Question

An express train is moving with a velocity $v_1$. Its driver finds that another train is moving on the same track in the same direction with velocity $v_2(v_1>v_2)$. To escape collision, driver applies break to produce retardation $‘a^{\prime }$ on the train. The minimum time to avoid collision will be

• A. $t=\frac{v_1-v_2}{a}$
• B. $t=\frac{v_1^2-v_2^2}{a}$
• C. $t=\frac{2(v_1-v_2)}{a}$
• D. None of the above

Answer 1

The correct option is C $t=\frac{2(v_1-v_2)}{a}$


Assuming the frame of reference at 2, we can say:
Relative velocity of 1 with respect to 2: $v_{12}=v_1-v_2$

We know,
$d=ut-\frac{1}{2}a{t}^2$
At limiting condition, $d=0$
$0=(v_1-v_2)t-\frac{1}{2}a{t}^2$
$\frac{1}{2}a{t}^2=(v_1-v_2)t$

$t=\frac{2(v_1-v_2)}{a}$