Answer:
1) 3.91 m
2) 7.73 m
Explanation:
For an object moving in a circular path, the velocity is tangential to the circle. If it's moving at constant speed, the object's acceleration is purely centripetal, so it points towards the center of the circle.
1) Originally, the particle's velocity is in the +y direction, and its acceleration is in the +x direction. So the particle is at the 9 o'clock position and moving clockwise.
Next, the particle's velocity is in the -x direction, and the acceleration is in the +y direction. So the particle is at the 6 o'clock position.
Assuming that the time interval is less than 1 period, that means the particle has moved 3/4 of a revolution.
At constant speed, distance = speed × time:
d = v t
3/4 (2π r) = (3.07 m/s) (9.29 s − 3.29 s)
3/2 π r = 18.42 m
r = 3.91 m
2) Using the same logic as the first problem, we know that originally, the particle starts at the 9 o'clock position and moves clockwise to the 6 o'clock position, covering 3/4 of the circle's circumference.
The radius of the circle is:
d = v t
3/4 (2π r) = (2.99 m/s) (9.27 s − 4.2 s)
3/2 π r = 15.16 m
r = 3.22 m
When the particle is originally at the 9 o'clock position, it is directly to the left of the center. So the x-coordinate of the center is:
x = 4.51 m + 3.22 m
x = 7.73 m
