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 2 - Kinematics in One Dimension - Exercises and Problems: 12

Answer

(a) $x = 6.0~m$ v = 4.0 m/s $a = 0$ (b) $x = 13.0~m$ v = 2.0 m/s $a = -2.0~m/s^2$

Work Step by Step

The area between the velocity versus time graph and the x-axis is equal to the displacement. We can read the velocity directly from the graph. The acceleration is equal to the slope of the velocity versus time graph. (a) From t = 0 to t = 1.0 s: $\Delta x = 4.0~m$ $x = x_0 + \Delta x$ $x = 2.0~m+4.0~m = 6.0~m$ v = 4.0 m/s $a = 0$ (b) From t = 1.0 s to t = 3.0 s: $\Delta x = 7.0~m$ $x = 6.0~m + \Delta x$ $x = 6.0~m+ 7.0~m = 13.0~m$ v = 2.0 m/s $a = \frac{-4.0~m/s}{2.0~s} = -2.0~m/s^2$
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