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,

$\dfrac{dy}{dx}=\dfrac{2x}{3}\ \ \&\ \ \dfrac{d^2y}{dx^2}=\dfrac{2}{3}$

Radius of curvature,

${\rho{}}_{x=3}=\dfrac{{\left[1+{\left(\dfrac{dy}{dx}\right)}^2\right]}^{3/2}}{\left\vert{}\ \dfrac{d^2y}{dx^{2\ \ }}\right\vert{}}\ at\ 3m$

$\therefore{}\rho{}=\dfrac{{\left[1+{\left(\dfrac{2\left(3\right)}{3}\right)}^2\right]}^{3/2}}{\left\vert{}\ \dfrac{2}{3}\ \right\vert{}}$

$\therefore{}\rho{}=16.77\ m$

Normal Acceleration,

$a_n=\dfrac{v^2}{\rho{}}$

$\therefore{}a_n=\dfrac{8^2}{16.77}$

$\therefore{}a_n=3.816\ m/s^2$

Tangential Acceleration,

$a_t=0\ as\ speed\ is\ constant.$

The acceleration of the point is thus-

$a=\sqrt{a_n^2+a_t^2}=3.816\ m/s^2$