The train approaching the station on the track of 900 meters has been reduced by 90 kilometers per hour to 36

The train approaching the station on the track of 900 meters has been reduced by 90 kilometers per hour to 36 kilometers per hour. Determine the acceleration with which the train was moving.

These tasks: S (the distance traveled by the train on the way to the station) = 900 m; V0 (initial speed) = 90 km / h (in SI V0 = 25 m / s); V (final speed) = 36 km / h (SI 10 m / s).

The acceleration with which this train was moving on its way to the station can be expressed from the formula: S = (V ^ 2 – V0 ^ 2) / 2a, whence a = (V ^ 2 – V0 ^ 2) / 2S.

Let’s perform the calculation: a = (10 ^ 2 – 25 ^ 2) / (2 * 900) = -0.292 m / s2.

Answer: On the way to the station, this train was moving with an acceleration of -0.292 m / s2.