# Uniform circular Motion Problem

1. Sep 18, 2009

### Gattz

1. The problem statement, all variables and given/known data
A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.80 s, it is at point (5.10 m, 5.60 m) with velocity (3.30 m/s) and acceleration in the positive x direction. At time t2 = 10.0 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.

2. Relevant equations
T=2pir/v a=v^2/r

3. The attempt at a solution
I tried finding the distance of the point, 5.1,5.6, along the circle, from the origin and got 7.57m. Not sure if this was needed, but I don't know where to go.

2. Sep 18, 2009

### Chewy0087

acceleration with uniform circular motion is always towards the center which means that

for the points 1 & 2, where is the position of the particle in relation to the center? you can find this out as it tells you where the acceleration is going towards. next, it says "assume that the particle is on the same orbit" which means that if you find the position of the particle at points 1 and 2, between time 3.8 and time 10 it has travelled from 1 to 2.

now, from that you can use your first equation to work out r.

when you have r you can see that at point 1 it is a distance 'r' from the centre in the direction of the acceleration.