Given, `t= sqrt x+ 3 or sqrt x = t -3 ` Squaring both the sides, we get
` x=(t-3) ^(2) =t^(2) -6 t +9`
Diferntiating it w.r.t. time (t) we get
velocity, ` v=(dx)/(dt) =2 t-6`
when `v=0` , then
`2 t -6 =0 or t=3 second`.
At ` t=3 `, distplacement is given by,
` x=t^(2) -6 t+ 9 = 3^(2) -6 xx 3+=0`.
Hence, displacement of the particle is zero when its velocity is zero.