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In a car race, car A takes a time t less...

In a car race, car A takes a time t less than car B at the finish and passes the finishing point with speed v more than that of the car B. Assuming that both the cars start from rest and travel with constant acceleration `a_1 and a_2` respectively. Show that `v=sqrt (a_1 a_2) t.`

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Let s be the distance covered by each car. Let the times taken by the two cars to complete the journey be `t_1` and `t_2` and their velocities at the finishing point be `v_1` and `v_2` respectively.
According to the given problem,
`v_1-v_2=v` and `t_2-t_1=t`
Now, `v/t=(v_1-v_2)/(t_2-t_1)=(sqrt(2a_1s)-sqrt(2a_2s))/(sqrt((2s)/a_2)-sqrt((2s)/a_1))=(sqrta_1-sqrta_2)/(sqrt(1/a_2)-sqrt(1/a_1))`
`therefore v/t =sqrt(a_1a_2)`
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