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 m s-2. Find how far the train will go before it is brought to rest.

Answers (1)

(We know, 1\ km = 1000\ m ; 1\ hr = 3600\ s)

Given, Initial speed of the train,u  = 90\ kmh^{-1} = \frac{90\times1000}{3600} = 25\ ms^{-1}

Acceleration of the train,a = -0.5\ ms^{-2}   (Negative sign implies retardation)

Since, the train has to be brought to rest, final speed of the train, v0\ ms^{-1}

We know, v^2 = u^2 + 2as

\\ \implies 0^2 = 25^2 + 2(-0.5)s \\ \implies 0 = 625 -s \\ \implies s = 625\ m

Therefore, the train travels a distance of 625\ m before coming to rest.

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