A particle is moving with velocity `v=t^(3)-6t^(2)+4`, where v is in `m//s` and t is in seconds. At what time will the velocity be maximum//minimum and what is it equal to?

Explanation

`v=t^(3)-6t^(2)+4`
For v to be maximum or minimum,
`a=(dv)/(dt)=3t^(2)-12t=3t(t-4)=0`
`t=0,4 s`
`(d^(2)v)/(dt^(2))=6t-12`
At `t=0, (d^(2)v)/(dt^(2))=-12lt0`, v is maximum
`v_(max)=4 m//s`
At `t=4 s, (d^(2)v)/(dt^(2))=6(4)-12=12gt0`, v is minimum
`v_(min)=(4)^(3)-6(4)^(2)+4=-28 m//s`