At the moment when the late passenger ran onto the platform, the penultimate carriage of the train

At the moment when the late passenger ran onto the platform, the penultimate carriage of the train passed by him in time t1 = 10 s. The last carriage passed the passenger in time t2 = 8 s. How long is the passenger late for the train departure? The train moves at uniform acceleration, the length of the carriage is the same.

These tasks: t1 (the duration of the passing of the late passenger of the penultimate carriage) = 10 s; t2 (duration of the last carriage passage) = 8 s.

1) Car length: L = Vн * t1 + a * t1 ^ 2/2 (penultimate car) or L = Vн1 * t ^ 2 + a * t2 ^ 2/2 (last car).

2) The speed of the train at the moment the passenger hits the platform: Vн = a * top.

3) The speed of the train at the moment of passing the last carriage: Vн1 = a * (t1 + top).

4) The resulting equality: a * top * t1 + a * t2 ^ 2/2 = a * (t1 + top) * t2 + a * t2 ^ 2/2.

top * t1 + t1 ^ 2/2 = (t1 + top) * t ^ 2 + t2 ^ 2/2.

top * 10 + 10 ^ 2/2 = (10 + top) * 8 + 8 ^ 2/2.

10top + 50 = 80 + 8top + 32.

2top = 62 and top = 62/2 = 31 s.

Answer: The passenger was 31 seconds late.