The train, accelerating on a track of 9 kilometers, spent 15 minutes and acquired a speed of 72 kilometers

The train, accelerating on a track of 9 kilometers, spent 15 minutes and acquired a speed of 72 kilometers an hour after the train’s uniform movement slowed down and stopped.The braking distance of the train was 1 kilometer during braking for 5 minutes, the average speed is 15 meters per second What is the time of uniform movement of the train.

Given:

S1 = 9 kilometers = 9000 meters – the first section of the route;

t1 = 15 minutes = 900 seconds – the time interval during which the train passed the first leg of the route;

S3 = 1 kilometer = 1000 meters – third section of the path (braking);

t3 = 5 minutes = 300 seconds – the third time interval;

v2 = 72 km / h = 20 m / s – the speed of the train on the second section of the track.

vav = 15 m / s is the average speed of the train along the entire route.

It is required to find t2 – the time interval during which the train moved uniformly.

vav = S total / t total = (S1 + S2 + S3) / (t1 + t2 + t3) =

= (S1 + v2 * t2 + S3) / (t1 + t2 + t3);

vav * t1 + vav * t2 + vav * t3 = S1 + v2 * t2 + S3;

vcr * t1 + vcr * t3 – S1 – S3 = v2 * t2 – vcr * t2;

vav * (t1 + t3) – (S1 + S3) = t2 * (v2 – vav);

t2 = (vav * (t1 + t3) – (S1 + S3)) / (v2 – vav) = (15 * (900 + 300) – (9000 + 1000)) / (20 – 15) =

= (15 * 1200 – 10000) / 5 = (18000 – 10000) / 5 = 8000/5 = 1600 seconds.

Answer: the train moved evenly for 1600 seconds.