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.

$600 m$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$725 m$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$625 m$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$500 m$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$625 m$

Here,
initial velocity, $u = 90 k m h^{- 1}$
$= \frac{90 \times 1000}{3600} m s^{- 1}$
$= 25 m s^{- 1}$
Acceleration, $a = - 0.5 m s^{- 2}$
Final Velocity, $v = 0$
Using $v^{2} = u^{2} + 2 a s,$
$0 = 25^{2} + (2 \times - 0.5 \times s)$
$\Rightarrow s = 625 m$
Hence, the distance covered is $625 m$

Suggest Corrections

Similar questions

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.

Join BYJU'S Learning Program

An express train going towards Delhi is travelling at a speed of 90 kmh−1. 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.

The correct option is C 625 m Here, initial velocity, u=90 kmh−1 =90×10003600 ms−1 =25 ms−1 Acceleration, a= –0.5 ms−2 Final Velocity, v=0 Using 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 $90 k m h^{- 1}$ . 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.