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 - Page 59: 11

Answer

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

Work Step by Step

(a) The area between the velocity versus time graph and the x-axis is equal to the displacement. The area under the curve from t = 0 to t = 2 is $\frac{6+2}{2.0}*2$ (because it is a trapezoid.) $\Delta x = 8.0~m$ $x = x_0+\Delta x$ $x = 2.0 + 8.0~m$ $x = 10.0~m$ (b) We can read the value of the velocity directly from the graph. Therefore, $v = 4.0~m/s$ (c) The acceleration is equal to the slope of the velocity versus time graph. Therefore, $a = \frac{-6.0~m/s}{3.0~s} = -2.0~m/s^2$
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