Find the components of velocity and normal acceleration when x = 2m.

$$y=5+0.3 x^2$$

$\therefore y'=\dfrac {dy}{dx}=0.6 x \space \text {and } \space y"=0.6\\ At \space \space x = 2, \\ y’ =0.6 \times 2 = 1.2 \text { and } y’’ = 0.6$

Radius of curvature = $\rho=\dfrac {(1+y^2)^{3/2}}{y"}=\dfrac {(1+1.2^2)^{3/2}}{0.6}=6.3523 m$

Given at $x=2,$ velocity $(v)=10m/s$ and since the velocity is constant, tangential acceleration $a_1 =0$

Now, Velocity is along the tangent to the curve.

Let velocity vector make an angle $\theta$ with the horizontal.

$\therefore \tan \theta =$ slope of tangent $=y’ = 1.2\\ \therefore \theta =50.1944^\circ \\ \therefore v_x =v\cos \theta =10\cos 50.1944 =4.4018 m/s \space and \space v_yv\sin \theta =10\sin 50.1944=7.6822 m/s $

Also, Normal acceleration = $a_n=\dfrac {v^2}{\rho} =\dfrac {10^2}{6.3523}=15.7422 m/s ^2 $

Angle made by normal acceleration with horizontal = $\theta + 90=50.1944 + 90= 140.1944^\circ \\ (a_n)_x =a_n\cos (90+\theta) 15.7422 \cos 140.1944=-4.9180 m/s ^2 \\ (a_n)_y =a_n\sin (90+\theta) 15.7422 \sin 140.1944=10.0779m/s ^2 $

Hence, When $X = 2$

X components of its velocity $v_x =6.4018m/s $

Y components of its velocity $v_y =7.6822m/s$

And,

Normal acceleration $= 15.7422m/s^2 \rightarrow 140.1944^\circ $

X components of its normal acceleration $(a_n)_x =-4.9180 m/s^2$

Y components of its normal acceleration $(a_n)_y =10.0779 m/s^2$