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

Answer

a

Explanation

`x=9t^(2)-t^(3)` ...........(i)
`therefore v=(dx)/(dt)=(d)/(dt)(9t^(2)-t^(3))=18-3t^(2)`
For speed u to be maximum, the first derivative shou ld be zero and the second derivative shou ld be negative.
`therefore (dv)/(dt)=18-6t` and `(d^(2)v)/(dt^(2))=-6`
=18-6t
t=3s
Thus, the speed will be maximum at t=3s. From eqn. (i).
`x=9(3)^(2)-(3)^(3)=54m`