$y = 5 + 0.3 x^2$ $\therefore y' = \dfrac {dy}{dx} = 0.6 x \ and \ y''=\dfrac {d^2 y}{dx^2} = 0.6$  At x=2, $y' = 0.6 \times 2 = 1.2 \ and \ y''=0.6$ ∴ Radius of curvature $= \rho = \dfrac {(1+ y'^2)^{3/2}}{y''} = \dfrac {(1+1.2^2)^{3/2}}{0.6} = 6.3523m$ Given at x=2, velocity (v)=10 m/s and since the velocity is constant, tangential acceleration at=0 Now, velocity is along the tangent to the curve. Let vlocity vector make an angle θ with the horizontal. ∴ tan θ=slope of tangent = y'=1.2 $\therefore \theta= 50.1994^\circ$ $\therefore v_x = v \cos \theta = 10 \cos 50.1944 = 6.4018 \ m/s \ and$ $v_y= v\sin \theta = 10 \sin 50.1944= 7.6822 \ m/s$ 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 = $\theta + 90$ $= 50.1944+90 = 140.1944^\circ$ $\therefore (a_n) _x = a_n \cos (90 + \theta ) = 15.7422 \cos 140.1944 = -4.9180 \ m/s^2 \ and$ $(a_n)_y = a_n \sin (90 + \theta)= 15.7422 \sin 140.1944 = 10.0779 \ m/s^2$ Hence, when x=2 X component of its velocity $v_x= 6.4018 \ m/s$ Y component of its velocity $v_y = 7.6822 \ m/s$ And, Normal accelerations $= 15.7422 \ m/s^2 \ \angle \ 140.1944^\circ$ X component of its Normal acceleration $(a_n)_x = - 4.9180 \ m/s^2$ Y component of its Normal acceleration $(a_n)_y = 10.0779 \ m/s^2$