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 $'v'$ more than that of 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} t $
• B. $\sqrt{2 a_1 a_2 } t$
• C. $\frac{2 a_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 $