Equation of curve is $y= x2/ 3 $

Differentiating with t on both sides

$dy/dt = [2x*(dx/dt)]/3 \\ but, dy/dt = vY \& dx/dt = vX \& x=3 \\ hence, vY =2 vX\text {____}(i) $

Also, net velocity(v)$= \sqrt{( V_x^2 + V_y^2)} =8\text {___}(ii) $

From (i) & (ii),

$vX = 3.58 m/s \& vY = 7.15 m/s \\ Now, dy/dx = 2x/3 \& d2y/dx2 = 2/3 \\ At, x=3... dy/dx = 2(3)/3 = 2 $

Hence. Finding radius of curvature

$\rho = {[1 + (dy/dx)2 ]3/2}/(2/3) \\ = (53/2)/(2/3) = 16.77 m $

Now, acceleration(a)$= v2/\rho = 64/16.77 = 3.816 m/s^2 $

x & y components of the velocity when $x=3$ are 3.58 m/s & 7.15 m/s respectively and the acceleration at x=3 is $3.816 m/s^2 $