A particle moving in parabola (Kinematics)

1. The problem statement, all variables and given/known data



A particle is moving along a parabola y = x² so that at any time v_x = 3 ms^-1. Calculate the magnitude and direction of velocity and acceleration of the particle at the point x = 2/3 m.



2. Relevant equations



Kinematics



3. The attempt at a solution



v_x = 3 (given)

y = x² so v_y = 2x



a_x = 0

a_y = 2v_x



When x = 2/3, v_x = 3 and v_y = 4/3

So v = √{3² + (4/3)²} = 3.3

Direction tan^-1{(4/3)/3} = tan^-1(4/6) = 33.7°



When x = 2/3, a_x = 0 and a_y = 2 x 3 = 6

So a = √{0² + 6²} = 6

Direction in +y direction



Above is how I tried to solve the problem. But then when I checked the answer, I found

v = 5, direction of velocity tan^-1(4/3) or 53°, a = 18 and direction of acceleration +y



Would someone please help me pointing out my mistakes?