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

Answer 1

Answer: (D)

For $A$ & $B$ let time taken by $A$ is $t_0$ from ques.
$v_A-v_B=v=\left(a_1-a_2\right)t_0-a_{2}t$ ....(i)
$x_B=x_A=\frac{1}{2}{a}_1{t}_0^2=\frac{1}{2}{a}_2{\left(t_0+t\right)}^2$
$\Rightarrow \sqrt{a_1{t}_0}=\sqrt{a_2}\left(t_0+t\right)$
$\Rightarrow \left(\sqrt{a_2}-\sqrt{a_2}\right)t_0=\sqrt{a_{2}t}$ ......(ii)
putting $t_0$ in equation
$v=\left(a_1-a_2\right)\frac{\sqrt{a_{2}t}}{\sqrt{a_1}-\sqrt{a_2}}-a_{2}t$
$=\left(\sqrt{a_1}+\sqrt{a_2}\right)\sqrt{a_{2}t}-a_{2}t\Rightarrow v=\sqrt{a_1{a}_{2}t}$
$\Rightarrow \sqrt{a_1{a}_{2}t}+a_{2}t-a_{2}t$