Sorry about that, I misunderstood what you were saying, I’m good now.
For part E I got average velocity to be the change in vector position divided by the t=5, so I get <(-3/5),(3/5),0>
However to convert that magnitude-angle format, I get <(3/5)rt(2),135degrees>… IS that correct?
One thing that is confusing me, it asks to find the instaneous velocity for part F, but I need a position function of t, r(t), in order to take the derivative. Do I have to somehow use the graph to generate a position function?
[quote]jdinatale wrote:
Sorry about that, I misunderstood what you were saying, I’m good now.
For part E I got average velocity to be the change in vector position divided by the t=5, so I get <(-3/5),(3/5),0>
However to convert that magnitude-angle format, I get <(3/5)rt(2),135degrees>… IS that correct?[/quote]
Yeah that looks good to me. For the instantaneous velocity, it would be the v that you calculated earlier on using the period and circumference. It says in the question speed is constant, which means the magnitude of the velocity vector will be constant, just in different directions. At the point (3,3,0) the velocity vector would be (0,v,0)m/s, and at (0,6,0), (-v,0,0)m/s. I don’t think there is any way to give that a direction wrt to the point O, but I may be mistaken.
I think
r(theta) = 3j + 3( sin(theta)i - cos(theta)j )
where theta is a function of time (didn’t have the patience to figure what it is).
Edit: Not sure of the +/-, didn’t check. Have to get in time to my own physics lecture 
Edit2: theta is (t/20s)2Pi