Question

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.`

Answer 1

`v_(P)-v_(Q)=v_(0)`
`t_(Q)-t_(P)=t_(0)`
`(v_(P)-v_(Q))/(t_(Q)-t_(P))=(sqrt(2alphad)-sqrt(2betad))/(sqrt((2d)/beta)-sqrt((2d)/alpha))=v_(0)/t_(0)`
`{:[("Car P": d=1/2alpha_P^2rArrt_P=sqrt((2d)/(alpha))),(v_P^2=2alphadrArrv_P=sqrt(2alphad)),("Similarly for car Q.")]:}`
`(sqrt(alpha)-sqrt(beta))/(1/sqrt(beta)-1/sqrt(alpha))=(sqrt(alpha)-sqrt(beta))/(((sqrt(alpha)-sqrt(beta))/sqrt(alpha beta)))=v_(0)/t_(0)`
`v_(0)=sqrt(alphabeta)t_(0)`