The position x of a particle with respect to time t along x-axis is given by `x=9t^(2)-t^(3)` where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the +x direction?

Options

A. 24 m
B. 32 m
C. 54 m
D. 81 m

Answer

C

Explanation

`x=9t^(2)-t^(3)`
`(dx)/(dt)=v=18 -3t^(2)`
At the instant where v is maximum
`(dv)/(dt)=0, (dv)/(dt) =18-6t=0`
`t=3, v ` is maximum
`x_(t=3) =9xx3^(2() -3^(3) =54m`