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Area under speed.

  1. Mar 31, 2005 #1
    Hi,

    Is the exact displacement = sum of ( speed * time ) ?

    thanks.
     
  2. jcsd
  3. Mar 31, 2005 #2
    Exact displacement is the area between a velocity curve and the x-axis.
     
  4. Mar 31, 2005 #3

    dextercioby

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    It's not a sum,but an integral...

    [tex] \int d\vec{r} =\int \vec{v} \ dt [/tex]

    Choose two moments of time [itex] t_{1} < t_{2} [/itex] and u'll find

    [tex]\vec{r}\left(t_{2}\right)-\vec{r}\left(t_{1}\right) =\int_{t_{1}}^{t_{2}} \vec{v} \ dt [/tex]

    Daniel.
     
  5. Mar 31, 2005 #4
    integral is sum...
     
  6. Mar 31, 2005 #5

    dextercioby

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    No.The integral is a limit of a Riemann/Darboux/... sum...

    Daniel.
     
  7. Mar 31, 2005 #6
    i forgot to mention that the time is very very very small... it is in measured in miliseconds
     
  8. Mar 31, 2005 #7

    dextercioby

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    The process of taking a limit is independent of the unit chosen for a physical quantity...

    Daniel.
     
  9. Mar 31, 2005 #8
    so how do I make a good estimation of the area under the curve given the speed and time in miliseconds?
     
  10. Mar 31, 2005 #9

    dextercioby

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    By evaluating that integral...Formulate the problem with its original text.Then we'll see what it needs to be done.

    Daniel.
     
  11. Mar 31, 2005 #10
    Just try to apply a basic discrete integration scheme, like rectangles :

    [tex] d=\sum_{i=1}^n v_i*\Delta t_i [/tex]

    I suppose in your case, the intervals [tex] \Delta t_i [/tex] are all the same.
    You can also use trapezoidal scheme.

    But you can apply better integration scheme, like Simpson (2n order interpolation), or even higher splines stuff, aso...
     
  12. Mar 31, 2005 #11
    what is the Simpson's method?
     
  13. Mar 31, 2005 #12
    I think in the Simpson method you interpolate 3 points with a parabola and integrate the obtained curve. This is equivalent to the ponderation : 2/3*rectangle+1/3*trapezes and gives :

    [tex] d=\sum_{i=1}^{N-1}(v(t_{i+1})+4v(\frac{t_i+t_{i+1}}{2})+v(t_{i})) \frac{\Delta t_i}{6} [/tex]

    It's of double order than rectangles....
     
  14. Mar 31, 2005 #13
    The easiest way is to find a fit to your curve, integrate that wrt time, and evaluate your endpoints. Best estimation you could get.
     
  15. Mar 31, 2005 #14
    If you are given intervals in time and speeds that remain constant (piecewise, presumably) over those intervals, then taking a sum over all the intervals of the time on the interval times the speed on the interval will give you the exact displacement, yes.
     
  16. Mar 31, 2005 #15

    Thank you.
     
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