[quote]A particle P travels with constant speed on a circle of radius r=3.00m (Figure) and completes one revolution in 20.0s. The particle passes through O at time t=0. State the following vectors in magnitude-angle notation (angle relative to the positive direction of x). With respect to O, find the particle’s position vector at times t of (a) 5.00 s, (b) 7.50 s, and (c) 10.0 s. (d) for the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle’s displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

http://img46.imageshack.us/img46/449/physicsu.png

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First I know that radius = 3.00m, and time to complete one cycle is 20.0 seconds. So I know that T=2(pi)r/v, so 20.0s=2pi(3.00m)/v, so speed = .94247 m/s

Then acceleration = (v^2)/r, so a = ((.94247m/s)^2)/(3.00m), so a=.29608 m/s

I know using Distance = rate x time, that at a time of 5 seconds, and at a speed of .94247 m/s, the object will be 4.71 meters along the circle

using the arclength formula, s=rÎ¸, 4.71=(3)Î¸, so the angle is 90 degrees.

So at 90, the position vector is 6 meters in the y direction, and zero meters in the x direction (At the very top of the circle)

before I proceed, is this correct?

So is the position vector <0,6,0>?