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.`

Options

A. `(2a_(1)a_(2))/(a_(1)+a_(2))t`
B. `tsqrt(2a_(1)a_(2))`
C. `(a_(1)+a_(2))/(2)t`
D. `tsqrt(a_(1)a_(2))`

Answer

d

Explanation

Let `t_(A)` and `t_(B)` are times taken by cars A and B to reach finish point.
`t_(A)=(t_(B)-t)`
`s=(1)/(2)a_(1)t_(A)^(2)` and `s=(1)/(2)a_(2)t_(B)^(2)`
Difference in velocity,
`v_(A)-v_(B)=(sqrt(2sa_(1))-sqrt(2sa_(2)))=v`
Difference in time
`(t_(A)-t_(B))=(sqrt((2s)/(a_(1)))-sqrt((2s)/(a_(2))))=t`
Solve equation i and ii we get
`v=tsqrt(a_(1)a_(2))`