A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If its equation of motion is y = bx² (b is a constant), its velocity component in the x-direction is:

$y=b{x}^2$

Differentiating w.r.t to t an both sides, we get

$$\begin{gathered} \frac{dy}{dt}=b2x\frac{dx}{dt} \\ \Rightarrow v_y=2bx{v}_x \end{gathered}$$

Again differentiating w.r.t to t on both sides we get

$\frac{d\left(v_y\right)}{dt}=2b{v}_x\frac{dx}{dt}+2bx\frac{d{v}_x}{dt}=2b{v_x}^2+0$

[$\frac{d{v}_x}{dt}$= 0, because the particle has constant acceleration along only y-direction]

Now, $\frac{d{v}_y}{dt}=a=2b{v_x}^2;$

$$\begin{gathered} {v_x}^2=\frac{a}{2b} \\ {v}_x=\sqrt{\frac{a}{2b}} \end{gathered}$$