# Magnitude of the average velocity vector (not the average speed)

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!

• A second hand goes how far in one second? – DJohnM Jan 26 '15 at 0:47
• Draw the velocity vector at the tip of the second hand ... how does it change with time? – Peter Diehr Feb 6 '16 at 2:45

Yes, you are missing something and I get the given answer. But this is a "homework" problem, so hints only.

The wording of the question is important. For instance the magnitude of the average velocity vector over 60 seconds would be exactly zero. Or to put it another way, if I travel at constant speed to a point, and then turn round and come back at the same speed, then the average speed is finite but the average velocity is zero.

i.e. you need to work out an expression for the velocity vector at an arbitrary angle and then find the average velocity vector over a 12 second (72 degree angle) period and only then find the magnitude of that.

Generally, the magnitude of an average of a vector (which is what this question asks for) and the average of the magnitude of this vector (which would be the average speed) are not the same. In equations:

$$\lvert\frac{1}{{t_a}}\int_0^{t_a}dt~\vec{v}\rvert\neq \frac{1}{{t_a}}\int_0^{t_a}dt~\lvert\vec{v}\rvert$$

Here you average over $t_a=12~\mathrm{s}$ and

$$\vec{v}=v\begin{pmatrix} \cos\omega t\\ -\sin\omega t \end{pmatrix}$$ where $v=2\pi R/T$ is the speed of the hand tip, $\omega=2\pi/T$ the angular frequency and $T=60~\mathrm{s}$ is the period of rotation.

In order to use these equations you need to proof or otherwise justify that the quantity you are interested in is independent of the time interval you integrate over, i.e. it does not matter whether you integrate from 0 to 12 s or from 3 to 15 s, etc.

$$\bar{v}=\lvert\frac{1}{t_a}\int_0^{t_a}dt~\vec{v}\rvert = \lvert\frac{v}{\omega{t_a}}\begin{pmatrix} \sin\omega {t_a}\\ \cos\omega {t_a}-1 \end{pmatrix}\rvert = \frac{v}{\omega{t_a}}\sqrt{2-2\cos\omega{t_a}}=\frac{R}{{t_a}}\sqrt{2-2\cos\omega{t_a}}$$
As for part b), the acceleration, $\vec{a}$, is a vector, so in the same way you have for the "the average acceleration of the hand tip over a 12 second interval":
$$\bar{a}=\lvert\frac{1}{t_a}\int_0^{t_a}dt~\vec{a}\rvert = ... = \frac{\bar{v}v}{R} = 0.0349~\mathrm{cm}/\mathrm{s}^2$$
You don't need any integral, even though it is nothing wrong in averaging the instantaneous values to get the average value. However, by definition, the average velocity is $\vec{v}_{ave}=\frac{\Delta \vec{r}}{\Delta t}$ or in words, average velocity is the ration between the displacement and the time taken for this displacement. So in your problem you need only to calculate the displacement of the tip between the two positions. This is not the arc between the position but the cord, the base of the triangle made by the two positions of the arm. You will need the angle between the two position and a little trigonometry to find it. Divide this by 12s and you have the magnitude of the average velocity.