a) the average speed is simply distance covered (half the circumference) over time given by .

b) the average velocity vector is displacement (the diameter) over time given by .

c) If the constant acceleration is w and initially the particle started with 0 speed then,

The final velocity of the particle at the end of the journey will be . This means the average acceleration during this time will be w given by the equation above.

## Monday, July 16, 2007

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## 3 comments:

Err, I don't think you can assume that the particle started with 0 velocity. nowhere is that mentioned. It is possible that the particle had been moving with that velocity all along and we are considering one (technically half) of it's infinite revolutions.

Thanks. :)

You are right. The answer holds correct only for two cases i) constant acceleration from 0 or ii) constant angular velocity.

In all other cases the acceleration vector can be something else. In fact in general it is (v1-v2)/tau where v1 and v2 are the velocity vectors in the two ends.

Consider for example that the particle starts at a 0 speed. Accelerates at a rate (4PiR/tau^2) for quarter circle - this takes tau/2 seconds. Then it decelerates at the same rate for tau/2 secs more until it reaches half circle point. So the particle covers half circle in tau seconds.

However, the average acceleration is 0 in this case since v1=v2=0.

Unfortunately, Irodov does not specify much in the problem :(, so I was not sure what to do.

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