Q 11-02-121JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
In a car race on straight road, car $A$ takes a time $t$ less than car $B$ at the finish and passes finishing point with a speed $v$ more than that of car $B$. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then $v$ is equal to:
Answer: (C) $\sqrt{a_1a_2}\,t$
Let the track length be $d$. Starting from rest, a car with acceleration $a$ takes $\sqrt{2d/a}$ and finishes with speed $\sqrt{2ad}$.
$$v = \sqrt{2d}\,(\sqrt{a_1}-\sqrt{a_2})$$
$$t = \sqrt{2d}\left(\frac{1}{\sqrt{a_2}}-\frac{1}{\sqrt{a_1}}\right) = \sqrt{2d}\,\frac{\sqrt{a_1}-\sqrt{a_2}}{\sqrt{a_1a_2}}$$
Dividing, $\dfrac{v}{t} = \sqrt{a_1a_2}$, so $v = \sqrt{a_1a_2}\,t$.
Solution by Sreeraj P, M.Sc Physics