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Integrating a squared velocity

  1. Feb 8, 2012 #1
    Okay I'm working on making a ballistics calculator and I need to know how to integrate this.

    gif.latex?\frac{d^{2}X}{dT^{2}}=-k(\frac{dX}{dT})^{2}.gif To get velocity, and ultimately to get time of flight. So that I can use that to determine drop with an angle of 0.
     
  2. jcsd
  3. Feb 8, 2012 #2
    Try to solve the equation in v first.
    Replace dX/DT by v (on both sides).
    You will have dv/dT=-kv^2. This can be solved by direct integration, after rearranging a little bit.
     
    Last edited: Feb 8, 2012
  4. Feb 8, 2012 #3

    rcgldr

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    Rearraging, this is what you would integrate, don't forget to include a constant after the integration (then solve for that constant based on initial conditions).

    dv / (-k v2) = dt
     
  5. Feb 8, 2012 #4
    ... that's just initial velocity is it not?
     
  6. Feb 8, 2012 #5

    rcgldr

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    The integral will produce a function of velocity versus time, so the constant would be added or subtracted from the function of velocity at time = zero to account for the initial velocity.
     
  7. Feb 9, 2012 #6
    So this gives me:

    1/kv+vi=T correct?

    From here I changed the V to dx/dt but since it is 1/v it becomes

    (1/k)*dt/dx+Vi=T

    Multiplying by dx I then get

    (1/k)dt+Vidx=Tdx

    Integrating yields

    T/k +ViX=T*X (accounting initial time and distance as zero)

    But that just seems wrong...

    Breaking it down all the way I get :

    (Vi*X)/(x-1/k)=T
     
    Last edited: Feb 9, 2012
  8. Feb 9, 2012 #7
    No. Look at the units.
    You add initial speed and inverse speed (times k).
    It would be more obvious if you put it like
    dv/v^2=-kdt
    The integration on both sides gives
    -1/v=-kt +C

    At t=0 v=vi so
    -1/vi=C or C=-1/vi
    Then you have
    1/v=1/vi+kt
    Solve this for v and you have v(t).
    Then you can integrate v(t) over time to get x(t).
     
  9. Feb 9, 2012 #8
    Cool, so I've solved for V:

    V= 1/(KT+1/Vi)

    Wolfram integrates it as

    Log(KTVi+1)/K=X

    That doesn't seem right,

    If I simply integrate KT+1/Vi with respect to X (because 1/v is dT/dX) I get

    KTX+X/VI=T

    Which becomes

    X/(Vi-KXVi)=T
     
    Last edited: Feb 9, 2012
  10. Feb 9, 2012 #9
    Why doesn't seem right?
    Of course, you should add the constant that takes into account the initial position.

    On the other hand, this does not seem right at all.
    If you write
    1/v=dt/dx=kt+1/vi
    you have mixed variables (t and x). You cannot integrate right away.
    After separation you have
    dx=dt/(kt+1/vi) which seems to be same as above.
     
    Last edited: Feb 9, 2012
  11. Feb 9, 2012 #10
    call it intuition. I'll have to do unit analysis though. K I think has a dimention but IIRC it's inverse mass I'll check after while.
     
  12. Feb 9, 2012 #11
    Sorry, I was editing my post while you posted your reply.
    See above.
     
  13. Feb 9, 2012 #12
    The dimension of k is inverse length.And kt is inverse speed.
     
  14. Feb 9, 2012 #13
    Problem is the solution with the log function is giving me ansers of several thousand years for a flight time

    28560fb9ccae7b5f811de11f965d5478.png This is where I get K, it's everything but the V squared term and divided by mass.

    Edit: hold on, it could be my drag co-efficient is f'ed up.
     
    Last edited: Feb 9, 2012
  15. Feb 9, 2012 #14
    Yup, it was the BC, or at least i think, I'll have to check it out some more. Thanks for the help.
     
  16. Feb 9, 2012 #15
    This may happen very well because you have only the drag force and no other forces.
    After some time (and distance traveled) the speed will decrease practically to zero.
    If your distance is much larger than the distance at which speed becomes very small, the time becomes very large. A plot of t(x) will show you better this behavior.
    For example, taking vi=10 m/s and k=1m^-1, the time to travel 2 m is 0.6 s but the time to travel 20 m is of the order of 10^7 s (about a year).
     
  17. Feb 9, 2012 #16
    Nah I was taking it at 100 meters. It was unit problem with the Drag coefficient converting from BC.
     
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