Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 3 - Vectors and Coordinate Systems - Exercises and Problems: 33

Answer

The magnitude of the net displacement is 7.5 meters.

Work Step by Step

Let $r$ be the displacement of the mouse. Then; $r_x = 5.0~m + 3.0~m~cos(45^{\circ})$ $r_x = 7.1~m$ $r_y = -3.0~m~sin(45^{\circ})$ $r_y = -2.1~m$ $r_z = -1.0~m$ We can find the magnitude of the displacement as: $r = \sqrt{(7.1~m)^2+(-2.1~m)^2+(-1.0~m)^2}$ $r = 7.5~m$ The magnitude of the net displacement is 7.5 meters.
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