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

Options

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

Answer

C

Explanation

`x=9t^(2)-t^(3)" ":. v=18t-3t^(2)`
`implies (dv)/(dt)=18-6t` for maximum speed `(dv)/(dt)=0 , and (d^(2)v)/(dt^(2))` negative so `18-6t=0 implies t=3s `
at `t=3s, x=9(3)^(2)-(3)^(3)=81-27=54m`