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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Solution

Given,
initial velocity, $u = 90 k m p h$ = $90 \times \frac{5}{18}$ $= 25 m s^{- 1}$
acceleration, $a = - 0.5 m s^{- 2}$
final velocity, $v = 0$ (brakes are applied)
Let 's' be the distance travelled.
From the third equation of motion,
$v^{2} = u^{2} + 2 a s$

$0 = 25^{2} + (2 \times - 0.5 \times s)$

$\Rightarrow s = 625 m$

Hence the distance covered is $625 m$ .

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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.

Q. 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.

An express train going towards Delhi is travelling at a speed of $90 k m h^{- 1}$ . 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.

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An express train going towards Delhi is travelling at a speed of 90 kmph. Brakes are applied so as to produce a uniform acceleration of –0.5m s−2 . Find how far the train will go before it is brought to rest.

Given, initial velocity, u=90 kmph = 90×518 =25 ms−1 acceleration, a= –0.5 ms−2 final velocity, v=0 (brakes are applied) Let 's' be the distance travelled. From the third equation of motion, v2=u2+2as 0=252+(2×−0.5×s) ⇒ s=625 m Hence the distance covered is 625 m.

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.