# Chase

Path

This animation shows many particles participating in a chase. A chases B, B chases C and C chases D and so on till finally the last particle is chasing the first particle. You can vary the number of particles participating in the chase.

Given the side of the polygon and the velocities of the particles (which are equal) it should be possible to find the time required for them to meet. This problem is a good case for the appreciation of the importance of a properly chosen reference frame to simplify the problem of analysis of motion. In this case the line joining the center of the polygon and a given particle is a good reference line for the analysis. The velocity vector always makes the same angle with this line. So motion relative to this line is simple enough to allow us to solve for time.

You should be able to notice that during the motion the velocity vector of any particle is always directed at an angle of with the line joining the center and the particle at that given moment where n is the number of particles participating in the chase. In other words though the velocity vector is changing direction, the line joining the particle with the center is also rotating resulting in a constant angle between the two. This should mean the length of the line decreases at a constant rate of V Cosθ. Since the initial length L of the line is known, the time taken to reach the center works out to be t = L/V Cosθ. In the animation if the checkbox for the path is unselected, the line joining the center and the particle and the velocity vector are shown during the motion. The yellow velocity vector is seen to make a constant angle with the red line. And that allows us to solve for time as mentioned above.