A train is travelling at a speed of 90 km h-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.

Given that

initial velocity (u) = 90 km/hour = 25 m.s-1

Terminal velocity (v) = 0 m.s-1

Acceleration (a) = -0.5 m.s-2

Find out

How far the train will go before it is brought to rest

Formula

As per the third motion equation, v2-u2=2as

Therefore,

distance traveled by the train (s) =(v2-u2)/2a

s = (02-252)/2(-0.5) meters

= 625 meters

The train must travel 625 meters at an acceleration of -0.5 ms-2 before it reaches the rest position.

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