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 a1 and a2 respectively. Then ‘v’ is equal to :

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$\sqrt{2 a_{1} a_{2} t}$

$\frac{2 a_{1} a_{2}}{a_{1} + a_{2}} t$

$\frac{a_{1} + a_{2}}{2} t$

$\sqrt{a_{1} a_{2} t}$

answer is D.

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Detailed Solution

For A & B let time taken by A is t₀

from question

V_A − V_B = V = (a₁ − a₂)t₀ − a₂t

x_B = x_A = 1/2a₁t₀² = 1/2a₂(t₀ + t)²

$\Rightarrow (\sqrt{a_{1}} - \sqrt{a_{2}} ) t_{0} = a_{2} t$

putting t₀ in equation

$v = (a_{1} - a_{2}) \frac{\sqrt{a_{2}} t}{\sqrt{a_{1}} - \sqrt{a_{2}}} - a 2 t$

$\sqrt{a_{1} a_{2} t}$

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