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Q.

In a car race, car A takes $t_{0}$ time less to finish than car B and passes the finishing point with a velocity $v_{0}$ more than car B. The cars start from rest and travel with constant accelerations $a_{1}$ and $a_{2} .$ Then the ratio $\frac{v_{0}}{t_{0}}$ is equal to

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a

$\frac{a_{1}^{2}}{a_{2}}$

b

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

c

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

d

$\frac{a_{2}^{2}}{a_{1}}$

answer is C.

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Detailed Solution

Let s be the distance travelled by each car.
$\sqrt{2 a_{1} s} - \sqrt{2 a_{2} s} = v_{0}$ and $\sqrt{\frac{2 s}{a_{2}}} - \sqrt{\frac{2 s}{a_{1}}} = t_{0}$
$∴ \frac{v_{0}}{t_{0}} = \frac{\sqrt{a_{1}} - \sqrt{a_{2}}}{\frac{1}{\sqrt{a_{2}}} - \frac{1}{\sqrt{a_{1}}}} = \sqrt{a_{1} a_{2}}$

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