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

Answer 1

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



Let time taken by $A$ to reach finishing point is $t_0$
$\therefore$ time taken by $B$ to reach finishing point $=t_0+t$
Using equation of motion
$v=u+at$
So velocity of car $A$ at finishing point-
$v_A=0+a_1{t}_0=a_1{t}_0$
So velocity of car $A$ at finishing point-
$v_B=0+a_2(t_0+t_1)=a_2(t_0+t_1)$

$v_A-v_b=v$ {Given}
$\Rightarrow v=a_1{t}_0-a_2(t_0+t)=(a_1-a_2)t_0-a_{2}t....\left(i\right)$
Distance travelled by both the cars-
$s=ut+\frac{1}{2}a{t}^2$
So,
$x_B=x_A=\frac{1}{2}{a}_1{t}_0^2=\frac{1}{2}{a}_2(t_0+t{)}^2$
$\Rightarrow t_0\sqrt{a_1}=(t_0+t)\sqrt{a_2}$
$\Rightarrow (\sqrt{a_1}-\sqrt{a_2})t_0=t\sqrt{a_2}$
$\Rightarrow t_0=\frac{t\sqrt{a_2}}{\sqrt{a_1}-\sqrt{a_2}}$

Putting this value of $t_0$ in equation $\left(i\right)$

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