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In a car race, car A takes less time to finish than car B and passes finishing point with a velocity $v_0$ more than velocity of car B and $t_0$ time earlier than car B. The cars starts from rest and travel with acceleration $a_1$ and $a_2$ respectively. Then ratio of $\large\frac{v_0}{t_0}$ is

\[(a)\;\frac{a_1^2}{a_2}\quad (b)\;\frac{a_1+a_2}{2} \quad (c)\;\frac{{a_2}^2}{a_1} \quad (d)\sqrt {a_1a_2}\]

1 Answer

Let s be the distance travelled by each car
$v_A=\sqrt {2 a_1 s}$
Also $v_B=\sqrt {2 a_2 s}$ 
$t_1=\sqrt {\large\frac{2s}{a_1}}$
$t_2=\sqrt {\large\frac{2s}{a_2}}$
$\sqrt {2 a_1 s}-\sqrt {2 a_2 s}=v_0$
$\sqrt {\large\frac{2s}{a_2}}-\sqrt {\large\frac{2s}{a_1}}=t_0$
$\large\frac{v_0}{t_0}=\frac{\sqrt {a_1}-\sqrt {a_2}}{\Large\frac{1}{\sqrt{a_2}}-\frac{1}{\sqrt{a_1}}}$$=\sqrt{a_1a_2}$
Hence d is the correct answer.


answered Jun 26, 2013 by meena.p
edited Jul 25, 2014 by thagee.vedartham

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