College Physics (4th Edition)

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

Chapter 3 - Problems - Page 115: 74


The speed of the current is $0.25~km/h$

Work Step by Step

Let $v_s$ be the speed of the ship and let $v_c$ be the speed of the water current. Let $d$ be the distance. We can find an expression for $v_s$: $d = (v_s-v_c)(20.8) = (v_s+v_c)(19.8)$ $v_s = v_c~(19.8+20.8)$ $v_s = 40.6~v_c$ We can use the trip upstream to find $v_c$: $d = (v_s-v_c)(20.8~h)$ $d = (40.6~v_c-v_c)(20.8~h)$ $d = (39.6~v_c)(20.8~h)$ $v_c = \frac{208~km}{(39.6)(20.8~h)}$ $v_c = 0.25~km/h$ The speed of the current is $0.25~km/h$
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