Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 3 - Vectors and Coordinates Systems - Exercises and Problems - Page 77: 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; $tan(\theta) = \frac{y}{x}$ $x = \frac{y}{tan(\theta)}$ $x = \frac{10~m/s}{tan(40^{\circ})}$ $x = 12~m/s$ The x-component of the vector is thus 12 m/s.
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