A particle moving in parabola (Kinematics)

[Adjoint]

Homework Statement

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.

Homework Equations

Kinematics

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?

 

[BiGyElLoWhAt]

when x = 2/3, vx =3 and vy/vx = 4/3

 

[BiGyElLoWhAt]

derivative is the slope, not the y value.

 

[Adjoint]

Oh I should have written v_y = 2xv_x
That was a super silly mistake. :shy:
Thanks!

 

[BiGyElLoWhAt]

No problemo XD