velocity=3.7
one fourth of a circle is covered in 11-4.8s, 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)
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.
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