College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 96: 3


(a) (3) Magnitude of the acceleration vector is between $4\,\mathrm{m/s^2}$ and $7\,\mathrm{m/s^2}$. (b) The acceleration has a magnitude of $5\,\mathrm{m/s^2}$, directed at angle of $53^{\circ}$ to the x-axis.

Work Step by Step

The magnitude of a vector is given by - $|a|=\sqrt{a_x^2+a_y^2}$ (a) This is analogous to the magnitude being the hypotenuse of a triangle with x- and y-components as the two other sides. Thus, the magnitude of the vector must be greater than the larger of the two components, which is 4 here. And, the magnitude must also be less than the sum of the components, which is 7 here. (b) Using the values of the components, $a_x=3$ and $a_y=4$ - $|a|=\sqrt{9+16}=5$. for direction at an angle $\theta$ to x-axis - $\tan{\theta}=\frac{a_y}{a_x}=\frac{4}{3}$ or, $\theta=53^{\circ}$.
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