In a car race, car `A` takes `4 s` less than can `B` at the finish and passes the finishing point with a velcity `v` more than the car `B` . Assumung that the cars start form restand travel with constant accleration `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_(2)^(1), S=(1)/(2) a_(2) t_(2)^(2)`
`v_(1) =a_(1)t_(1), v_(2)= a_(2)t_(2) rArr v_(2)+v=a_(1)t_(1)`
`rArr a_(2)t_(2) +v=a_(1)t_(1) =a_(1)t_(2) rArr t_(2) =(v+a_(1)+a_(1)t)/(a_(1)-a_(2))`
`sqrt((a_(2))/(a_(1))) =(t_(1))/(t_(2)) =1 -(1)/(t_(1)) rArr sqrt((a_(2))/(a_(1))) =1 -(t(a_(1)) -a_(2))/((v+a_(1)a_(1)t))`
`rArr (sqrt(a_(2)))/(sqrt a_(1))=(v+a_(2)t)/(v+a_(1)t)rArr sqrt(a_(2)) v+a_(1)sqrta_(2)t=vsqrta_(1)+a_(1)sqrta_(1)t`
` rArr v=(sqrt(a_(1)a_(2)))t=8 ms^(-1)`.