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 $v_2$ $(v_2<v_1)$. To escape collision, driver applies a retardation a on the train. The minimum time to escape the collision is

• A. $\frac{(v_1-v_2)}{a}$
• B. $\frac{(v_1^2-v_2^2)}{2}$
• C. $\frac{\left(v_2\right)}{a}$
• D. $\frac{(v_1+v_2)}{a}$

Answer 1

The correct option is A $\frac{(v_1-v_2)}{a}$

As the trains moving in the same direction so the initial relative speed
$(v_1-v_2)$ and by applying retardation final relative speed become zero.
Initial relative velocity, $u_{rel}=v_1-v_2$
Final relative velocity, $v_{rel}=0$
Relative acceleration, $a_{rel}=-a-0=-a$
Using first equation of motion,
$v_{rel}=u_{rel}+a_{rel}t$
$0=(v_1-v_2)-at$
$\Rightarrow t=\frac{(v_1-v_2)}{a}$