1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Displacement after Time T in a Velocity Field

  1. Mar 29, 2012 #1
    I have a velocity field that is static in time. At every location, x, there is a corresponding velocity vector.

    I'm trying to work out the displacement of a particle after T seconds if I drop it into the velocity field at time t=0 and at location x_0.

    I was thinking something along the lines of ∫v(x_0)dt between the limits of t=0 and t=T but it doesn't seem right because the integrand doesn't seem to depend on t at all (in fact v(x_0) is a constant so it can't be right). Maybe there's a double integral involved...

    Anyone know what a suitable form for this equation would be?
  2. jcsd
  3. Mar 29, 2012 #2
    For simplicity, you could consider a one dimensional case:


    This is a separable differential equation, so:

    [itex]\int_{x_0}^{x_f}\frac{dx}{v(x)}=\int_0^T dt[/itex]

    But it gets you a solution for the final time, T.

    [itex]T = \int_{x_0}^{x_f}\frac{dx}{v(x)} [/itex]

    If for a fixed x0, you can suppose that the total time is a function of the ending position.

    [itex]T = f(x_f) [/itex], where this function is given by the right hand side of the equation above.

    So, to get the ending position for a given total time, you would need to take the inverse of that function

    [itex]x_f = f^{-1}(T)[/itex]

    In order to get a closed form for this, you need to plug in the velocity field. I don't think there's a way to get a general expression for an arbitrary velocity function.

    Now, for a two dimensional problem you need to do this for each component of the position vector.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook