Question

An express train is moving with a velocity $v_1$. Its driver finds another train is moving on the same track in the same direction with velocity $v_2$. To escape collision driver applies a retardation $a$ on the train. The minimum time of escaping collision be

• A. $t=\frac{(v_1-v_2)}{a}$
• B. $t=\frac{({v_1}^2-{v_2}^2)}{2}$
• C. none
• D. both

Answer 1

The correct option is A

$t=\frac{(v_1-v_2)}{a}$

Given:

The velocity of the express train = $v_1$

The velocity of the other train = $v_2$

Retardation of the express train = $a$

Both the trains are moving in the same direction.

Calculating the relative velocity between the two trains:

Initial relative velocity will be $=v_1–v_2$

Now, the express train gets retarded since the driver applies brakes. So both the trains will move with the same velocity i.e $=v_1–v_2$

So, the final velocity between them, $v_1–v_2=0$

Finding the minimum time for escaping collision:

Applying the equation of motion

$v=u+at$

Here, final relative velocity between the two trains $=0$

Initial relative velocity, $u=v_1-v_2$

Now, substituting the values

$0=(v_1–v_2)+at$

Since the train is retarded, put negative sign for $a$

$0=(v_1–v_2)-at$

$at=(v_1–v_2)$

We get,

$t=\frac{(v_1–v_2)}{a}$

Hence, the answer is (A) $t=\frac{(v_1–v_2)}{a}$