The train, having a speed of 70 km / h, began to move with equal slowness and after 10 seconds reduced

The train, having a speed of 70 km / h, began to move with equal slowness and after 10 seconds reduced the speed to 52 km / h. With what acceleration did the train move on this section? What way did he go?

Given: V0 (train speed before the start of equally slow motion) = 70 km / h (in SI V0 = 19.44 m / s); t (time) = 10 s; V (final speed) = 52 km / h (in SI V = 14.44 m / s).

1) Acceleration in a section with equally slow motion: a = (V – V0) / t = (14.44 – 19.44) / 10 = -5 / 10 = -0.5 m / s2.

2) Path: S = (V + V0) * t / 2 = (19.44 + 14.44) * 10/2 = 169.4 m or S = (V ^ 2 – V0 ^ 2) / 2a = ( 14.44 ^ 2 – 19.44 ^ 2) / (2 * (-0.5)) = 169.4 m.

Answer: The acceleration of the train was -0.5 m / s2; the distance covered is 169.4 m.