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: 10

Answer

At t = 5 s, the magnitude of the acceleration is $2.2~m/s^2$

Work Step by Step

The acceleration is the slope of the velocity versus time graph. We can find $a_x$: $a_x = \frac{\Delta v_x}{\Delta t}$ $a_x = \frac{20~m/s}{10~s}$ $a_x = 2.0~m/s^2$ We can find $a_y$: $a_y = \frac{\Delta v_y}{\Delta t}$ $a_y = \frac{10~m/s}{10~s}$ $a_y = 1.0~m/s^2$ We can find the magnitude of the puck's acceleration; $a = \sqrt{(a_x)^2+(a_y)^2}$ $a = \sqrt{(2.0~m/s^2)^2+(1.0~m/s^2)^2}$ $a = 2.2~m/s^2$ The magnitude of the acceleration is constant and it is equal to $2.2~m/s^2$. At t = 5 s, the magnitude of the acceleration is $2.2~m/s^2$.
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