University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 1 - Units, Physical Quantities, and Vectors - Problems - Exercises - Page 31: 1.75

Answer

The dog should run 29.6 m at an angle of $71.4^{\circ}$ south of east.

Work Step by Step

We can find the east component $d_x$ of the direction $d$. $d_x + 12.0~m - 28.0~sin(50.0^{\circ})~m = 0$ $d_x = -12.0~m + 28.0~sin(50.0^{\circ})~m$ $d_x = 9.45~m$ We can find the south component $d_y$ of the direction $d$. $d_y - 28.0~cos(50.0^{\circ})~m = 10.0~m$ $d_y = 10.0~m + 28.0~cos(50.0^{\circ})~m$ $d_y = 28.00~m$ We can use $d_x$ and $d_y$ to find the magnitude of the distance $d$. $d = \sqrt{(d_x)^2+(d_y)^2}$ $d = \sqrt{(9.45~m)^2+(28.00~m)^2}$ $d = 29.6~m$ We can find the angle south of east. $tan(\theta) = \frac{28.00}{9.45}$ $\theta = tan^{-1}(\frac{28.00}{9.45}) = 71.4^{\circ}$ The dog should run 29.6 m at an angle of $71.4^{\circ}$ south of east.
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