Basic Velocity -> Speed -> Position of particle

[oddjobmj]

Homework Statement

A particle's trajectory is described by x=((1/2)*(t^3)-2t^2)m and y=((1/2)*(t^2)-2t)m, where t is in s.

a) What are the particle's position and speed at t=0s and t=4s?

b) What is the particle's direction of motion, measured as an angle from the x-axis, at t=0s and t=4s?

Homework Equations

Listed above

The Attempt at a Solution

I feel like I am over-thinking the problem. As I understand it the equations provided are velocities in the directions of the x and y axies. So, the integral of the equation for 'x' would give me the speed. Integrating it a second time would give me position. I can work through the integrals but something doesn't feel right about the relationship between the work I'm doing and the material in the chapter.

integral of x = 1/8 t^4 - 2/3 t^3 and second integral of x = t^5/40 - t^4/6

integral of y = 1/6 t^3 - t^2 and second integral of y = t^4/24 - t^3/3

Plugging in the values of 0 and 4:

The position at 0 seconds = (0,0)
The speed at 0 seconds = 0 m/s

The position at 4 seconds = (-17 1/15, -10 2/3)
The speed at 4 seconds = -10 2/3 m/s

The second part I can simply use the triangles created to find the angle in relation to x.

Am I on the right track?

 

[lewando]

So, the integral of the equation for 'x' would give me the speed. Integrating it a second time would give me position.

Workin' in the wrong direction (differentiate, not integrate). dx/dt = v, dv/dt = a

 

[oddjobmj]

Bah, of course. Thank you!