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Question

A train is travelling at a speed of 90 kmph. Brakes are applied so as to produce a uniform acceleration of 0.5 ms2. Find how far the train will go before it is brought to rest.

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Solution

Given:
Initial speed of the train, u=90 kmph=90×518=25 ms1
Final speed of the train, v=0 (as the train finally comes to rest)
Acceleration of the train, a =0.5 ms2

Solving for the distance covered by the train:
Let s be the distance covered.
According to the third equation of motion, v2=u2+2as
(0)2=(25)2+(2×0.5×s)
s=2522×0.5=625 m

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