# The train moving at a speed of 20 M / s stopped 30 seconds after the start of braking.

The train moving at a speed of 20 M / s stopped 30 seconds after the start of braking. What is the braking distance of the train if the acceleration during braking was equal to 0.5 m / s ^ 2?

V0 = 20 m / s.

t = 30 s.

a = 0.5 m / s2.

V = 0 m / s.

S -?

With uniformly accelerated motion, when the body decreases its speed (decelerates), the path of the body S is determined by the formula: S = (V0 ^ 2 – V ^ 2) / 2 * a, where V, V0 are the final and initial speed of motion, a is the acceleration of the body when driving.

Since the train has stopped, its final speed is V = 0 m / s.

S = V0 ^ 2/2 * a.

S = (20 m / s) ^ 2/2 * 0.5 m / s2 = 400 m.

The braking distance can be expressed by another formula: S = V0 * t – a * t ^ 2/2.

S = 20 m / s * 30 s – 0.5 m / s2 * (30 s) ^ 2/2 = 375 m.

a = V / t.

Under such conditions, the acceleration of the train should be, a = 20 m / s / 30 s = 0.66 m / s2

This should not be the case, there was a mistake in the problem statement. 