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 4 - Kinematics in Two Dimensions - Exercises and Problems: 6

Answer

The boat is moving with a speed of 9.2 m/s at an angle of $19.6^{\circ}$ north of east.

Work Step by Step

We can find the eastward component of velocity after the 6.0 second acceleration period. $v_x = v_{0x}+a_xt$ $v_x = 5.0~m/s+(0.80~m/s^2)~cos(40^{\circ})(6.0~s)$ $v_x = 8.68~m/s$ We can find the northward component of velocity after the 6.0 second acceleration period. $v_y = v_{0y}+a_yt$ $v_y = 0+(0.80~m/s^2)~sin(40^{\circ})(6.0~s)$ $v_y = 3.09~m/s$ We can find the boat's speed. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{(8.68~m/s)^2+(3.09~m/s)^2}$ $v = 9.2~m/s$ We can find the angle north of east. $tan(\theta) = \frac{v_y}{v_x}$ $\theta = arctan(\frac{3.09~m/s}{8.68~m/s})$ $\theta = 19.6^{\circ}$ The boat is moving with a speed of 9.2 m/s at an angle of $19.6^{\circ}$ north of east.
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