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 is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?

Given that,

$x=9t^2-t^3$.....(I)

We know that,

$v=\frac{dx}{dt}$

The maximum speed is

$v=18t-3t^2$

$v$ maximum, at $\frac{dv}{dt}=0$

$\frac{dv}{dt}=18-6t$

The time will be

$0=18-6t$

$t=3sec$

The position of the particle at 3 sec

Put the value of t in equation (I)

$x=9\times 9-27$

$x=81-27$

$x=54m$

The position of the particle will be $54m$.

Hence, Option A is correct