Physics - Kinematics Question with Solution | TestHub

Given, 
Initially both car is at rest, so $u_1=u_2=0$
Acceleration of car $A \& B$ is $a_1 \&\hspace{0.17em}{a}_2$

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
$u_{1}t+\frac{1}{2}{a}_1{t}_1^2=u_{2}t+\frac{1}{2}{a}_2{t_2}^2$
$\Rightarrow 0\times t+\frac{1}{2}{a}_1{t}_0^2=0\times t+\frac{1}{2}{a}_2{\left(t_0+t\right)}^2$
$\Rightarrow \sqrt{\frac{a_1}{a_2}}{t}_0=t_0+t$
$\Rightarrow t_0=\frac{t}{\sqrt{\frac{a_1}{a_2}}-1}$

From first equation of motion, we have
$\begin{array}{cc}{v}_1=a_1{t}_0 \&& v_2=a_2\left(t_0+t\right)\end{array}$
$\Rightarrow v=v_1-v_2=\left(a_1-a_2\right)t_0-a_{2}t$
$\Rightarrow v=\left(a_1-a_2\right)\frac{t}{\sqrt{\frac{a_1}{a_2}}-1}-a_{2}t$
$\Rightarrow v=t\left(\frac{a_1\sqrt{a_2}}{{\sqrt{a}}_1-{\sqrt{a}}_2}-\frac{a_2{\sqrt{a}}_2}{{\sqrt{a}}_1-{\sqrt{a}}_2}-a_2\right)$
$\Rightarrow v=t\left(\frac{a_1\sqrt{a_2}-a_2{\sqrt{a}}_1}{{\sqrt{a}}_1-{\sqrt{a}}_2}\right)$
$\Rightarrow v=\sqrt{a_1.a_2} t$