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.

Options

A. `29 ms^(-1)`
B. `5 ms^(-1)`
C. `6 ms^(-1)`
D. `12 ms^(-1)`

Answer

C

Explanation

`s=t^(3)-6t^(2)+18t+9`
`upsilon=(ds)/(dt)=3t^(2)-12t+18`
For `upsilon` be minimum or maximum
`a=(d upsilon)/(dt)=6t-12=0 rArr t=2s`
At t = 2s
`(d^(2)upsilon)/(dt^(2))=6gt 0,` i.e., `upsilon` is minimum at t = 2 s
`upsilon_(min)=3(2)^(2)-12(2)+18`
= 6 m/s