# A particle moving in parabola (Kinematics)

1. Jul 2, 2014

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

A particle is moving along a parabola y = x2 so that at any time vx = 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

vx = 3 (given)
y = x2 so vy = 2x

ax = 0
ay = 2vx

When x = 2/3, vx = 3 and vy = 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, ax = 0 and ay = 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

2. Jul 2, 2014

### BiGyElLoWhAt

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

3. Jul 2, 2014

### BiGyElLoWhAt

derivative is the slope, not the y value.

4. Jul 2, 2014

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

5. Jul 2, 2014

### BiGyElLoWhAt

No problemo XD