Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 59: 10

speeding up: $a = 16~m/s^2$ slowing down: $a = -5.33~m/s^2$

The acceleration is equal to the slope of the velocity versus time graph. Therefore, speeding up: $a = \frac{\Delta v}{\Delta t} = \frac{0.8~m/s}{0.05~s}$ $a = 16~m/s^2$ slowing down: $a = \frac{\Delta v}{\Delta t} = \frac{-0.8~m/s}{0.15~s}$ $a = -5.33~m/s^2$