The position x of a particle moving in one dimension is related to time t by the equation $t =\sqrt{x}+3$ where x is in metres and t in seconds. The position of the particle when its velocity is zero is 

Given $t=\sqrt{x}+3$ Squaring, we have

  $x=t^2-6t+9$                             (i) 

 velocity   $v=\frac{dx}{dt}=\frac{d}{dt}(t^2 - 6t + 9) = 2 t - 6$       (ii) 

 Find t from Eq. (ii) when v = 0. ,  2t - 6= 0 

Use this value of t is Eq. (i).     so  $t=3$s

at   $t=3$,    $x=3^2-6\left(3\right)+9=0$

Hence, the answer is 0