The train passed the distance between the two stations at an average speed of Vav = 72 km / h in t = 20 min.

The train passed the distance between the two stations at an average speed of Vav = 72 km / h in t = 20 min. Acceleration and deceleration together lasted t1 = 4min, and the rest of the time the train moved uniformly. What was the speed V of the train with uniform motion?

Vav = 72 km / h = 20 m / s – the average speed of the train along the entire route;

t = 20 minutes = 1200 seconds – the entire time of the train movement;

t1 = 4 minutes = 240 seconds – the time taken for acceleration and deceleration.

It is required to determine V (m / s) – the speed of uniform movement of the train.

Let’s find the common path that the train traveled:

S = Vav * t = 20 * 1200 = 24000 meters.

Let V be the speed of uniform movement of the train, t11 is the time spent on acceleration, t12 is the time spent on uniform movement, t13 is the time spent on braking. Then:

S = S1 + S2 + S3 (where S1 is the acceleration path, S2 is the steady motion path, S3 is the braking path).

S = V * t11 / 2 + V * t12 + V * t13 / 2;

S = (V * t11 + 2 * V * t12 + V * t13) / 2;

S = V * (t11 + 2 * t12 * t13) / 2.

Since t11 + t13 = t1, and t12 = t – t1, we get:

S = V * (t1 + 2 * (t – t1)) / 2;

S = V * (t1 + 2 * t – 2 * t1) / 2;

S = V * (2 * t – t1) / 2, hence:

V = 2 * S / (2 * t – t1) = 2 * 24000 / (2 * 1200 – 240) = 48000/2160 = 22.2 m / s.

Answer: the speed of the train with uniform movement was 22.2 m / s.