Position (in m) of a particle moving on a straight line varies with time (in sec) as `x=t^(3)//3-3t^(2)+8t+4 (m)`. Consider the motion of the particle from t=0 to t=5 sec. `S_(1)` is the total distance travelled and `S_(2)` is the distance travelled during retardation. if `s_(1)//s_(2)=((3alpha+2))/11` the find `alpha`.

Explanation

`x=t^(3)//3-3t^(2)+8t+4`
`v=t^(2)-6t+8=(t-2)(t-4)`
`a=2(t-3)`

`S_(1)=((32)/(3)xx4)+((32)/(3)-(28)/(3))+((32)/(3)-(28)/(3))=(20)/(3)+(8)/(3)=(28)/(3)m`.

`S_(2)=((32)/(3)-4)+(10-(28)/(3))=(20)/(3)+(2)/(3)=(22)/(3)m`
`(S_(1))/(S_(2))=(28)/(22)=(14)/(11)=(3alpha+2)/(11)rArr alpha=4`