In a car race on straight road, car $A$ takes a time $t$ less than car $B$ at the finish and passes…
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:
Solution
Given,
Initially both cars are at rest, so $u_1 = u_2 = 0$
Acceleration of car $A$ and $B$ is $a_1$ and $a_2$
Let us assume,
Time of reach to destination of car $A$ is, $t_1 = t_0$
Time of reach to destination of car $B$ is, $t_2 = t_0 + t$
Using second equation of motion, we have
$u_1 t + \frac{1}{2} a_1 t_1^2 = u_2 t + \frac{1}{2} a_2 t_2^2$
$\Rightarrow 0 \times t + \frac{1}{2} a_1 t_0^2 = 0 \times t + \frac{1}{2} a_2 (t_0 + t)^2$
$\Rightarrow \sqrt{\frac{a_1}{a_2}} t_0 = t_0 + t$
$\Rightarrow t_0 = \frac{t}{\sqrt{\frac{a_1}{a_2}} - 1}$
From first equation of motion, we have
$v_1 = a_1 t_0$ and $v_2 = a_2 (t_0 + t)$
$\Rightarrow v = v_1 - v_2 = (a_1 - a_2) t_0 - a_2 t$
$\Rightarrow v = (a_1 - a_2) \frac{t}{\sqrt{\frac{a_1}{a_2}} - 1} - a_2 t$
$\Rightarrow v = t \left( \frac{a_1 \sqrt{a_2} - a_2 \sqrt{a_1}}{\sqrt{a_1} - \sqrt{a_2}} \right)$
$\Rightarrow v = t \left( \frac{a_1 \sqrt{a_2} - a_2 \sqrt{a_1}}{\sqrt{a_1} - \sqrt{a_2}} \right)$
$\Rightarrow v = \sqrt{a_1 a_2} t$