# The train stopped after 3 minutes of its movement, having covered a distance of 1.8 km.

July 27, 2021 | education

| **The train stopped after 3 minutes of its movement, having covered a distance of 1.8 km. find the initial speed and acceleration of the train.**

Task data: t (deceleration duration) = 3 min (in SI t = 180 s); S (distance traveled by the train when braking) = 1.8 km (in SI S = 1800 m).

1) Determine the initial speed of the train: S = (V + V0) * t / 2 = (0 + V0) * t / 2 and V0 = 2S / t = 2 * 1800/180 = 20 m / s.

2) Define the acceleration: a = (V – V0) / t = (0 – 20) / 180 = -0.11 (1) m / s2.

Answer: Acceleration during deceleration was -0.11 (1) m / s2; the initial speed of the train was 20 m / s.

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