A train is traveling south at 25.6 m/s when the brakes are applied. It slows down with a constant acceleration to a speed of 6.90 m/s in a time of 8.38 s. What is the acceleration of the train during the 8.38-s interval?

Answer 1

a = ∆v/∆t
= (6.90-25.6)m/s / (8.38-0)s
= (-18.7 m/s)/8.38s
= -2.23 m/s^2