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. `v_(0) = sqrt((alpha^(2)-beta^(2))/(2)t_(0)`
B. `v_(0) = ((alpha+beta)/(2))t_(0)`
C. `v_(0) = sqrt(alpha beta)t_(0)`
D. `v_(0) = sqrt((alpha^(2)+beta^(2))/(2)t_(0)`

Answer

C

Explanation

In a car race ……….

P.given `t_(Q) - t_(p) = t_(0) - 0 V_(p) - V_(Q) = V_(0)` ………(2)
`x =0 xxt_(p) + (1)/(2)alpha t_(p)^(2) implies t_(p) = sqrt((2x)/(alpha))`
`v_(p)^(2) = 0^(2) + 2alpha x implies V_(p) = sqrt(2alpha x)`
`Q. x = 0 xxt_(Q) + (1)/(2)beta t_(Q)^(2)`
`t_(Q) = sqrt((2x)/(beta))`
now ` (V_(p) - V_(Q))/(t_(Q) - t_(p)) = (v_(0))/(t_(0)) implies (sqrt(2alpha x)-sqrt(2beta x))/(sqrt((2x)/(beta))-sqrt((2x)/(alpha))) = (v_(0))/(t_(0))`
`implies v_(0) = sqrt(alpha beta n t_(0))`