In a car race on straight road, car A takes a time t less than car B at the finish and passes 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 :

Explanation

In the car race, the cars start from zero velocity an daccilerate with constant accelerations. Here we' ll discuss an imprtant concept of uniformly accelerated motion. If a body starts from rest, i.e., with zero initial velocity an daccelerates with acceleration (a) travellin distance `s`, its velocity can be given by speed equation `v^(2)u^(2)+2as`.
As we have `u=0 rArr v=sqrt2as`
For the time taken to travel this distance `s`, we use eqution as
` s=ut+(1)/(2)at^(2)`
Hence again we have `u=0. s(1)/(2)at^(2)` or `t=sqrt((2s)/(a))`
In the questions given, car `A` starts with acceleration `a_(1)` and car `B` withe speed `u`, car `A` will reach at time `T-t` and with speed `u+v` as given in the question.
As both the cars start from rest and cover same distance, say `s`, we have
For car A : `v+u=sqrta_(1)s` and `T-t=sqrt((2s)/(q_(1))`
for car B: `u=sqrt(a_(2))s` and `T=sdqrt((2s)/(a_(2))`
From above equations, elinatint the terms of `u` and `T`, we get
`v=sqrt(2a_(1)s)- sqrt(2a_(2)s)`
`t= sqrt((2s)/(a_(2)))-sqrt((2s)/(a_(1)))`.
Diveding the above equations , we get ` v= sqrt(a_(1) a_(2)t)`.