A train running along a straight track at 32.90 m/s is slowed at a rate of 0.783 m/s/s to a stop how far did the train travel before coming to rest ?

The distance traveled by the train is 691.2 m

Explanation:

Given;

initial velocity of the train, u = 32.9 m/s

deceleration of the train, a = - 0.783 m/s²

final velocity of the train when brought to a stop, v = 0

The distance traveled by the train is given by;

v² = u² + 2as

0 = (32.9)² + 2(-0.783)s

0 = 1082.41 - 1.566s

1.566s = 1082.41

s = 1082.41 / 1.566

s = 691.2 m

Therefore, the distance traveled by the train is 691.2 m