The passenger train braked, changing its speed from 54 km / h to 5 m / s.

| September 10, 2021 | education

The passenger train braked, changing its speed from 54 km / h to 5 m / s. Determine the distance that the train traveled when braking if the acceleration of the train was constant and equal to 4 m / s2.

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

V = 5 m / s.

a = 4 m / s2.

S -?

For uniformly accelerated movement of a passenger train, the distance traveled by it during braking is expressed by the formula: S = V0 * t – a * t ^ 2/2, where V0 is the initial speed of the train, a is the acceleration of the train during braking, t is the braking time.

We express the braking time of a passenger train t from the definition of the body acceleration: а = (V0 – V) / t.

t = (V0 – V) / a.

t = (15 m / s – 5 m / s) / 4 m / s2 = 2.5 s.

S = 15 m / s * 2.5 s – 4 m / s ^ 2 * (2.5 s) ^ 2/2 = 25 m.

Answer: the stopping distance of the passenger train was S = 25 m.



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