Question

A train is travelling at a speed of 90 kmh^-1 Brakes are applied so as to produce a uniform acceleration of -0.5 ms^-2. Find how far the train will go before it is brought to rest.

Answer 1

Let the initial speed of the train be u = 90 Km/h

= 25 m/s.

Final speed of the train, v = 0 (train comes to rest)

acceleration a = 0.5 ms^-2

As per 3^rd law of motion

v² = u² + 2as

(0)² = (25)² + 2(0.5)s

s = train travelled distance

∴ s = \(\frac{(25)^2}{2(0.5)}\) = 625 m

Train will travel 625 km before it is brought to rest.