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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 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:
Explanation
Let time taken by A to reach finishing point is $t_{0} \therefore$ Time taken by B to reach finishing point = $t_{0} + t$ $v_{A} - v_{B} = v$ $\Rightarrow v = a_{1}t_{0} - a_{2}(t_{0} + t) = (a_{1} - a_{2})t_{0} - a_{2}t$ $x_{B} = x_{A} = \frac{1}{2}a_{1}t_{0}^{2} = \frac{1}{2}a_{2}(t_{0} + t)^{2}$ $\Rightarrow \sqrt{a}_{1} t_{0} = \sqrt{a}_{2}(t_{0} + t) \Rightarrow (\sqrt{a}_{1} - \sqrt{a}_{2})t_{0} = \sqrt{a}_{2} t$ $\Rightarrow t_{0} = \sqrt{a}_{2} t/(\sqrt{a}_{1} - \sqrt{a}_{2})$ Putting this value of $t_{0}$ in equation (i) $v = (a_{1} - a_{2}) \sqrt{a}_{2} t/(\sqrt{a}_{1} - \sqrt{a}_{2}) - a_{2}t$ = $(\sqrt{a}_{1} + \sqrt{a}_{2})\sqrt{a}_{2} t - a_{2}t = \sqrt{a_{1}a_{2}} t + a_{2}t - a_{2}t$ or, $v = \sqrt{a_{1}a_{2}} t$