In a car race on a straight road,car A takes | vedclass

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(C) Let the distance of the race be $L$. For car $A$,$L = \frac{1}{2}a_1 t_1^2$,so $t_1 = \sqrt{\frac{2L}{a_1}}$.For car $B$,$L = \frac{1}{2}a_2 t_2^2

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$\sqrt{a_1a_2}t$

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(C) Let the distance of the race be $L$. For car $A$,$L = \frac{1}{2}a_1 t_1^2$,so $t_1 = \sqrt{\frac{2L}{a_1}}$.For car $B$,$L = \frac{1}{2}a_2 t_2^2$,so $t_2 = \sqrt{\frac{2L}{a_2}}$.Given $t_2 - t_1 = t$,so $\sqrt{2L} \left( \frac{1}{\sqrt{a_2}} - \frac{1}{\sqrt{a_1}} \right) = t \Rightarrow \sqrt{2L} \left( \frac{\sqrt{a_1} - \sqrt{a_2}}{\sqrt{a_1 a_2}} \right) = t$.Final speeds are $v_A = a_1 t_1 = a_1 \sqrt{\frac{2L}{a_1}} = \sqrt{2a_1 L}$ and $v_B = a_2 t_2 = a_2 \sqrt{\frac{2L}{a_2}} = \sqrt{2a_2 L}$.Given $v_A - v_B = v$,so $\sqrt{2L} (\sqrt{a_1} - \sqrt{a_2}) = v$.Dividing the two equations: $\frac{v}{t} = \frac{\sqrt{2L}(\sqrt{a_1} - \sqrt{a_2})}{\sqrt{2L} \frac{(\sqrt{a_1} - \sqrt{a_2})}{\sqrt{a_1 a_2}}} = \sqrt{a_1 a_2}$.Therefore,$v = \sqrt{a_1 a_2} t$.

In a car race on a straight road,car $A$ takes a time $t$ less than car $B$ at the finish and passes the finishing point with a speed $v$ more than that 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

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