In a car race, car `A` takes `4 s` less than can `B` at the finish and passes the finishing point with a velocity `v` more than the car `B` . Assuming that the cars start form rest and travel with constant acceleration `a_(1)=4 m s^(-2)` and `a_(2) =1 m s^(-2)` respectively, find the velocity of `v` in m `s^(-1)`.

Answer

8

Explanation

`t_(1)=t_(2)-t,v_(1)v_(2)=v,S=(1)/(2)a_(1)t_(1)^(2),S=(1)/(2)a_(2)t_(2)^(2)`
`v_(1)=a_(1)t_(1),v_(2)=a_(2)t_(2)impliesv_(2)+v=a_(1)t_(1)`
`impliesa_(2)t_(2)+v=a_(1)t_(1)=a_(1)t_(2)impliest_(2)=(v+a_(1)t)/(a_(1)-a_(2))`
`sqrt((a_(2))/(a_(1)))=(t_(1))/(t_(2))=1-(t)/(t_(2))rArr sqrt((a_(2))/(a_(1)))=1-(t(a_(1)-a_(2)))/((v+a_(1)t))`
`implies(sqrt(a_(2)))/(sqrt(a_(1)))=(v+a_(2)t)/(v+a_(1)t)impliessqrt(a_(2))v+a_(1)sqrt(a_(2)t)=vsqrt(a_(1))+a_(2)sqrt(a_(1)t)`
`impliesv=(sqrt(a_(1)a_(2)))t=8ms^(-1)`