College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Review & Synthesis: Chapters 6-8 - MCAT Review - Page 327: 9


The correct answer is: D. $50~m/s^2$

Work Step by Step

We can find the tangential speed $v$ of a point on the wheel at a distance of $0.3~m$ from the center: $v = \frac{d}{t} = \frac{4800~m}{(20)(60~s)} = 4.0~m/s$ The can find the radial acceleration: $a_r = \frac{v^2}{r} = \frac{(4.0~m/s)^2}{0.3~m} = 53.3~m/s^2$ This value is closest to $50~m/s^2$ The correct answer is: D. $50~m/s^2$
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