College Physics for AP® Courses 2e

# Chapter 2

### Problems & Exercises

1.

(a) 7 m

(b) 7 m

(c) $+7 m+7 m$

3.

a.

b.

c.

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) $−2.50m/s2−2.50m/s2$

20.

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

(b) 21.

38.9 m/s (about 87 miles per hour)

23.

(a) $16.5 s16.5 s$

(b) $13.5 s13.5 s$

(c) $−2.68 m/s2−2.68 m/s2$

25.

(a) $20.0 m20.0 m$

(b) $−1.00 m/s−1.00 m/s$

(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) $−80.4m/s2−80.4m/s2$

(b) $9.33×10−2 s9.33×10−2 s$

35.

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

(b) $−15×102m/s2−15×102m/s2$. 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 $v2=v02+2ay−y0v2=v02+2ay−y0$ 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. (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. 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)

Figure 2.67

### Test Prep for AP® Courses

1.

(a)

3.

(a) Use tape to mark off two distances on the track — one for cart A before the collision and one for the combined carts after the collision. Push cart A to give it an initial speed. Use a stopwatch to measure the time it takes for the cart(s) to cross the marked distances. The speeds are the distances divided by the times.

(b) If the measurement errors are of the same magnitude, they will have a greater effect after the collision. The speed of the combined carts will be less than the initial speed of cart A. As a result, these errors will be a greater percentage of the actual velocity value after the collision occurs. (Note: Other arguments could properly be made for ‘more error before the collision' and error that ‘equally affects both sets of measurement.')

5.

The position vs. time graph should be represented with a positively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a positive y-intercept and a negative slope. Because the final velocity of the book is zero, the line should finish on the x-axis.

The position vs. time graph should be represented with a negatively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a negative y-intercept and a positive slope. Because the final velocity of the book is zero, the line should finish on the x-axis.]

7.

(c) Do you know how you learn best?