# [Physics] Magnitude of the average velocity vector (not the average speed)

angular velocityhomework-and-exerciseskinematics

Thank you ahead of time for taking to look at this. For this following problem we were given an answer however I am almost positive the given answer is wrong. It doesn't even make sense. So here is the question:

The sweep-second hand of a clock is 3.4 cm long. What are the magnitudes of:

a) the magnitude of the average velocity vector [not the average speed!] of the hand tip over a 12 second interval?

Firstly isn't the magnitude of average velocity = the average speed? I don't see how those could be different but the question seems to indicate this

The answer given is $$.333 cm/s$$. This seems highly improbable given that the circumference is $$21.36 cm$$ so in $$12s$$ the second hand tip would have only traversed $$3.96 cm$$ around the circumference in 12 seconds considering a second hand completes a complete revolution every minute I don't see how this could be right.

Instead I would say the magnitude of the velocity vector = $$\frac{(2\pi r)}{t}$$

Where r is radius and t is time to complete revolution so:

$$v = \frac{2\pi 3.4}{1} = 21.36 cm/s$$

Which seems a lot more realistic

Part b) further confuses me as they state:

the average acceleration of the hand tip over a 12 second interval?

and they give an answer of $$0.0349 cm/s^2$$

It is my understand that $$a = \frac{v^2}{r}$$ so even using their own value for velocity

$$a = \frac{0.333^2}{3.4} = 0.0326$$ which is a different answer.

Am I missing something?

Thanks!