A particle located at $$x=0$$  at time $$t=0,$$  starts moving along with the positive $$x$$-direction with a velocity $$'v\,'$$ that varies as $$v = \alpha \sqrt x .$$   The displacement of the particle varies with time as-

Solution :
$$\eqalign{ & v = \alpha \sqrt x ,\,\,\frac{{dx}}{{dt}} = \alpha \sqrt x \Rightarrow \frac{{dx}}{{\sqrt x }} = \alpha \,dt \cr & \int\limits_0^x {\frac{{dx}}{{\sqrt x }}} = \alpha \int\limits_0^t {dt} \cr & \Rightarrow \left[ {\frac{{2\sqrt x }}{1}} \right]_0^x = \alpha \left[ t \right]_0^t \cr & \Rightarrow 2\sqrt x = \alpha t \cr & \Rightarrow x = \frac{{{\alpha ^2}}}{4}{t^2} \cr} $$