A train is travelling at a speed of $108 k m / h$ . Brakes are applied so as to produce a uniform retardation of $1 m / s^{2}$ . Find how far the train will go before it is brought to rest :

800 m

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450 m

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1000 m

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1050 m

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Solution

The correct option is B $450 m$
Given,
$u = 108 k m / h = \frac{108 \times 1000}{60 \times 60} = 30 m / s$
$a = - 1 m / s^{2}$
$v = 0 m / s$
From 3rd equation of motion,
$2 a s = v^{2} - u^{2}$
$- 2 \times 1 \times s = 0^{2} - 30 \times 30$
$s = \frac{- 30 \times 30}{- 2 \times 1}$
$s = 450 m$
The correct option is B.

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An express train going towards Delhi is travelling at a speed of $90 k m p h$ . 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.

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A train is travelling at a speed of 108 km/h. Brakes are applied so as to produce a uniform retardation of 1 m/s2. Find how far the train will go before it is brought to rest :

The correct option is B 450 mGiven,u=108km/h=108×100060×60=30m/sa=−1m/s2v=0m/sFrom 3rd equation of motion,2as=v2−u2−2×1×s=02−30×30s=−30×30−2×1s=450mThe correct option is B.

A train is travelling at a speed of $108 k m / h$ . Brakes are applied so as to produce a uniform retardation of $1 m / s^{2}$ . Find how far the train will go before it is brought to rest :