Posted by **Anonymous** on Tuesday, September 22, 2009 at 1:24am.

A particle P travels with constant speed in a circle of radius 8.6 m and completes one revolution in 12.0 s. The particle passes through O at t = 0 s. What is the magnitude of the average velocity during the interval from t = 5.4 s and t = 9.7 s?

What is the magnitude of the instantaneous velocity at t = 5.4 s?

What is the magnitude of the instantaneous acceleration at t = 9.7 s?

I have attempted this problem several times and cannot seem to arrive at the correct answers, so if someone could give me the answers, I would greatly appreciate it, and I could work backwards using my formulas. Thanks

- Physics -
**drwls**, Tuesday, September 22, 2009 at 6:23am
Between 5.4 and 9.7s, the particle completes 4.3/12.0 = 35.8% of a revolution, or 129 degrees of travel around the circle. Displacement during that interval is 2R*sin(129/2)= 15.52 m

Average velocity during the interval is that displacement divided by 4.3s.

For instantaneous velocity, multiply R by the angular speed w = 2 pi/12 = 0.5236 rad/s

Instantaneous velocity magnitude is

a = R w^2.

It is centripetal only. There is no tangential acceleration along the circle

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