Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises - Page 104: 7

Answer

$\textbf{(a)}$ The angle is $$\theta=26.6^\circ$$ west of north. $\textbf{(b)}$ The speed is $$v_2=1.07\text{ m/s}.$$

Work Step by Step

$\textbf{(a)}$ The angle $\theta$ is the same as the angle at the 'upper' vertex of the triangle in the figure. Further, since the angle at the point $1$ is $45^\circ$, Then the altitude of the triangle splits the base of the length of $1.5$ km into two pieces, of $1.0$ km and $0.5$ km. Now we have $$\tan\theta=\frac{0.5\text{ km}}{1.0\text{ km}}=0.5$$ so we get $$\theta=\arctan 0.5 =26.6^\circ.$$ $\textbf{(b)}$ From the Pythagorean theorem we get that the distance that canoeist $2$ must cover is $d_2=\sqrt{(1.0\text{ km})^2+(0.5\text{ km})^2} = \sqrt{1.25}\text{ km}=1.12\text{ km}.$ Similarly, the distance that the canoeist $1$ has to cover is $d_1=\sqrt{(1.0\text{ km})^2+(1.0\text{ km})^2}=1.41\text{ km}.$ Equating the required travel times, which is simply distance over speed, we get: $$\frac{d_1}{v_1}=\frac{d_2}{v_2}\Rightarrow v_2=\frac{d_2}{d_1}v_1=\frac{1.12\text{ km}}{1.41\text{ km}}\cdot1.35\text{ m/s}=1.07\text{ m/s}.$$
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