Question

When a driver of Shatabdi Express, running with velocity 108 km/hr, sights a goods train

going ahead of him at a distance 50 m in the same direction on the same track, running

with velocity 72 km/hr, he applies brakes. In order to avoid an accident, what should be the

magnitude of the deceleration produced by the brakes ?

(A)5 ms^–2 (B)1 ms^–2 (C)3 ms^–2 (D)None of these​

Answer 1

Answer:

its d caus the numerical i not showing the option

Answer 2

Answer:

The magnitude of the deceleration produced by the brakes should be equal to 1ms⁻² in order to avoid an accident.

Explanation:

Given,

The velocity of Shatabdi express, $v_{1}=108km/hr$

The velocity of goods train $v_{2}=72km/hr$

Now, the relative initial velocity Shatabdi Express of  will be:

v₀ = v₁₂ = v₁ - v₂ = 108 - 72

v₀ = 36km/hr

Now change km/hr into m/s:

$v_{o}=(\frac{36*1000}{3600} )ms^{-1}$

$v_{o}=10ms^{-1}$

The distance between the two trains $S=50m$

The final velocity of Shatabdi express will be zero, $v=0$

From the second equation of motion:

$v^{2}=u^{2}+2aS$

Here, $v^{2}=v_{o}^{2}+2aS$                                          ....................(1)

Put the values in Eq.(1);

$0=(10)^{2}+2a(50)$

$100a=-100$

$a=-1ms^{-2}$

The deceleration produced by the brakes will be equal to 1ms⁻². The negative sign shows the retardation.