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 4 - Kinematics in Two Dimensions - Exercises and Problems: 7

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 $a_x$. $a_x = \frac{\Delta v_x}{\Delta t}$ $a_x = \frac{20~m/s}{10~s}$ $a_x = 2~m/s^2$ We can find $a_y$. $a_y = \frac{\Delta v_y}{\Delta t}$ $a_y = \frac{10~m/s}{10~s}$ $a_y = 1~m/s^2$ We can find the magnitude of the puck's acceleration. $a = \sqrt{(a_x)^2+(a_y)^2}$ $a = \sqrt{(2~m/s^2)^2+(1~m/s^2)^2}$ $a = 2.2~m/s^2$ 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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