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 59: 2.12

Answer

(a) (i) $1.7 ~m/s^2$ (ii) $-1.7 ~m/s^2$ (iii) $0 ~m/s^2$ (iv) $0 ~m/s^2$ (b) (i) $0 ~m/s^2$ (ii) $-1.7 ~m/s^2$

Work Step by Step

$v = (60 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 16.7 ~m/s$ (a) The average acceleration is $\frac{\Delta ~v}{\Delta ~t}$ (i) $\frac{\Delta ~v}{\Delta ~t} = \frac{16.7 ~m/s}{10 ~s} = 1.7 ~m/s^2$ (ii) $\frac{\Delta ~v}{\Delta ~t} = \frac{-16.7 ~m/s}{10 ~s} = -1.7 ~m/s^2$ (iii) $\frac{\Delta ~v}{\Delta ~t} = \frac{0 ~m/s}{20 ~s} = 0 ~m/s^2$ (iv) $\frac{\Delta ~v}{\Delta ~t} = \frac{0 ~m/s}{40 ~s} = 0 ~m/s^2$ (b) The instantaneous acceleration is the slope of the velocity versus time graph. (i) At t = 20 s, the slope of the velocity versus time graph is zero. Therefore the instantaneous acceleration is $0 ~m/s^2$ (ii) At t = 35 s, the slope of the velocity versus time graph is $\frac{-16.7 ~m/s}{10 ~s} = -1.7 ~m/s^2$. Therefore the instantaneous acceleration is $-1.7 ~m/s^2$
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