# A freight train weighing 2,000 tons began to slow down. Under the action of a braking force of 200 kN

A freight train weighing 2,000 tons began to slow down. Under the action of a braking force of 200 kN, its braking distance to a stop was 500 meters. What was the initial speed of the train?

Initial data: m (mass of a freight train) = 2000 t (2000 * 10 ^ 3 kg); F (value of the braking force) = 200 kN (200 * 10 ^ 3 N); S (the path that the freight train traveled to the stop) = 500 m.

1) The acceleration with which the freight train was moving: a = F / m = 200 * 10 ^ 3 / (2000 * 10 ^ 3) = 0 m / s.

2) Initial speed of the commodity train: S = (V2 – V0 ^ 2) / 2 (-a) = V0 ^ 2 / 2a, whence V0 = √ (2 * a * S) = √ (2 * 0.1 * 500 ) = 10 m / s.

Answer: The initial speed of the freight train was 10 m / s.

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