The relation $\sqrt{3x}=3t-6$ describes the position of a particle in one direction where ‘x’ is in meters and t in sec . The displacement when velocity is zero, is……….

Given that  $\sqrt{3x}=3t-\mathrm{6..........}\left(1\right)$
$\Rightarrow 3x=9t^2+36-36t$ 
Differentiate on both sides with respect to time

$$\begin{gathered} \begin{array}{l}\therefore 3\left(\frac{dx}{dt}\right)=18t+0-36\\ \Rightarrow 3V=18t-36\\ \text{ If }V=0\text{ then }18t-36=0\\ \Rightarrow 18t=36\\ \Rightarrow t=2S\dots \dots \dots \dots ..\left(2\right)\end{array} \\ as v=6t-12 \\ acceleration of the particle is a=\frac{dv}{dt }=6 m{s}^{-2}=cons\tan t \\ at t=o velocity of the particle is v=-12 m{s}^{-1} \\ displacement of the body s=ut +\frac{a{t}^2}{2} \\ by the time t=2 sec velocity becomes zero \\ s=-12 \times 2+\frac{6\times 4}{2}=-12 m \end{gathered}$$

$\therefore$ from equation …….(1)
At  $t=0$ the initial position of the particle  $x_1=12m$
At $t=2S$ the final position of the partial  $x_2=0$
$\therefore$ Displacement  S=change in position
$=x_2-x_1=-12m$