1.

In a car race, `A` takes a time of `t` s, less than car `B` at the finish and passes the finishing point with a velocity `v` more than car `B`. Assuming that the cars start from rest and travel with constant accelerations `a_(1)` and `a_(2)`. Respectively, show that `v=sqrt(a_(1) a_(2)t)`.

Answer» Let (S) be the distance to be covered by each car. Let ` t_(1), t_(2)` be the times taken by cars ` A and B` to complete their journey and ` v_(1), v_(2)` be their velocitues at the finishing pont. According to given problem `
` v_(1) -v_(2) =v and ` t_(2)-t_(1) =t`
As, distance travelled =average velcity x time
interval
when ` u=0, so`
` S= (0+ v_(1))/2 t_(1) =(0+v_(2))/2 t_(2) or S=(v_(1) t_(1)/2 =(v_(2) t_(2)/2`
or ` v_(1) =2 S//t_(1) and v_(2) a_(2) t_(2)^(2)`
or ` t_(1)=sqrt(2 S)/a_(1) and t_(2) =sqrt (2 S)/a_(2)`
Now `v/t =(v_(1)-v_(2))/(t_(2)-t_(1))=((2 S//t_(1)) -(2 S//t_(2)))/(t)(2)-t_(1) =(2 S)/ t_(1) t_(2)`
` (2 S)/(sqrt(2 S)/a_(1) xx sqrt92 S)/a_(2) = sqrta_(1) a_(2)`
or `v=t sqer a_(1) a_(2)`.


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