Question

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 1

Let A takes `t_(1)` second, then according to the given problem B will take `(t_(1)+t)` seconds. Further, let `v_(1)` be the velocity of B at finishing point, then velocity of A will be `(v_(1)+v)`. Writing equations of motion for A and B.
`v_(1)+v=a_(1)t_(1)` ...(i)
`v_(1)=a_(2)(t_(1)+t)` ...(ii)
From these two equations, we get
`v=(a_(1)-a_(2))t_(1)-a_(2)t` ...(iii)
Total distance travelled by both the cars is equal.
or `s_(A)=s_(B)`
or `(1)/(2)a_(1)t_(1)^(2)=(1)/(2)a_(2)(t_(1)+t)^(2)` or `t_(1)=(sqrt(a_(2))t)/(sqrt(a_(1))-sqrt(a_(2)))`
Substituting this value of `t_(1)` in Eq. (iii), we get the desired result.
or `v=(sqrt(a_(1)a_(2)))t`