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

A particle is moving along a parabola y = x^{2} 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 = x^{2} 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 = √{3^{2} + (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 = √{0^{2} + 6^{2}} = 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?

http://ift.tt/1iWhj1O