Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 119: 113


The duck's displacement has a magnitude of $78.5cm$ and is directed at $21.8^o$ south of east.

Work Step by Step

1) We take eastward to be $+i$ direction and northward to be $+j$ direction. - Reaction force on the duck as it paddles: $$\vec{A}=(0.1N)i$$ - The current force: $$\vec{B}=(0.2\cos52N)i-(0.2\sin52N)j$$ $$\vec{B}=(0.12N)i-(0.16N)j$$ The total force that acts on the duck: $$\sum \vec{F}=\vec{A}+\vec{B}=(0.22N)i-(0.16N)j$$ The duck has mass $m=2.5kg$. We can find its acceleration $\vec{a}$ $$\vec{a}=\frac{\sum \vec{F}}{m}=(0.088m/s^2)i-(0.064m/s^2)j$$ 2) We have $\vec{v}_0=(0.11m/s)i$, $t=3s$. The displacement vector would be $$\vec{d}=\vec{v}_0t+\frac{1}{2}\vec{a}t^2=(0.33m)i+4.5\Big((0.088m/s^2)i-(0.064m/s^2)j\Big)$$ $$\vec{d}=(0.73m)i-(0.29m)j$$ - Magnitude: $d=\sqrt{0.73^2+0.29^2}=0.785m=78.5cm$ - Direction: Take $\theta$ to be the angle of $\vec{d}$ relative to due east $$\tan\theta=\frac{-0.29}{0.73}=-0.4$$ $$\theta=-21.8^o$$ which means $21.8^o$ south of east
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