Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - General Problems - Page 46: 64

Answer

The average acceleration is $0.038 ~m/s^2$

Work Step by Step

First, we find: $v_1 = (15 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 4.2 ~m/s$ $v_2 = (65 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 18 ~m/s$ We can use $v_1$ and $v_2$ to find the average acceleration. $a = \frac{v_2^2 - v_1^2}{2 ~\Delta x} = \frac{(18 ~m/s)^2 - (4.2 ~m/s)^2}{(2)(4000 ~m)} = 0.038 ~m/s^2$ Therefore, the average acceleration is $0.038 ~m/s^2$
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