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

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