College Physics (4th Edition)

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

Review & Synthesis: Chapters 1-5 - Review Exercises - Page 193: 19

Answer

The trawler moved with a speed of $28.75~km/h$

Work Step by Step

In $4.0~h$, the fish swim $20.0~km$ to the northwest. We can find the fish's displacement to the north: $(20.0~km)~sin~45^{\circ} = 14.14~km$ We can find the fish's displacement to the west: $(20.0~km)~cos~45^{\circ} = 14.14~km$ To intercept the fish, the trawler needs to move $114.14~km$ to the north and $14.14~km$ to the west. We can find the distance $d$ that the trawler must travel: $d = \sqrt{(114.14~km)^2+(14.14~km)^2} = 115~km$ We can find the trawler's speed: $v = \frac{d}{t} = \frac{115~km}{4.0~h} = 28.75~km/h$ The trawler moved with a speed of $28.75~km/h$.
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