In a car race on straight road, car A takes a time t less than car B… | JEE Main 2019 (9 Jan, Shift 2)

Q31JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Equations of Kinematics

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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Answer: (c)

Same track length $d$ , from rest: $t_i = \sqrt{\frac{2d}{a_i}}$ , $v_i = \sqrt{2da_i}$ .
$v = v_1 - v_2 = \sqrt{2d}(\sqrt{a_1} - \sqrt{a_2})$
$t = t_2 - t_1 = \sqrt{2d}\,\frac{\sqrt{a_1} - \sqrt{a_2}}{\sqrt{a_1a_2}}$
$\frac{v}{t} = \sqrt{a_1a_2} \Rightarrow v = \sqrt{a_1a_2}\,t$
So the answer is (c) $\sqrt{a_1a_2}\,t$ .

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Equations of Kinematics question from JEE Main 2019 (9 Jan, Shift 2)

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