College Physics (4th Edition)

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

Chapter 3 - Problems - Page 115: 77

Answer

$t = 38.8~s$

Work Step by Step

Let $d$ be the distance the person moves along the escalator. We can find the person's climbing velocity $v_p$ on the stalled elevator: $v_p = \frac{d}{94~s}$ We can find the escalator's moving velocity $v_e$: $v_e = \frac{d}{66~s}$ We can find the person's velocity $v$ when walking along the moving escalator: $v = v_p+v_e = \frac{d}{94~s}+\frac{d}{66~s}$ We can find the time it would take to walk up the moving escalator: $t = \frac{d}{v} = \frac{d}{\frac{d}{94~s}+\frac{d}{66~s}}$ $t = \frac{1}{\frac{66~s+94~s}{(94~s)(66~s)}}$ $t = \frac{(94~s)~(66~s)}{66~s+94~s}$ $t = 38.8~s$
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