A car, starting from rest, accelerates at the rate f through a distance S, then contunes at constant speed for time t and then come to rest with retardation `(f)/(2)` if the total distance travelled is 15sS then calculate the value of S in term of f and t.

Explanation

Let constant speed be `v_(m)`
for time `t_(1), v_(m)= ft_(1) and S=(1)/(2)ft_(1)^(2)`
for time, `t_(2) O=v_(m) -(f)(2) t_(2)rArr t_(1)=2t_(1)`
`S-(3)=(1)/(2)((f)/(2))t_(2)^(2)=((f)/(4))(4t_(1)^(2)) =ft_(1)^(2)=2S`
Therefore `S+v_(m)t+2S=15SrArrv_(m)t=12SrArr ft_(1)t=12S`
`rArrf((2S)/(f))^(1//2)t=12S rArr 2Sf= (144S^(2))/(t^(2)) rArr S=(ft^(2))/(72)`