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Modelling Acceleration to Constant Velocity

  1. Oct 29, 2013 #1
    Hi there,

    I'm trying to create a computer model which takes into account output power in a car due to acceleration, drag and rolling resistance. The car will be accelerating for a set amount of time and I am able to model the power required for this with respect to time, as well as the power required for the rolling resistance and drag as it is accelerating.
    However, after it reaches a certain velocity, its velocity will remain constant (the acceleration power will drop to zero and the drag and rolling resistance will then be constant). Is there anyway of fitting all of this into one formula for programming purposes so that when it reaches a certain velocity it will then remain constant?

    Thanks for any help.
     
  2. jcsd
  3. Oct 29, 2013 #2

    arildno

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    Hmm..how precisely, have you modelled the acceleration?
     
  4. Oct 29, 2013 #3
    Not very precisely at all. Just using basic equation of motions at a constant acceleration. I'm at a very basic level of the model so far and my brain isn't really working today. Basically, i'm trying to find a way to model the power output with respect to time incorporating both the accelerating and constant velocity sections using one equation. Will I need to use a differential equation?
     
  5. Oct 29, 2013 #4

    arildno

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    If you just want a look-alike development (rather than something realistic), why not just model your final velocity as a constant?
     
  6. Oct 29, 2013 #5
    I suppose I could do that. It's just i'm running it in matlab in respect to time (using a loop). So say I want to graph the output velocity at the end, it would be handy to have something that would give me a constant velocity after a certain number of iterations of the loop..
     
  7. Oct 29, 2013 #6

    arildno

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    Well, use a play formula like the following then:
    [tex]v(t)=v_{final}(1-e^{-kt})[/tex]
    where "k" is a constant>0 modelling the effects of drag&resistance.
     
  8. Oct 29, 2013 #7
    Looks good, will give that a go. Thank you..
     
  9. Oct 29, 2013 #8

    arildno

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    Note that this equation is, basically, the solution of the following differential equation:
    [tex]\frac{dv}{dt}=F/m-kv[/tex]
    where F/m is applied force per unit mass (considered for simplicity as a constant), and -kv a force per unit mass resistance term acting oppositely to the velocity.
    We then have the relation:
    v_final=F/(m*k)
     
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