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Velocity from acceleration if acceleration is a function of space

  1. Feb 22, 2012 #1
    Hello Forum,

    usually the acceleration a is a function of time: a(t)= d v(t) /dt

    to find v(t) se simply integrate v(t)= integral a(t) dt

    What if the acceleration was a function of space, i.e. a(x)?

    what would we get by doing integral a(x) dx? The velocity as a function of space, v(x)?

    but a(x) is not defined to be v(x)/dx or is it? maybe some chain rule is involved..


  2. jcsd
  3. Feb 22, 2012 #2


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    You use chain rule

    a(x) = dv/dt = f(x)

    multiply dv/dt by dx/dx:

    (dv/dt)(dx/dx) = (dx/dt)(dv/dx) = v dv/dx = f(x)

    multiply both sides by dx

    v dv = f(x) dx

    If f(x) can be integrated and the integral of f(x) = g(x), then

    1/2 v2 = g(x) + c

    This results in an equation that relates velocity and position.

    v = sqrt(2 (g(x) + c))

    To get time versus position, you start with

    v = dx/dt = sqrt(2 (g(x) + c))

    and then integrate

    dx/(sqrt(2 (g(x) + c))) = dt

    assuming h(x) is the integral of dx/(sqrt(2 (g(x) + c))), you get

    t = h(x) + d

    where d is constant of integration. It may be possible to solve this equation to get x = some function of t.

    One simple case is a(x) = -x, which results in x(t) = e(sin(t+f)), where e and f are constants.
    Last edited: Feb 22, 2012
  4. Feb 23, 2012 #3
    Thanks rgcldr!

    great explanation and example!

  5. Mar 2, 2012 #4
    Hi rcgldr,

    one question:

    when you get v dv = f(x) dx, is v a function of t or of x, i.e. is v equal to v(t) or v(x)?

    It looks like it would be a function of x, v(x), since 1/2 v^2 = g(x) + c results in v(x) = sqrt(2 (g(x) + c)).

    I am confused because when we write v=dx/dt I always assume that v must be a function of time t, i.e. v(t).
    When in your first step you write a(x) = dv/dt I tend to think that the v must be function of t since dv/dt is a derivative with respect to time.

    So which one is correct? a(x)= dv(x)/dt or a(t)=dv(t)/dt. Acceleration is always the time derivative of the velocity vector but the velocity vector can be a function of any dependent variable: v(t), v(x), v(F), where F is force, etc....

  6. Mar 2, 2012 #5
    I guess my point is: if a function, like a, is defined as a time derivative of another function, dv/dt, does the differentiated function need to be a function if time?
  7. Mar 2, 2012 #6


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    Homework Helper

    At this point, it's just a relationship betwen v and x.

    It is, but I was only able to take advantage of that in the last step where I find t as a function of x.

    True, but I used chain rule to get rid of the dt and end up with v dv = f(x) dx.

    Both are correct, but if acceleration is defined as a function of x, then it may not be possible to find an equation for acceleration as a function of time. In the simple example I gave, a(x) = -x, you will be able to find both a(x) and a(t). I ended up finding x(t) = e sin(t + f). You can take the derivative of this to find v(t), and the derivative of v(t) to find a(t), so a(x) = -x, and a(t) = -e sin(t+f). For other situations, you may not be able to solve for x(t), v(t), or a(t), in which case numerical integration will be required.
  8. Mar 3, 2012 #7
    So it is mathematically legal and ok to write a derivative of a function even if
    the independent variable a of the function and the differentiation variable b are not the same: f(a) and db to create df(a)/db...

    I guess that is all the chain rule is about....
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