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Three approaches to magnitude of average velocity

  1. Feb 21, 2012 #1
    Is there such a thing as magnitude of average velocity? If so, is it one of these?

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

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

    ([itex]1_{x}+2.2_{y}[/itex]) - ([itex]2_{x}+0_{y}[/itex]) = ([itex]-1_{x}+2.2_{y}[/itex])


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

    and so magnitude of the velocity components is 1.2m/s
    __________________________________________________

    However, what if I did this:

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

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

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


    Which would be correct? And, perhaps, of the approaches that don't work, why not?
     
  2. jcsd
  3. Feb 21, 2012 #2
    __________________________________________________

    However, what if I did this:

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

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

    Where are you getting these results from?
     
  4. Feb 21, 2012 #3
    Oh, I probably miscalculated.



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


    equals


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


    So the magnitude of the difference of two vectors, then divided by time, lΔ[itex]\stackrel{\rightarrow}{r}[/itex]l/2 produces the same number as Δr/2, then finding the magnitude?
     
  5. Feb 21, 2012 #4
    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
     
  6. Feb 21, 2012 #5
     
  7. Feb 22, 2012 #6
    Okay, I think I'm starting to get the idea a little bit.

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


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

    so, now that the components of Δ[itex]\stackrel{\rightarrow}{r}[/itex] 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[itex]\stackrel{\rightarrow}{r}[/itex] 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.
     
    Last edited: Feb 22, 2012
  8. Feb 22, 2012 #7
    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.
     
  9. Feb 22, 2012 #8

    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?
     
  10. Feb 22, 2012 #9
    Wait a minute, taking a step back for a second, I am seeing a contradiction here:



    [itex]\frac{lΔ\stackrel{\rightarrow}{r}l}{Δt}[/itex] must always equal l[itex]\frac{Δ\stackrel{\rightarrow}{r}}{Δt}[/itex]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??
     
    Last edited: Feb 22, 2012
  11. Feb 22, 2012 #10
     
  12. Feb 22, 2012 #11
    Okay, I think I'm starting to get the idea a little bit.

    So on a postion vs position graph, the l[itex]\stackrel{\rightarrow}{r}[/itex]l-l[itex]\stackrel{\rightarrow}{r}[/itex]l does not give magnitude of Δ[itex]\stackrel{\rightarrow}{r}[/itex] and thus (l[itex]\stackrel{\rightarrow}{r}[/itex]l-l[itex]\stackrel{\rightarrow}{r}[/itex]l)/2 does not give Δl[itex]\stackrel{\rightarrow}{r}[/itex]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.
     
  13. Feb 22, 2012 #12
    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 )
     
  14. Feb 22, 2012 #13
     
  15. Feb 22, 2012 #14
    Here's an illustration of what i'm trying to understand more clearly.

    graph.png


    What is incorrect here, if anything?
     
  16. Feb 22, 2012 #15
    I dont 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 dont understand where the last line of working out came from and its use??
     
  17. Feb 23, 2012 #16


    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.
     
  18. Feb 23, 2012 #17
    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.
     
  19. Feb 23, 2012 #18
    So you'd confirm that everything in that illustration (besides the negative signs within the distance formula) is accurate?
     
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