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Chapter 2 Solutions

*2.1

(a) (b)

– v = 2.30 m/s ∆x 57.5 m – 9.20 m – v = = = 16.1 m/s ∆t 3.00 s ∆x 57.5 m – 0 m – v = = = 11.5 m/s ∆t 5.00 s Displacement = (8.50 × 104 m/h)  x = (49.6 + 130) × 103 m = 180 km 35.0  h + 130 × 103 m 60.0 

(c)

2.2

(a)

(b)

Average velocity =

displacement 180 km = = 63.4 km/h time (35.0 + 15.0)  60.0 + 2.00 h  

2.3

(a)

vav =∆x 10 m = = 5 m/s ∆t 2s 5m = 1.2 m/s 4s x2 – x1 5 m – 10 m = = –2.5 m/s t2 – t1 4s–2s x2 – x1 –5 m – 5 m = = –3.3 m/s t2 – t1 7s–4s x2 – x1 0–0 = = 0 m/s t2 – t1 8–0

(b)

vav =

(c)

vav =

(d)

vav =

(e) 2.4

vav =

x = 10t2 For t(s) = 2.0 x(m) = 40 (a) 2.1 3.0

44.1 90

∆x 50 m – v = = = 50.0 m/s ∆t 1.0 s ∆x 4.1 m – v = = = 41.0 m/s ∆t 0.1 s

(b)

© 2000 byHarcourt College Publishers. All rights reserved.

2

Chapter 2 Solutions

2.5

(a)

Let d represent the distance between A and B. Let t1 be the time for which the walker d has the higher speed in 5.00 m/s = . Let t2 represent the longer time for the return trip t1 d d d in –3.00 m/s = – . Then the times are t1 = and t2 = . The average t2 (5.00 m/s) (3.00 m/s) speed is: Total distance – v = =Total time = (8.00 m/s)d d d (5.00 m/s) + (3.00 m/s) (15.0 m2/s2) d+d 2d

2(15.0 m2/s2) – v = = 3.75 m/s 8.00 m/s (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0

2.6

(a)

Total distance – v = Total time Let d be the distance from A to B. Then the time required is d d + . v1 v2 2v 1 v 2 2d = d d v1 + v2 + v1 v2

– And the average speedis v =

(b) 2.7 (a)

With total displacement zero, her average velocity is 0 .

x (m) 5

0 2 —5 4 6

t (s)

(b)

v = slope =

5.00 m – (–3.00 m) 8.00 m = = 1.60 m/s (6.00 s – 1.00 s) 5.00 s

© 2000 by Harcourt College Publishers. All rights reserved.

Chapter 2 Solutions
2.8 (a) At any time, t, the displacement is given by x = (3.00 m/s2)t2. Thus, at ti = 3.00 s: (b) xi =(3.00 m/s2)(3.00 s)2 = 27.0 m

3

At tf = 3.00 s + ∆t : xf = (3.00 m/s2)(3.00 s + ∆t)2, or xf = 27.0 m + (18.0 m/s)∆t + (3.00 m/s2)(∆t)2

(c)

The instantaneous velocity at t = 3.00 s is: x f – x i v = lim  = lim [(18.0 m/s) + (3.00 m/s2)∆t], or ∆t → o  ∆ t  ∆t → 0 v = 18.0 m/s

2.9

(a)

at ti = 1.5 s, xi = 8.0 m (Point A) at tf = 4.0 s, xf = 2.0 m (Point B) (2.0 – 8.0) m 6.0 m –xf – xi v = = =– = –2.4 m/s tf – ti (4 – 1.5) s 2.5 s

x (m) 12 10 8 6 4 2 0 0 1 2 D 3 4 5 6 B t (s) C A

(b)

The slope of the tangent line is found from points C and D. (tC = 1.0 s, xC = 9.5 m) and (tD = 3.5 s, xD = 0), v ≅ –3.8 m/s

(c) 2.10 (b)

The velocity is zero when x is a minimum. This is at t ≈ 4 s . At t = 5.0 s, the slope is v ≅ At t = 4.0 s, the slope is v ≅ At t = 3.0 s,the slope is v ≅ At t = 2.0 s, the slope is v ≅ 58 m ≅ 23 m/s 2.5 s 54 m ≅ 18 m/s 3s 49 m ≅ 14 m/s 3.4 s 36 m ≅ 9.0 m/s 4.0 s
v (m/s) 20

x (m)

60

40

20

0 0 2 4

t (s)

(c) (d)

∆v 23 m/s – a = ≅ ≅ 4.6 m/s2 ∆t 5.0 s Initial velocity of the car was zero .
0

t (s) 0 2 4

© 2000 by Harcourt College Publishers. All rights reserved.

4

Chapter 2 Solutions

2.11

(a)v=

(5 – 0) m = 5 m/s (1 – 0) s (5 – 10) m = –2.5 m/s (4 – 2) s

x (m) 10 8 6 4 2 0 −2 −4 −6 1 2 3 4 5 6 7 8 t (s)

(b)

v=

(c)

(5 m – 5 m) v= = 0 (5 s – 4 s) 0 – (–5 m) v= = +5 m/s (8 s – 7 s)

(d)

2.12

vf – vi 0 – 60.0 m/s – a = = = – 4.00 m/s2 tf – ti 15.0 s – 0 The negative sign in the result shows that the acceleration is in the negative x direction.

*2.13

Choosethe positive direction to be the outward perpendicular to the wall. v = vi + at a= ∆v 22.0 m/s – (–25.0 m/s) = = 1.34 × 104 m/s2 ∆t 3.50 × 10–3 s

2.14

(a)

Acceleration is constant over the first ten seconds, so at the end v = vi + at = 0 + (2.00 m/s2)(10.0 s) = 20.0 m/s Then a = 0 so v is constant from t = 10.0 s to t = 15.0 s. And over the last five seconds the velocity changes to v =...
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