Question 2
A train 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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Solution

Given, initial speed of the train, $u = 90 k m p h = 90 \times \frac{5}{18} = 25 m s^{- 1}$ , final speed of the train, $v = 0$ (Finally the train comes to rest)
and acceleration $= - 0.5 m s^{- 2}$
Let $s$ be the distance covered.
According to the third equation of motion, $v^{2} = u^{2} + 2 a s$
$(0)^{2} = (25)^{2} + (2 \times - 0.5 \times s)$
$\Rightarrow$ $s = \frac{25^{2}}{2 \times 0.5} = 625 m$

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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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Question 2 A train is travelling at a speed of 90 kmph. 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, initial speed of the train, u=90 kmph=90×518=25 ms−1, final speed of the train, v=0 (Finally the train comes to rest) and acceleration =−0.5 ms−2 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

Question 2
A train 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.