## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 64: 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 under the velocity versus time graph is equal to the displacement. We can read the velocity directly from the velocity versus time graph. The acceleration is equal to the slope of the velocity versus time graph. (a) From t = 0 to t = 1.0 s: The area under the velocity versus time graph is 4.0 m. $\Delta x = 4.0~m$ At t = 1.0 s: $x = x_0 + \Delta x$ $x = 2.0~m+4.0~m = 6.0~m$ At t = 1.0 s: $v = 4.0 ~m/s$ At t = 1.0 s, the slope of the velocity versus time graph is 0. Therefore, $a = 0$. (b) From t = 1.0 s to t = 3.0 s: The area under the velocity versus time graph is 7.0 m. $\Delta x = 7.0~m$ At t = 3.0 s: $x = 6.0~m + \Delta x$ $x = 6.0~m+ 7.0~m = 13.0~m$ At t = 3.0 s: $v = 2.0~m/s$ At t = 3.0 s, the slope of the velocity versus time graph is $\frac{-4.0~m/s}{2.0~s} = -2.0~m/s^2$ Therefore, $a = -2.0~m/s^2$.

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