A particle is moving such that `s=t^(3)-6t^(2)+18t+9`, where s is in meters and t is in meters and t is in seconds. Find the minimum velocity attained by the particle.

Explanation

`s=t^(3)-6t^(2)+18t+9`
`v=(ds)/(dt)=3t^(2)-12t+18`
For v to be minimum or maximum,
`a=(ds)/(dt)=6t-12=0impliest=2 s`
At `t=2 s`,
`(d^(2)v)/(dt^(2))=6gt0`, i.e. v is minimum at `t=2 s`
`v_(min)=3(2)^(2)-12(2)+18=6 m//s`