College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 60: 11

Answer

(a) (3) (d) Net displacement is $44.72\,\mathrm{m}$.

Work Step by Step

(a) In the figure, $d$ is the net displacement - given by the hypotenuse of the right triangle formed, as shown. Since we have one side of the triangle as $40\,\mathrm{m}$, the net displacement is (3) between $40\,\mathrm{m}$ and $60\,\mathrm{m}$. (b) Net displacement, $d=\sqrt{40^2+20^2}=44.72\,\mathrm{m}$
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