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Average Acceleration of a Accel vs Speed Graph

  1. Dec 12, 2012 #1
    Hi All

    So I have an Accel vs Speed graph and data.
    Goes from 0 to 80 mph.
    This is all the info I have.

    I would like to figure out what Average Acceleration would give me the same distance, of an acceleration from 0 MPH to 80 MPH.
    AKA, using the formula Distance = (Vfinale^2)/(2*Acceleration)

    Any ideas on how to figure this out?
     
  2. jcsd
  3. Dec 12, 2012 #2

    mfb

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    If you plot 1/acceleration versus speed, your function value is proportional to the time spent at a specific velocity (assuming your acceleration is always positive).
    If you plot speed/acceleration versus speed, the area below the function is proportional to the distance travelled.
     
  4. Dec 14, 2012 #3
    Sorry, but that still confuses me as the units don't add up. Also when I attempt to plot this out in excel and add up the area under the curve, the resulting numbers cannot be right
     
  5. Dec 14, 2012 #4

    sophiecentaur

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    The area under the v/t graph should give total distance, whatever the instantaneous velocity and accelerations are. Does that hold for an (a/v)/v graph too? Is there not a constant needed after the integration?
     
  6. Dec 14, 2012 #5

    mfb

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    A simple example: v(t)=at+c with constant a and c.
    Distance after time T is ##\frac{1}{2}aT^2 + cT##.

    v/a = t + c/a
    t=(v-c)/a
    Therefore, v/a = (v-c)/a + c/a = v/a (trivial in this example, as a is constant)
    Speed increases from c to c+aT. If we integrate v/a from c to c+aT, we get ##\int \frac{v}{a}dv = \frac{1}{2a}((c+aT)^2-c^2) = cT+\frac{1}{2}aT^2## - the same as above.
     
  7. Dec 14, 2012 #6

    sophiecentaur

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    Wine and gin make this less accessible but the sums seem right in this case. It that also OK for non constant acceleration? I guess it could be - piecewise. Good one. I love it when someone else does the algebra and I can follow it.
     
  8. Dec 14, 2012 #7
    Not just proportional to the distance traveled. It is equal to the distance traveled. Another way to do this is to plot 1/(2a) as a function of v2. This will also be equal to the distance traveled.
     
  9. Dec 15, 2012 #8

    mfb

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    I did not check the prefactor when I wrote my first post here, but it is 1, right.

    It works there as well, but the integration can get more difficult.
     
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