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

Answer

C

Explanation

`x=9t^(2)-t^(3) therefore v=18t-3t^(2)`
`rArr=(dv)/(dt)=18-6t`
`rArr(dv)/(dt)=18-6t`
for maximum speed `(dv)/(dt)=0`, and `(d^(2)v)/(dt^(2))=` negative
so `18-6t=0 rArrt=3s`
`x_(t=3s)=9(3)^(2)-(3)^(3)=81-27=54m`