The position x of a particle with respect to time t along the x...

Let us first find speed.

Speed, $v = \dfrac{dx}{dt} = \dfrac{d}{dt} ( 9{t}^{2} - {t}^{3} ) = 18t - 3{t}^{2}$

For maximum speed, using $\dfrac{dv}{dt} = 0$

$\dfrac{dv}{dt} = 0 ⇒ 18 - 6t = 0 ⇒ t = 3$ ,

Now finding displacement at that time, we have :

⇒ {x}_\max = 81 - 27 = 54m