Q. 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 :

Solution:

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 $

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