# Don't know where to start-Uniform Circular Motion

1. Sep 17, 2007

### prelic

At t$$_{}1$$=2.00s, the acceleration of a particle in counter-clockwise circular motion is (6.00m/s$$^{}2$$)i + (4.00m/s$$^{}2$$)j. It moves at constant speed. At time t$$_{}2$$=5.00s, its acceleration is (4.00m/s$$^{}2$$)i + (-6m/s$$^{}2$$)j. What is the radius of the path taken by the particle if t$$_{}2$$ - t$$_{}1$$ is less than one period?

I really don't even know where to start. I know that the speed is constant, and that I'm probably going to have to use a=v$$^{}2$$/r, but I really don't know where to even begin with this.

2. Sep 17, 2007

### Staff: Mentor

Hint: Since the motion is uniform circular motion, where must those acceleration vectors point?

3. Sep 17, 2007

### prelic

the acceleration vectors always point towards the center, but I still don't understand what that means...does that mean something nets to 0?

4. Sep 17, 2007

one way to tackle this is draw yourself a simple diagram: you have two acceleration vectors at two points of the circular path. Can you work out the angle between the vectors?

You know how long it takes to traverse this angle... so from that you can work out something which will lead you straight to radius given constant speed.

5. Sep 17, 2007

### prelic

ok im starting to understand...before I do all the work for no reason, should i use a dot product with these 2 vectors to figure out the angle?

6. Sep 17, 2007

### Staff: Mentor

Sure. Good idea.