13.0s-3.7s=9.3s
9.3s*3.10m/s=28.83m
c=pi*2*r
.75c=.75*pi*2*r
28.83m=.75*pi*2*r
r=6.1179m
x=4.3m+6.1179m=10.4179m
(To see if this is within the 2% margin of error for my other answer. If it is, this is also wrong.
10.4179*.02=.208
10.4179+.208=10.626
This does NOT encompass my...
Additional Info on Problem (enough wrong answers allows you a hint): At each time, consider where the center of the circle is, and where the particle is along its path around the circle. From the known speed and times, the distance the particle travels can be found. This allows you to calculate...
Ohhh. So I could only use w*r=v if v at point 1 was 3.10 and if it was then different at point 2?
Is there any way to solve this numerically, or is it purely a logic-out question like what I did for y? It doesn't give me the location of the 2nd point, so I don't know x2, and therefore can't...
Homework Statement
A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.70 s, it is at point (4.30 m, 5.50 m) with velocity (3.10 m/s)j and acceleration in the positive x direction. At time t2 = 13.0 s, it has velocity (–3.10 m/s)i...