Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems: 27

Answer

a) 2.2 radians per second b) 2.8 seconds c) .63 meters

Work Step by Step

We find: a) $\omega = \frac{a_{max}}{v_{max}}=\frac{3.1}{1.4}=\fbox{2.2 seconds}$ b) $ T = \frac{2\pi }{\omega} = \frac{2\pi }{2.2}= \fbox {2.8 seconds}$ c) We simplify the equation for the maximum possible speed to find: $A = \frac{v_{max}}{\omega}= \frac{1.4}{2.2}=\fbox{.63 meters}$
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