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Linear acceleration as a fx of displacement

  1. Sep 14, 2008 #1
    Hi everyone. I just began my Calculus based physics course 3 weeks ago and I really love it. I've been going ahead of the class though and came across the following problem:

    As g, the acceleration due to gravity, is constant for only a limited range of height distances, a better approximation would be the linear equation

    a(h) = g - hg'

    where h is the distance from the ground and g' is a small constant of proportional dimensions.

    My problem is that I now need to derive velocity as a function of distance, and then as a function of time.

    I've tried everything I can think of, but I just can't seem to wrap my head around how I can express height as a function of time, or vice versa, without knowing either acceleration as a function of time or velocity as a function of time, but I can't get either of those without knowing one of the other two!!!!

    Any help would be greatly appreciated. Thanks in advance.
     
  2. jcsd
  3. Sep 15, 2008 #2

    HallsofIvy

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    Actually, a(h)= GM/(h+ R)2 where M is the mass of the earth, R is the radius of the earth and G is the "universal gravitational constant" but yes, that is a better approximation for a(h) than just a constant.

    With continuously varying quantities, acceleration is the derivative of v, with respect to time: a= dv/dt. By the chain rule, then a= (dv/dx)(dx/dt)= v(dv/dx). To find the acceleration as a function of height, x, you would need to solve the differential equation v(dv/dx)= g- g' x. That can be "separated' as dv= (g- g'x)dx and integrated directly.
     
  4. Sep 15, 2008 #3
    I was able to get to this point myself, but I'm confused as to where the v goes in this equation. It seems to me that the equation would be v(dv) = (g0 - g'x)(dx), and I don't quite understand how the v(dv) would represent velocity as a function of height...??
     
  5. Sep 15, 2008 #4

    HallsofIvy

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    You integrate, as I said. [tex]\int vdv= \int (g_0- g' x)dx[/tex]. You will get
    [tex]\frac{1}{2}v^2= g_0x- \frac{g'}{2}x^2+ C[/itex]. The "C" will depend upon the initial height and velocity.
     
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