The position x of a particle with respect to time t along X-axis is given by $\mathrm{x} = 9{\mathrm{t}}^2 - {\mathrm{t}}^3,$ where x is in metres and t in second. What will he the position of this particle when it achieves maximum speed along the positive x-direction?

The position x of a particle w.r.t. time t along X-axis,

                 $\mathrm{x} = 9{\mathrm{t}}^2 - {\mathrm{t}}^3$                                                      …(i)

Differentiating Eq. (i) w.r.t. time, we get speed, i.e.

                $v=\frac{dx}{dt}=\frac{d}{dt}\left(9t^2-t^3\right)$

$\Rightarrow$           $v = 18t - 3t^2$                                                     …(ii)

Again, differentiating Eq. (ii) w.r.t. time, we get acceleration, i.e.

                 $a=\frac{dv}{dt}=\frac{d}{dt}\left(18t-3t^2\right)$

$\Rightarrow$            $a = 18 - 6t$                                                        …(iii)

Now, when speed of particle is maximum, its acceleration is zero.

                  a = 0

$\Rightarrow$         $18 - 6t = 0$

$\Rightarrow$                     t = 3 s

Putting in Eq. (i), we obtain position of particle at that time

                      $x=9(3{)}^2-(3{)}^3=9\left(9\right)-27$

                         $= 81 - 27 = 54 \mathrm{m}$