College Physics (4th Edition)

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

Chapter 3 - Problems - Page 111: 4


The length of the displacement vector is 0.283 mi and the direction is $45^{\circ}$ north of west (which is directly northwest).

Work Step by Step

The west component of the displacement is 0.500 mi - 0.300 mi = 0.200 mi west. The north component of the displacement is 0.200 mi north. We can find the length of the displacement: $\sqrt{(0.200~mi)^2+(0.200~mi)^2} = 0.283~mi$ We can find the direction north of west: $tan~\theta = \frac{0.200~mi}{0.200~mi}$ $\theta = tan^{-1}(\frac{0.200~mi}{0.200~mi})$ $\theta = tan^{-1} (1)$ $\theta = 45^{\circ}$
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