## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 3 - Vectors and Coordinates Systems - Exercises and Problems: 31

#### Answer

(a) $d = (-62\hat{i}-65\hat{j})~mi$ (b) The magnitude of the displacement vector is 90 miles and it is directed at an angle of $46.4^{\circ}$ south of west.

#### Work Step by Step

We can find a displacement vector from the origin to Ward's home as; $W = (100\hat{i}+50\hat{j})~mi$ We can find Ruth's displacement vector from the origin as; $R = (38\hat{i}-15\hat{j})~mi$ (a) We can find the displacement vector $d$ that Ward needs to fly. $W+d = R$ $d = R-W$ $d = (38\hat{i}-15\hat{j})~mi-(100\hat{i}+50\hat{j})~mi$ $d = (-62\hat{i}-65\hat{j})~mi$ (b) We then find the magnitude of $d$: $d = \sqrt{(-62~mi)^2+(-65~mi)^2}$ $d = 90~mi$ We can find the angle $\theta$ south of west; $tan(\theta) = \frac{65}{62}$ $\theta = arctan(\frac{65}{62})$ $\theta = 46.4^{\circ}$ The magnitude of the displacement vector is 90 miles and it is directed at an angle of $46.4^{\circ}$ south of west.

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