A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 4.60 s, it is at point (4.30 m, 6.20 m) with velocity (3.30 m/s) and acceleration in the positive x direction. At time t2 = 13.0 s, it has velocity (–3.30 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.
To find the coordinates of the center of the circular path, we need to look at the information provided about the particle's velocity and acceleration at two different times.
Let's break down the problem step by step:
Step 1: Analyze velocity at t1
At time t1 = 4.60 s, the particle has a velocity of 3.30 m/s and an acceleration in the positive x direction. This means that at this point, the particle is moving in the positive x direction and is at the far right of the circular path. Therefore, the center of the circular path must be located on the left side.
Step 2: Analyze velocity at t2
At time t2 = 13.0 s, the particle has a velocity of -3.30 m/s and an acceleration in the positive y direction. This means that at this point, the particle is moving in the negative y direction and is at the bottom of the circular path. Therefore, the center of the circular path must be located at the top.
Step 3: Determine x-coordinate of the center
Based on the information from steps 1 and 2, we can conclude that the x-coordinate of the center of the circular path is 4.30 m. This is because the particle starts at (4.30 m, 6.20 m) at t1 and moves in the positive x direction.
Step 4: Determine y-coordinate of the center
Based on the information from steps 1 and 2, we can conclude that the y-coordinate of the center of the circular path is 6.20 m. This is because the particle starts at (4.30 m, 6.20 m) at t1 and moves in the positive y direction, which means the center of the circle must be above it.
Therefore, the (a) x-coordinate of the center of the circular path is 4.30 m, and the (b) y-coordinate of the center of the circular path is 6.20 m.
To find the coordinates of the center of the circular path, we can use the fact that the velocity and acceleration vectors of a particle moving in a circular path are always perpendicular to each other.
Given that at time t1 = 4.60 s, the velocity vector is in the positive x direction and the acceleration vector is also in the positive x direction, we can conclude that at this point, the particle is moving horizontally along the x-axis.
Since the velocity vector is in the positive x direction, we can determine that the particle is moving in a counterclockwise direction.
Therefore, the x-coordinate of the center of the circular path at time t1 is simply the x-coordinate of the particle, which is 4.30 m.
Now, at time t2 = 13.0 s, the velocity vector is in the negative y direction and the acceleration vector is in the positive y direction. This indicates that the particle is moving vertically along the negative y-axis.
Again, since the velocity vector is in the negative y direction, we can determine that the particle is moving in a counterclockwise direction.
Therefore, the y-coordinate of the center of the circular path at time t2 is the negative of the y-coordinate of the particle, which is -6.20 m.
To summarize:
(a) The x-coordinate of the center of the circular path is 4.30 m.
(b) The y-coordinate of the center of the circular path is -6.20 m.