Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 3 - Vectors and Coordinate Systems - Exercises and Problems - Page 82: 4


The x-component of the vector is 12 m/s

Work Step by Step

Since the $40^{\circ}$ angle is below the positive x-axis, the x-component of the vector is positive. We can find the x-component. Let $y$ be the magnitude of the y-component. Therefore; $tan(\theta) = \frac{y}{x}$ $x = \frac{y}{tan(\theta)}$ $x = \frac{10~m/s}{tan(40^{\circ})}$ $x = 12~m/s$ The x-component of the vector is 12 m/s.
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