Question

In a car race on straight road, car $A$ takes a time $t$ less than car $B$ at the finishing point and passes the finishing point with a speed ${}^{\prime }{v}^{\prime }$ more than that car $B$. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then $v$ is equal to

• A. $\frac{a_1+a_2}{2}$
• 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

The correct option is D $\sqrt{a_1{a}_2}t$

Let the total time taken by car $A$ is $t_0$.


From question,

$v_A-v_B=v=a_1{t}_0-(t+t_0)a_2$

$\Rightarrow v=(a_1-a_2)t_0-a_{2}t....\left(i\right)$

And,

$x_B=x_A$

$\Rightarrow \frac{1}{2}{a}_1{t}_0^2=\frac{1}{2}{a}_2(t_0+t{)}^2$

$\sqrt{a_1}{t}_0=\sqrt{a_2}(t_0+t)$

$({\sqrt{a}}_1-\sqrt{a_2})t_0=\sqrt{a_2}t$

$t_0=\frac{\sqrt{a_2}t}{({\sqrt{a}}_1-\sqrt{a_2})}....\left(ii\right)$

Putting $t_0$ in equation $\left(i\right)$,

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

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

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

Hence, $\text{(}\text{D})$ is the correct answer.