University Physics with Modern Physics (14th Edition)

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

Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 95: 3.33

Answer

The velocity of the canoe relative to the river is 0.36 m/s at an angle of $51.8^{\circ}$ south of west.

Work Step by Step

Let $E$ be the Earth. Let $R$ be the river. Let $C$ be the canoe. $v_{C/E} = v_{C/R}+v_{R/E}$ $v_{C/R} = v_{C/E} - v_{R/E}$ We can find the east component of $v_{C/R}$ $v_{C/R} = v_{C/E} - v_{R/E}$ $v_{C/R} = (0.40~m/s)~cos(45^{\circ}) - 0.50~m/s$ $v_{C/R} = -0.22~m/s$ Note that the negative sign means that the relative velocity is 0.22 m/s toward the west. We can find the south component of $v_{C/R}$ $v_{C/R} = v_{C/E} - v_{R/E}$ $v_{C/R} = (0.40~m/s)~sin(45^{\circ}) - 0$ $v_{C/R} = 0.28~m/s$ toward the south We can find the magnitude of $v_{C/R}$. $v_{C/R} = \sqrt{(0.22~m/s)^2+(0.28~m/s)^2}$ $v_{C/R} = 0.36~m/s$ We can find the angle south of west. $tan(\theta) = \frac{0.28~m/s}{0.22~m/s}$ $\theta = tan^{-1}(\frac{0.28}{0.22}) = 51.8^{\circ}$ The velocity of the canoe relative to the river is 0.36 m/s at an angle of $51.8^{\circ}$ south of west.
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