University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 60: 2.29

Answer

(a) (i) The average acceleration is $5.59~m/s^2$ (ii) The average acceleration is $7.74~m/s^2$ (b) (i) The rocket travels 179 meters in the first 8.00 seconds. (ii) The rocket travels 12.8 kilometers between 8.00 seconds and 1.00 minute.

Work Step by Step

$v_1 = (161~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 44.7~m/s$ $v_2 = (1610~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 447~m/s$ (a) average acceleration = $\frac{\Delta v}{\Delta t}$ (i) From t = 0 to t = 8.00 s: $\frac{\Delta v}{\Delta t} = \frac{44.7~m/s- 0 }{8.00~s} = 5.59~m/s^2$ (ii) From t = 8.00 s to t = 1.00 min: $\frac{\Delta v}{\Delta t} = \frac{447~m/s-44.7~m/s}{52.00~s} = 7.74~m/s^2$ (b) (i) $y = \frac{1}{2}at^2$ $y = \frac{1}{2}(5.59~m/s^2)(8.00~s)^2$ $y = 179~m$ (ii) $y = v_0t+\frac{1}{2}at^2$ $y = (44.7~m/s)(52.00~s)+\frac{1}{2}(7.74~m/s^2)(52.00~s)^2$ $y = 12,800~m = 12.8~km$
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