# The freight train is moving at a constant speed V1 = 36 km / h. It is followed in the same direction

**The freight train is moving at a constant speed V1 = 36 km / h. It is followed in the same direction by a passenger train at a speed of V2 = 72 km / h. The driver of the passenger train noticed a freight train going in front when the distance between them became L = 500m. Turning on emergency braking, he gave his composition an acceleration of a = 0.5m / s ^ 2. Will the collision be avoided?**

V1 = 36 km / h = 10 m / s.

V02 = 72 km / h = 20 m / s.

L = 500 m.

a = 0.5 m / s2.

S1 -?

S2 -?

Let us find the braking path of the passenger train S2 according to the formula: S2 = (V02 ^ 2 – V2 ^ 2) / 2 * a, where V2, V2 are the final and initial speed of the passenger train, and is the acceleration during braking of the train.

Since the train stops, its final speed V2 = 0 m / s.

S1 = V02 ^ 2/2 * a.

S1 = (20 m / s) ^ 2/2 * 0.5 m / s2 = 400 m.

Let us find the braking time of a passenger train t: t = (V0 ^ 2 – V2) / a.

t = (20 m / s – 0 m / s) / 0.5 m / s2 = 40 s.

Let’s find the distance that the freight train S1 will travel during this time using the formula: S1 = V1 * t.

S1 = 10 m / s * 40 s = 400 m.

When braking a passenger train, its braking distance S2 is equal to the distance traveled by a freight train S1: S2 = S1.

And since there was a distance L = 500 m between the trains at the beginning of braking, the collision can be avoided.

Answer: there will be no train collisions.