1. Not finding help here? Sign up for a free 30min 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!

Integration problem

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle is moving along a straight line such that its acceleration is defined as a=(-2v)m/s^2.
    If v=20 m/s when s=0 and t=0, determine the particle's velocity as a function of position and the distance the particle moves before it stops.

    The answer is v=(20-2s)m/s ; s=10m

    2. Relevant equations

    a=[tex]\frac{dv}{ds}[/tex]*[tex]\frac{ds}{dt}[/tex] which becomes...


    3. The attempt at a solution
    given: a=(-2v) m/s
    v=20 m/s
    solution attempt:




    [tex]\int dt[/tex] = [tex]\int\frac{dv}{-2v}[/tex]

    That's as far as I can get. If I evaluate the left hand side from 0 to t, I get t.
    If I evaluate the left, I get messed up. If I pull out the -[tex]\frac{1}{2}[/tex],
    I am left with [tex]\frac{1}{v}[/tex]. I think the integral of that is ln v.
    I need some direction on this one please.
  2. jcsd
  3. Jan 17, 2009 #2
    This problem is about damping. You have an object moving against a force proportional to its velocity, but directed opposite to it: F = -bv. The solution is an exponentially decaying one. This is what you got!

    -2t = ln v + C ===> [tex]v = v_0 e^{-2t}[/tex]
  4. Jan 17, 2009 #3
    I need velocity as a function of position.
    How about this?
    Can I use the equation v^2=v0^2+2a([tex]\Delta s)[/tex]

    then I get v= [tex]\sqrt{200-4s}[/tex]

    which would give me v=20-2s which is the answer in the book.

    My concern here is the "s" and the value for "a" is -2v not just -2.
    How do I deal with this under the square root sign?
  5. Jan 17, 2009 #4
    You can't use that equation because that only applies when one has constant acceleration.

    (1) [tex] v= v_0 e^{-t/\tau}[/tex]

    (2) [tex] x = v_0\int_0^t e^{-\frac{t'}{\tau}} dt' = v_0\tau(1 - e^{-\frac{t}{\tau}})[/tex]

    Solve for time in second equation (2). Then you can substitute that into (1). It will get hairy, but you should get an answer like the one in your book. Remember: [tex]e^{\ln{x}} = x[/tex].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Integration problem
  1. Integral problem (Replies: 2)

  2. Integration problem (Replies: 8)

  3. Integral problem. (Replies: 2)

  4. Integration problem (Replies: 4)

  5. Integration problem (Replies: 3)