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Calculate an acceleratin in a racing (drag) application

  1. Feb 20, 2008 #1
    I am trying to calculate an acceleratin in a racing (drag)application. I am an engineer at a medical implant manufacturing company so i know a bit about calulations but it has been a while for most (computers,,) Here is the scenario, we race from 150' to 300'. the quickest time determines the winner. I would like to know what type of speeds are achieved. an example would be 180 foot track with a time of 3.3 seconds from a dead stop. In most cases, the acceleration curve would continue past the 200 or 300ft if it would be continued. do you have any suggestions?
  2. jcsd
  3. Feb 20, 2008 #2

    Doc Al

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    Staff: Mentor

    You can find the average speed by simply taking Distance/time. In your example, 180 ft/3.3 seconds = 54.5 f/s = 37.2 mph. Assuming you accelerated uniformly from 0, your speed at the 180 ft mark would be twice your average speed = 74.4 mph.

    Hopefully some knowledgeable racing fan might be able to give you more practical advice about determining top speed in a real race.
  4. Feb 20, 2008 #3
    erm i was thinking maybe if you knew the relation between the acceleration and time, then you could integrate and plot out the graph for the velocity.
  5. Feb 20, 2008 #4

    Doc Al

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    What do you mean by acceleration curve? Using your data from before (180 ft in 3.3 seconds) you can calculate your average acceleration during that interval using:

    [tex]a_{ave} = \frac{2d}{t^2}[/tex]

    This gives you an average acceleration of about 33 ft/s^2.

    Again, assuming that your acceleration is constant from the start, you can calculate your speed at any point using:
    [tex]v = \sqrt{2a_{ave}d}[/tex]

    at 200 ft: v = 115 ft/s = 78 mph
    at 300 ft: v = 141 ft/s = 96 mph

    On the other hand, if the acceleration varies with time, you'd have to factor that into your calculation (as Oerg suggests). If you knew how it varied.
  6. Feb 20, 2008 #5
    A racing car would initialy accelerate with aproximately constant acceleration (limited by friction), resulting in the covered distance:


    a=g*k (k is coefficient of friction)

    After a time certain time the speed becomes large enough that the power of the engine becomes the limit:

    a=P/(m*v) -> mv^2/2=P*(t-to)+W(to) ,

    The second formula replaces the first one when the second acceleration becomes
    smaller than the first one.

    If you are interested in covered distance, then integrate v(t):

    s(t)=Integral(v(t')dt') from 0 to t
  7. Feb 20, 2008 #6


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    I certainly hope that doesn't mean that there's a Hemi-powered colonoscope in my future. :eek:
    However brief they might be, your acceleration curve will also be determined by your shift points. Even with a tweaked automatic, there'll be a tiny 'hiccup' for each gear.
  8. Feb 20, 2008 #7
    what i meant by the acceleration curve would continue is that most cases acceleration does not plateau in the distance of the race. Thanks for all your replies, of course this if just a i wonder,,,,

    oh, luckily we only make spine implants
    Last edited: Feb 20, 2008
  9. Feb 20, 2008 #8


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    Glad to hear it.
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