Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems - Page 88: 69c

Answer

$a = 13~m/s^2$

Work Step by Step

Relative to the crew member standing on the ground, the cheetah is moving $50~km/h$ west and then $45~km/h$ east The magnitude of the change in velocity is $95~km/h$ We can express $95~km/h$ in units of $m/s$: $(95~km/h)(\frac{1~h}{3600~s})(\frac{1000~m}{1~km}) = 26.4~m/s$ We can find the magnitude of the acceleration according to the nervous crew member: $a = \frac{\Delta v}{\Delta t}$ $a = \frac{26.4~m/s}{2.0~s}$ $a = 13~m/s^2$
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