Question

In a car race on straight road, car A takes a times t less than car B at the finish and passes finishing point with a speed 'v' more than that of car B. Both the car start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then 'v' is equal to

• A. $\frac{a_1+a_2}{2}t$
• 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)

$\sqrt{a_1{a}_{2}t}$

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