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

The train's acceleration is $-1.0~m/s^2$
In the 2.0 seconds that the train moved at a constant speed, the train traveled 4.0 meters. Since the train comes to a stop a distance of 6.0 meters from the point of release, the train decelerates over a distance of 2.0 meters. We can find the rate of deceleration as: $a = \frac{v^2-v_0^2}{2d}$ $a = \frac{0-(2.0~m/s)^2}{(2)(2.0~m)}$ $a = -1.0~m/s^2$ The train's acceleration is $-1.0~m/s^2$.