The train, moving after the start of braking with an acceleration of 0.40 m / s 2, stopped 25 s later

The train, moving after the start of braking with an acceleration of 0.40 m / s 2, stopped 25 s later. Find the speed at the start of braking and the braking distance.

Given: a (acceleration of a given train after the start of braking) = -0.4 m / s2; t (deceleration duration) = 25 s; V (final speed) = 0 m / s.

1) Let’s calculate the speed of a given train at the moment of braking start: V0 = V – a * t = 0 – (-0.4 * 25) = 10 m / s.

2) Determine the braking distance: S = (V + V0) * t / 2 = (0 + 10) * 25/2 = 125 m.

Check: S = V0 * t + a * t ^ 2/2 = 10 * 25 + (-0.4) * 25 ^ 2/2 = 250 – 125 = 125 m (correct).

Answer: Before braking, the train had a speed of 10 m / s; the braking distance is 125 m.