The train was moving at a speed of v = 54 km / h. Approaching the station, it began to slow down evenly

The train was moving at a speed of v = 54 km / h. Approaching the station, it began to slow down evenly and, after a time t = 3.5 min, stopped. Determine the number of cars on the train if its stopping distance is seven times longer than the length of the entire train. Take the length of the electric locomotive and each carriage equal to l = 15 m.

V0 = 54 km / h = 15 m / s.

t = 3.5 min = 210 s.

V = 0 m / s.

S = 7 * L.

l = 15 m.

n -?

For uniformly accelerated motion, the following formulas are valid: a = (V – V0) / t, S = (V ^ 2 – V0 ^ 2) / 2 * a, where V, V0 are the final and initial speed of motion, t is the time of motion, S is braking distance, a – acceleration during movement.

S = (V ^ 2 – V0 ^ 2) * t / 2 * (V – V0) = (V – V0) * (V + V0) * t / 2 * (V – V0) = (V + V0) * t / 2.

S = (15 m / s + 0 m / s) * 210 s / 2 = 1575 m.

S = 7 * L.

Since the length of the entire train L is determined by the formula: L = n * l, where n is the number of cars, l is the length of 1 car.

S = 7 * n * l.

n = S / 7 * l.

n = 1575 m / 7 * 15 m = 15.

Answer: the train consists of n = 15 cars.