¯

5.

(a) 3.0 ×104 m/s3.0 ×104 m/s

(b) 0 m/s

7.

2 × 10 7 years 2 × 10 7 years

9.

34 . 689 m/s = 124 . 88 km/h 34 . 689 m/s = 124 . 88 km/h

11.

(a) 40.0 km/h40.0 km/h

(b) 34.3 km/h, 25º S of E.25º S of E.

(c) average speed=3.20 km/h,v-=0.average speed=3.20 km/h,v-=0.

13.

384,000 km

15.

(a) 6.61×1015rev/s6.61×1015rev/s

(b) 0 m/s

16.

4 . 29 m/s 2 4 . 29 m/s 2

18.

(a) 1.43 s1.43 s

(b) 2.50m/s22.50m/s2

20.

(a) 10.8m/s10.8m/s

(b)

Line graph of position in meters versus time in seconds. The line begins at the origin and is concave up, with its slope increasing over time.
21.

38.9 m/s (about 87 miles per hour)

23.

(a) 16.5 s16.5 s

(b) 13.5 s13.5 s

(c) 2.68 m/s22.68 m/s2

25.

(a) 20.0 m20.0 m

(b) 1.00 m/s1.00 m/s

(c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at 2.00 m/s22.00 m/s2, then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.

27.

0 . 799 m 0 . 799 m

29.

(a) 28.0 m/s28.0 m/s

(b) 50.9 s50.9 s

(c) 7.68 km to accelerate and 713 m to decelerate

31.

(a) 51.4m51.4m

(b) 17.1 s17.1 s

33.

(a) 80.4m/s280.4m/s2

(b) 9.33×102 s9.33×102 s

35.

(a) 7.7 m/s7.7 m/s

(b) 15×102m/s215×102m/s2. This is about 3 times the deceleration of the pilots, who were falling from thousands of meters high!

37.

(a) 32.6 m/s232.6 m/s2

(b) 162 m/s162 m/s

(c) v>vmaxv>vmax, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32.6 m/s232.6 m/s2 during the last few meters, but substantially less, and the final velocity would be less than 162 m/s.

39.

104 s

40.

(a) v=12.2 m/sv=12.2 m/s; a=4.07 m/s2a=4.07 m/s2

(b) v=11.2 m/sv=11.2 m/s

41.

(a) y1=6.28 my1=6.28 m; v1=10.1 m/sv1=10.1 m/s

(b) y2=10.1 my2=10.1 m; v2=5.20 m/sv2=5.20 m/s

(c) y3=11.5 my3=11.5 m; v3=0.300 m/sv3=0.300 m/s

(d) y4=10.4 my4=10.4 m; v4=4.60 m/sv4=4.60 m/s

43.

v 0 = 4 . 95 m/s v 0 = 4 . 95 m/s

45.

(a) a=9.80 m/s2a=9.80 m/s2; v0=13.0 m/sv0=13.0 m/s; y0=0 my0=0 m

(b) v=0m/sv=0m/s. Unknown is distance yy to top of trajectory, where velocity is zero. Use equation v2=v02+2ayy0v2=v02+2ayy0 because it contains all known values except for yy, so we can solve for yy. Solving for yy gives

v 2 v 0 2 = 2a y y 0 v 2 v 0 2 2a = y y 0 y = y 0 + v 2 v 0 2 2a = 0 m + 0 m/s 2 13.0 m/s 2 2 9.80 m /s 2 = 8.62 m v 2 v 0 2 = 2a y y 0 v 2 v 0 2 2a = y y 0 y = y 0 + v 2 v 0 2 2a = 0 m + 0 m/s 2 13.0 m/s 2 2 9.80 m /s 2 = 8.62 m

Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.

(c) 2.65 s2.65 s

47.
Path of a rock being thrown off of cliff. The rock moves up from the cliff top, reaches a transition point, and then falls down to the ground.

(a) 8.26 m

(b) 0.717 s

49.

1.91 s

51.

(a) 94.0 m

(b) 3.13 s

53.

(a) -70.0 m/s (downward)

(b) 6.10 s

55.

(a) 19.6 m19.6 m

(b) 18.5 m18.5 m

57.

(a) 305 m

(b) 262 m, -29.2 m/s

(c) 8.91 s

59.

(a) 115 m/s115 m/s

(b) 5.0 m/s25.0 m/s2

61.
v = ( 11.7 6.95 ) × 10 3 m ( 40 . 0 – 20 .0 ) s = 238 m/s v = ( 11.7 6.95 ) × 10 3 m ( 40 . 0 – 20 .0 ) s = 238 m/s
63.
Line graph of position versus time. Line begins with a slight positive slope. It then kinks to a much greater positive slope.
65.

(a) 6 m/s

(b) 12 m/s

(c) 3 m/s23 m/s2

(d) 10 s

67.

(a) Car A is traveling faster at the checkpoint because it must go past the speed of car B to reach the same distance.

(b) i. Yes, the equation is consistent with the answer because the speed of car A is only a constant away from the correct answer. ii. Yes, the equation makes sense because V=2 V 0 V=2 V 0 .

(c)

The figure shows velocity v graphed as a function of position x. The plot of V sub A is linear, with a positive slope, starting at the origin and ending at a maximum velocity of 2 times v sub 0 at velocity v sub zero. The plot of V sub B is a constant at velocity v sub zero.
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