Step 1: Find the Velocity ($v$): Velocity is the first derivative of position with respect to time:
$$v = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3)$$
$$v = 18t - 3t^2$$
Step 2: Find the time for Maximum Speed: Speed is maximum when the acceleration ($a$) is zero. Acceleration is the derivative of velocity:
$$a = \frac{dv}{dt} = \frac{d}{dt}(18t - 3t^2)$$
$$a = 18 - 6t$$
Setting $a = 0$:
$$18 - 6t = 0 \implies 6t = 18 \implies t = 3 \text{ s}$$
Step 3: Calculate Position ($x$) at $t = 3 \text{ s}$: Substitute $t = 3$ back into the original position equation:
$$x = 9(3)^2 - (3)^3$$
$$x = 9(9) - 27$$
$$x = 81 - 27 = 54 \text{ m}$$
Therefore, the particle is at a position of $54 \text{ m}$ when it reaches its maximum speed.