Three approaches to magnitude of average velocity

[LearninDaMath]

Is there such a thing as magnitude of average velocity? If so, is it one of these?

Suppose I have to position vectors. \stackrel{\rightarrow}{r}= 1_x + 2.2_y and \stackrel{\rightarrow}{r}_2 = 2_x + 0_y

And the distance between the vectors is Δ\stackrel{\rightarrow}{r}= \stackrel{\rightarrow}{r}_1 - \stackrel{\rightarrow}{r}_2

(1_{x}+2.2_{y}) - (2_{x}+0_{y}) = (-1_{x}+2.2_{y})


Velocity = Δr/Δt (-1_{x}+2.2_{y})/2 = (1/2)_{x} + (2.2/2)_{y}

and so magnitude of the velocity components is 1.2m/s
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However, what if I did this:

magnitude of (-1_{x}+2.2_{y}) is 3.84m

and the velocity is 3.84/2 = 1.92m/s
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OR, if I got the magnitude of each vector first: l\stackrel{\rightarrow}{r}l - l\stackrel{\rightarrow}{r}l = 2.41 - 2 = .41 and then .41/2 = .205m/s
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Which would be correct? And, perhaps, of the approaches that don't work, why not?

 

[sg001]

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However, what if I did this:

magnitude of (-1_{x}+2.2_{y}) is 3.84m

and the velocity is 3.84/2 = 1.92m/s
__________________________________________________

Where are you getting these results from?

 

[LearninDaMath]

Oh, I probably miscalculated.



So \sqrt{1_{x}^2+2.2_{y}^2}/2


equals


\sqrt{(1/2)_{x}^{2}+(2.2/2)_{y}^{2}}


So the magnitude of the difference of two vectors, then divided by time, lΔ\stackrel{\rightarrow}{r}l/2 produces the same number as Δr/2, then finding the magnitude?

 

[genericusrnme]

OR, if I got the magnitude of each vector first: l\stackrel{\rightarrow}{r}l - l\stackrel{\rightarrow}{r}l = 2.41 - 2 = .41 and then .41/2 = .205m/s

That will only work if the two vectors point in the same direction
Take a vector (1,0) and a vector (0,1)
These both have magnitude 1, so doing what you did would get me 0 and so the average velocity between those two points should be 0, which is untrue

If I take (1,0) away from (0,1) I get (-1,1) which has magnitude √2 ≠ 0

 

[sg001]

Oh, I probably miscalculated.



So \sqrt{1_{x}^2+2.2_{y}^2}/2


equals


\sqrt{(1/2)_{x}^{2}+(2.2/2)_{y}^{2}}

This would be the correct way of doing it.



However, what if I did this:

magnitude of (-1_{x}+2.2_{y}) is 3.84m

and the velocity is 3.84/2 = 1.92m/s

You can't use this because your not finding the magnitude of the vector.
You have to use the distance formula.

Adding the average of x & y coordinates does not give you the magnitude of a vector.
What you have actually done, is found the midpoint of the vector by dividing x & y by 2 (although that is amount of time), in this particular problem.

ie (x1 + x2)/2 = x(m)idpoint
& (y1 + y2)/2 = y(m)idpoint

Now if I add the new x & y coords, this does not give me a magnitude of the vector, as these are just points.

ie. Consider the point (2,1), this could lie on any given size vector ie 2 units or 2000 units
Likewise, the point (500,10), this could lie on a vector of magnitude 1 unit or 400 units.
and so on...

 

[LearninDaMath]

Okay, I think I'm starting to get the idea a little bit.

So on a postion vs position graph, the l\stackrel{\rightarrow}{r}l-l\stackrel{\rightarrow}{r}l does not give magnitude of Δ\stackrel{\rightarrow}{r} and thus (l\stackrel{\rightarrow}{r}l-l\stackrel{\rightarrow}{r}l)/2 does not give Δl\stackrel{\rightarrow}{r}l/Δt nor a midpoint magnitude. It is nothing more than a subtraction of two numbers on a number line.


\stackrel{\rightarrow}{r}-\stackrel{\rightarrow}{r} = (\stackrel{\rightarrow}{r}_x+\stackrel{\rightarrow}{r}y) - (\stackrel{\rightarrow}{r}_x+\stackrel{\rightarrow}{r}y) = (\stackrel{\rightarrow}{r}-\stackrel{\rightarrow}{r}_{x}) + (\stackrel{\rightarrow}{r}-\stackrel{\rightarrow}{r}_{y}) = Δr

so, now that the components of Δ\stackrel{\rightarrow}{r} are produced, the order in which magnitude is taken dictates the meaning of the end results. So dividing the components by time = 2, such that Δr/Δt = components of velocity, and then finding the magnitude will result in the magnitude of average velocity.

However, finding the magnitude of Δ\stackrel{\rightarrow}{r} first, then dividing by a number, simply gives a fraction of that magnitude of the vector. So dividing by time = 3 seconds will not produce an average velocity. And that is why finding the midpoint by dividing by a number 2 simply ends up as the same numerical result as if velocity at time = 2 were were evaluated, by coincidence, but not because they mean the same thing.

Thus average velocity at time = 3 seconds will be a different value than simply dividing the magnitude of Δr by the number three.

And that is basically all saying that the distance formula divided by 2 produces the midpoint value, but that is not equal to the Δr components divided by 2 seconds all square rooted, even though the numerical value is the same in both cases.

 

[sg001]

And that is basically all saying that the distance formula divided by 2 produces the midpoint value, but that is not equal to the Δr components divided by 2 seconds all square rooted, even though the numerical value is the same in both cases.[/QUOTE]

Just to clarify the distance formula is

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\,

This is a distance formula and does NOT give you the midpoint of a given line.
It gives you the distance of a line, or the magnitude of a single vector.

 

[LearninDaMath]

Just to clarify the distance formula is

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\,

This is a distance formula and does NOT give you the midpoint of a given line.
It gives you the distance of a line, or the magnitude of a single vector.

right, it gives the distance of the line, and only when I divide that distance afterwords will it give me the magnitude of half that distance (or midpoint).

How about instead of a position vs position graph, suppose we were looking at a position vs time graph. And there were two vectors. The difference between the y's over the difference between the x's would represent the slope or average velocity. So what would it mean if I found the magnitude of the numerator and denominator of the slope? Would that also represent magnitude of average velocity? Or would the magnitude of the slope on the position vs time graph not mean anything?

 

[LearninDaMath]

Wait a minute, taking a step back for a second, I am seeing a contradiction here:



\frac{lΔ\stackrel{\rightarrow}{r}l}{Δt} must always equal l\frac{Δ\stackrel{\rightarrow}{r}}{Δt}l

I have tried this with multiple numbers representing time (or number in the denominator). When time = 2, yes I get lΔr/Δtl which is magnitude of average velocity. And it is also equal to the midpoint, which is lΔrl/Δt.

However, I think I was under the impression that this was only numerically equal to each other for the specific case of time = 2. However, the two processes seem to produce the same number for any given denominator.

For instance, if I take the magnitude of Δr, then divide the result by 10, I get some value "A." Or, if I first find the value of Δr/10, then take the magnitude of the whole thing, I still get an end value of "A."

So it should be able to be said that:

The distance of the difference between two position vectors (the distance between two points) is equal to the magnitude of velocity when time = 1 second.

or

The distance of the difference between two position vectors divided by 50 (the distance between two points divided by 50) is equal to the magnitude of velocity when time = 50 seconds.

Is this correct??

 

[sg001]

right, it gives the distance of the line, and only when I divide that distance afterwords will it give me the magnitude of half that distance (or midpoint).

I think you maybe getting mixed up with what the midpoint is.
The midpoint is a POINT in a given space, and does NOT equal the magnitude of half a given distance.
ie, think about if you had a 1m string and cut it in half, the midPOINT of the original string would be at 50 cm. once cut in half the string would be 50cm, so the mid point would now lie on one end of the new string.
However, this does not equal the total length or "magnitude" of the new string, as it is just a Point on the string.

 

[sg001]

Okay, I think I'm starting to get the idea a little bit.

So on a postion vs position graph, the l\stackrel{\rightarrow}{r}l-l\stackrel{\rightarrow}{r}l does not give magnitude of Δ\stackrel{\rightarrow}{r} and thus (l\stackrel{\rightarrow}{r}l-l\stackrel{\rightarrow}{r}l)/2 does not give Δl\stackrel{\rightarrow}{r}l/Δt nor a midpoint magnitude. It is nothing more than a subtraction of two numbers on a number line.

As mentioned earlier by 'genericusrnme' this will only work if the two vectors point in the same direction.

 

[LearninDaMath]

I think you maybe getting mixed up with what the midpoint is.
The midpoint is a POINT in a given space, and does NOT equal the magnitude of half a given distance.
ie, think about if you had a 1m string and cut it in half, the midPOINT of the original string would be at 50 cm. once cut in half the string would be 50cm, so the mid point would now lie on one end of the new string.
However, this does not equal the total length or "magnitude" of the new string, as it is just a Point on the string.

Okay, i am no longer under the assumption that lΔrl-lΔrl = is the same as lΔrl.
Also, I know that the midpoint is just the coordinates of a point midway between two points.

I am confused by post number 5 where u said that I was finding the midpoint. I don't think I was finding the midpoint then. I think I was finding the distance/2 ..or half the distance between the points (in otherwords, lΔrl/2 ). And that distance/2 produces the same exact number as finding the velocity components and then taking the magnitude of the whole thing (in otherwords, l(Δr/2)l )

 

[sg001]

Oh, I probably miscalculated.



So \sqrt{1_{x}^2+2.2_{y}^2}/2


equals


\sqrt{(1/2)_{x}^{2}+(2.2/2)_{y}^{2}}

This would be the correct way of doing it.



However, what if I did this:

magnitude of (-1_{x}+2.2_{y}) is 3.84m

and the velocity is 3.84/2 = 1.92m/s

I was implying that by not using the distance formula, between two points whose vectors are not parallel to each other (ie point in the same direction), as you have done here

However, what if I did this:

magnitude of (-1_{x}+2.2_{y}) is 3.84m

and the velocity is 3.84/2 = 1.92m/s

Does not give you the magnitude of the vector, you have simply found the eqn of the line (- the slope of course). Because in your case the vectors ane not parallel to each other (or are they?) you can not use this approach as it does not lead anywhere.

By the way my mistake, you were not finding the midpoint as you did not add the two cood's together you subtracted.

The first eqn using the distance formula is correct then dividing by time.

Does this clear things up?

 

[LearninDaMath]

Here's an illustration of what I'm trying to understand more clearly.

What is incorrect here, if anything?

 

[sg001]

I don't understand where your getting the last line from??

The top two lines of working are correct (besides the -5, it should be +5. but since your squaring does not matter.) and are the same thing and describes the resultant vector.
If you want to find half the resultant vector then just divide by two. I don't understand where the last line of working out came from and its use??

 

[LearninDaMath]

I don't understand where your getting the last line from??

The top two lines of working are correct (besides the -5, it should be +5. but since your squaring does not matter.) and are the same thing and describes the resultant vector.
If you want to find half the resultant vector then just divide by two. I don't understand where the last line of working out came from and its use??

and yea, -5 and 5 and -6 and 6 are switched because I wrote it in the first equation one way, but subtracted components the other way while I was typing it into microsoft paint...and ur right, same thing since I'm squaring.


And regarding the last equation, lrl - lrl is from √(5^2 + 12^2) - √(10^2 + 6^2) difference in the magnitude of each vector... which I would suppose would not be applicable to finding average velocity.

 

[sg001]

And regarding the last equation, lrl - lrl is from √(5^2 + 12^2) - √(10^2 + 6^2) difference in the magnitude of each vector... which I would suppose would not be applicable to finding average velocity.

No because its not the distance formula. to make this work you would have √ ((10)^2 + (12)^2) - √ ((6)^2 + (5)^2) & then dividing by two.

Just the same as using the distance formula, it is always the bigger x value - the samller x value, and same goes for y values.

 

[LearninDaMath]

No because its not the distance formula. to make this work you would have √ ((10)^2 + (12)^2) - √ ((6)^2 + (5)^2) & then dividing by two.

Just the same as using the distance formula, it is always the bigger x value - the samller x value, and same goes for y values.

So you'd confirm that everything in that illustration (besides the negative signs within the distance formula) is accurate?