Posted by Nelson on Monday, October 3, 2011 at 9:03pm.
A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 4.80 s, it is at point (4.40 m, 6.20 m) with velocity (3.70 m/s) and acceleration in the positive x direction. At time t2 = 11.0 s, it has velocity (–3.70 m/s) and acceleration in the positive y direction. What are the (a)x and (b)y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.

Physics  bobpursley, Monday, October 3, 2011 at 9:15pm
velocity=3.7
one fourth of a circle is covered in 114.8s, or 6.2 seconds. So period = 24.8 seconds.
velocity= distance/time
3.7m/s=2PI*r/2.48
solve for r.
Now, at point one, acceleration is in the direction of the center, so center must be (4.40+r, 6.20)
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