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Velocity for t->infinity with given equation of motion

  1. Dec 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A body of unit mass, whose position is x(t) is subject to a velocity-dependent force of the form
    F=a(dx/dt)-b(dx/dt)^2
    where a and b are positive constants and the positive x direction is to the right.
    a) Write down the equation of motion
    b) If the motion is initially to the right, what would be the velocity for t->infinity?

    2. Relevant equations


    3. The attempt at a solution
    Part a is straight forward. Since m is a unit mass, we can set m=1. Then we have F=(1)x''=ax'-b(x')^2 =>x''-ax'+b(x')^2=0.
    But I do not know how to start with part b.
    The solution says "the equation has fixed point at v=a/b so it will asymptotically reach a/b as t->infinity" but I am not sure how this came about. Why set x''=0 to find the velocity at t->infinity?
    Thanks!
     
  2. jcsd
  3. Dec 12, 2015 #2

    nrqed

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    As t goes to infinity, the velocity approaches an asymptotic value which, by definition, is a constant. In other words, the velocity approaches a certain value and does not change in time anymore, so its derivative (the acceleration) is then zero. This is the trick used to find, for example, the terminal velocity of a falling object: upon reaching the terminal speed, the acceleration is zero.
     
  4. Dec 12, 2015 #3

    PeroK

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    It's probably a good idea to check for yourself why the asymptotic argument works in this case. If you draw a graph of acceleration against velocity (##acc = av - bv^2##) then you'll notice that for ##0 < v < a/b## acceleration is positive, for ##v = a/b## acceleration is 0 and for ##v > a/b## acceleration is negative.

    If ##v_0 > 0##, then there are two cases: ##0 < v_0 < a/b## and ##v_0 > a/b##. You should try to figure out what happens to ##v## in those two cases, as ##t## increases.
     
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