Question

In a car race, car `A` takes `t_0` time less to finish than car `B` and passes the finishing point with a velocity `v_0` more than car `B`. The cars start from rest and travel with constant accelerations `a_1 and a_2`. Then the ratio `(v_0)/(t_0)` is equal to.

Answer 1

The distance covered by both cars is same Thus, `s_(1)=s_(2)=s`
If the cars take time `t_(1)` and `t_(2)` for the race and their velocities at the end of race be `v_(1)` and `v_(2)`, then it is given that
`v_(1)-v_(2)=v` and `t_(2)-t_(1)=t`
Now, `(v)/(t)=(v_(1)-v_(2))/(t_(2)-t_(1))=(sqrt(2a_(1)s)-sqrt(2a_(2)s))/(sqrt((2s)/(a_(2)))-sqrt((2s)/(a_(1))))`
`=(sqrta_(1)-a_(2))/(sqrt((1)/(a_(2))-sqrt((1)/(a_(1))))) :. =sqrt(a_(1)a_(2))`