Question

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 ${}^{\prime }{v}^{\prime }$ more than that of car $B$. Both the cars start from rest and travel with constant acceleration$a_1$ and $a_2$ respectively. Then ${}^{\prime }{v}^{\prime }$ is equal to

• A. $\sqrt{a_1{a}_{2}t}$
• B. $\sqrt{2a_1{a}_{2}t}$
• C. $\frac{2a_1{a}_2}{a_1+a_2}t$
• D. $\frac{a_1+a_2}{2}t$

Answer 1

The correct option is A $\sqrt{a_1{a}_{2}t}$

For $A$ and $B$ let time taken by $A$ is $t_0$

Given that
$v_A-v_B=v=(a_1-a_2)t_0-a_{2}t\dots \left(i\right)$
$x_B=x_A=\frac{1}{2}{a}_1{t}_0^2=\frac{1}{2}{a}_2(t_0+t{)}^2$
$\sqrt{a_1{t}_0}=\sqrt{a_2}(t_0+t)$
$\left(\sqrt{a_1}\right)-\sqrt{a_2})t_0=\sqrt{a_{2}t}\dots (ii)$

Putting $t_0$in equation
$v=(a_1-a_2)\frac{\sqrt{a_{2}t}}{\sqrt{a_1}-\sqrt{a_2}}-a_{2}t$

$=(\sqrt{a_1}+\sqrt{a_2})\sqrt{a_{2}t}-a_{{}_2}t$

$=\sqrt{a_1{a}_2}t+a_{2}t-a_{2}t$

$v=\sqrt{a_1{a}_2}t$