Given,

Path Equation: $y=\dfrac{x^2}{3}$ and $\left\vert{}\bar{V}\right\vert{}=8\dfrac{m}{s}$

Differentiate equation w.r.t to time,

$\dfrac{dy}{dt}=\dfrac{2x}{3}\times{}\dfrac{dx}{dt}$

(\therefore{}V_y=\dfrac{2}{3}x.V_x\ ...\ ... ①)

Now when x=3m, y=3m (from path equation)

$V_y=\dfrac{2}{3}\left(3\right)V_x$

$\therefore{}V_y=2V_x$

Here magnitude of velocity is,

$\left\vert{}\bar{V}\right\vert{}=\sqrt{{\left(V_x\right)}^2+{\left(V_y\right)}^2}$

${\left(8\right)}^2={\left(V_x\right)}^2+{\left(V_y\right)}^2$

$64={\left(V_x\right)}^2+{\left(2V_x\right)}^2$

$64=5V_x{\ }^2\ $

$\therefore{}V_x=3.58\ m/s$

$\&\ V_y=2\left(3.58\right)=7.16\ m/s$

To find acceleration of the point,

$\dfrac{dy}{dx}=\dfrac{2x}{3}\ \ \&\ …