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 - Search and Learn - Page 48: 3

Answer

(a) see graph (b) t = 21.9 s (c) $a = 3.54 m/s^2$ (d) v = 77.5 m/s

Work Step by Step

(a) see graph (b) We can convert the speeder's speed to m/s. $(140~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 38.9~m/s$ $t = \frac{d}{v} = \frac{850~m}{38.9~m/s} = 21.9~s$ (c) $x = \frac{1}{2}at^2$ $a = \frac{2x}{t^2} = \frac{(2)(850~m)}{(21.9~s)^2}= 3.54~m/s^2$ (d) $v = at = (3.54~m/s^2)(21.9~s) = 77.5~m/s$
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