The train, moving downhill with an acceleration of 0.2 m / s2, covered a distance of 340 m

The train, moving downhill with an acceleration of 0.2 m / s2, covered a distance of 340 m and developed a speed of 19 m / s. How long did the train move and what was its speed at the beginning of the countdown?

These tasks: a (the acceleration with which the train was moving down the mountain) = 0.2 m / s2; S (distance traveled) = 340 m; V2 (the speed that this train developed) = 19 m / s.
1) Determine the train speed at the origin: S = (V2 ^ 2 – V1 ^ 2) / 2a, whence V1 ^ 2 = V2 ^ 2 – 2a * S and V1 = √ (V2 ^ 2 – 2a * S) = √ (19 ^ 2 – 2 * 0.2 * 340) = √225 = 15 m / s.
2) The duration of the movement of this train: t = (V2 -V1) / a = (19 – 15) / 0.2 = 4 / 0.2 = 20 s.
Answer: The speed at the origin is 15 m / s, the movement time is 20 s.