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: 1.77

Answer

The distance between Karl's tent and Joe's tent is 28.2 meters.

Work Step by Step

Let west and south be positive directions. Let $K$ be the vector from your tent to Karl's tent. Let $J$ be the vector from your tent to Joe's tent. Let $d$ be the vector from Karl's tent to Joe's tent. $K+d = J$ $d = J - K$ We can find the west component $d_x$ of $d$. $d_x = J_x - K_x$ $d_x = -21.0~cos(23.0^{\circ})~m - (-32.0)~cos(37.0^{\circ})~m$ $d_x = 6.23~m$ west We can find the south component $d_y$ of $d$. $d_y = J_y - K_y$ $d_y = 21.0~sin(23.0^{\circ})~m - (-32.0)~sin(37.0^{\circ})~m$ $d_y = 27.5~m$ south We can use $d_x$ and $d_y$ to find the magnitude of $d$. $d = \sqrt{(d_x)^2+(d_y)^2}$ $d = \sqrt{(6.23~m)^2+(27.5~m)^2}$ $d = 28.2~m$ The distance between Karl's tent and Joe's tent is 28.2 meters.
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