The position x of a particle w.r.t. time t along x-axis is given by a `x = 9 t^(2)-t^(3)`, where x is in metre and t in sec. 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`