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Question: The driver of a train moving with a speed v<sub>1</sub>sights another train at a distance s, ahead o...

The driver of a train moving with a speed v1sights another train at a distance s, ahead of him moving in the same direction with a slower speed v2,. He applies the brakes and gives a constant deceleration a to his train. For no collision, s is

A

=(v2−v1)22a= \frac{\left( v_{2} - v_{1} \right)^{2}}{2a}

B

>(v1−v2)22a> \frac{\left( v_{1} - v_{2} \right)^{2}}{2a}

C

<(v1−v2)2a< \frac{\left( v_{1} - v_{2} \right)}{2a}

D

<v1−v22a< \frac{v_{1} - v_{2}}{2a}

Answer

>(v1−v2)22a> \frac{\left( v_{1} - v_{2} \right)^{2}}{2a}

Explanation

Solution

To avoid collision, the faster train should come to rest after covering a distance d. Using v2 – u2 = 2aS we get

0 – Vn2 = 2(-a)d

(where Vn = v1 – v2)

or d = Vn22a=(v1−v2)22a\frac{V_{n}^{2}}{2a} = \frac{\left( v_{1} - v_{2} \right)^{2}}{2a}

Collision can be avoided if

d > (v1−v2)22a\frac{\left( v_{1} - v_{2} \right)^{2}}{2a}