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Acceleration as a function of time and position

  1. Apr 26, 2012 #1
    I'm quite curious as to whether there is an equation (whether it be in scalar, vector, or tensor form) that defines the acceleration of a particle as a function of the three space coordinates and time.

    My curiosity arose when I was thinking of the one dimensional equation:
    a = v*(dv/dx); where the acceleration and velocity are only defined in the x-direction.

    I thought the generalization to three dimensions would be:
    a = vx*(dvx/dx)i + vy*(dvy/dy)j + vz*(dvz/dz)k

    To include the time variable I thought it might just be the material derivative of acceleration (like in fluid mechanics).

    These assumptions are just based on my gut feeling (I have done no derivation whatsoever). Is there an actual equation?
     
  2. jcsd
  3. Apr 26, 2012 #2
    Well I don't know if there is a general equation (which there probably is), but in one dimension you are just looking for the second derivative of the equation of motion of the body.

    In more than one dimension you can just split it up into an arbitrary amount of simultaneous parametric equations, and once again its the second derivative of each individual equation.

    Once again there might be a general rule, but what I just said above will definitely work.
     
  4. Apr 26, 2012 #3
    In one dimension we can say dV/dt=(dx/dt)(dV/dx)=Vx(dV/dx) (but the sub x notation isn't needed because there is only one velocity, and its in x)

    but for 3 dimensions, instead of x we use R

    so dV/dt=(dR/dt)(dV/dR)=V*(dV/dR)

    and

    dV/dR = [itex]( \partial V/ \partial x )i +( \partial V/ \partial y )j +( \partial V/ \partial z )k [/itex]

    so

    a = |V|*[itex]\left[( \partial V/ \partial x )i +( \partial V/ \partial y )j +( \partial V/ \partial z )k \right] [/itex]

    I believe you have to use the magnitude of V, rather than dotting V into dV/dR, for some reason. I can't explain/remember why, but I think this is how it is supposed to be. Could be wrong, idk. but that's my 2 cents, hope it helps
     
  5. Apr 27, 2012 #4
    - The equation seems right, but then again why wouldn't the material derivative of acceleration apply (as in fluid mechanics). Is there an assumption I am missing that doesn't apply to ordinary particles? The equation is:

    [itex]\vec{a}[/itex] = [itex]\frac{D\vec{v}}{Dt}[/itex] = [itex]\vec{v}[/itex][itex]\cdot[/itex]([itex]\vec{\nabla}[/itex][itex]\vec{v}[/itex]) + [itex]\frac{\partial \vec{v}}{\partial t}[/itex]
     
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