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Homework Help: Drag force F=kv^a

  1. Dec 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Only force acting on body is drag force $$F=kv^a$$ where k is constant, a coefficient bigger than zero and v is velocity. Body is given initial velocity u.
    Find a such that body travels infinitely large distance.

    2. Relevant equations

    3. The attempt at a solution
    Since distance is infinite I think that v should never reach zero.
    From $$-m \frac{dv}{dt}=kv^a$$ I found v(t) :
    $$v=(u^{-a+1}- \frac{kt}{m}(1-a))^{\frac{1}{-a+1}}$$
    Here I looked at cases a<1,a=1,a>1 and im not sure but i think that a=1 is good.
    Is this ok?
    Last edited: Dec 28, 2014
  2. jcsd
  3. Dec 28, 2014 #2
    Unless I'm wrong, if ##a=1## then ## v(t) = ue^{-\frac{k}{m}t} ## and it tends to 0 rather quickly, so it won't travel infinitely large distance.
  4. Dec 28, 2014 #3
    How did you get this?
  5. Dec 28, 2014 #4
    Ok I see why a=1 doesn't work now...
  6. Dec 28, 2014 #5


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    Good, but you're still not at the answer. How v behaves with time is useful, but you need to get to how distance depends on time.
    That's not a correct deduction.
  7. Dec 28, 2014 #6
    Actually, it is a correct deduction.

  8. Dec 28, 2014 #7


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    It may be a correct statement, but it is not correct as a deduction.
  9. Dec 28, 2014 #8
    Right. They would have to show it mathematically.

  10. Dec 28, 2014 #9
    I understand Haruspex's remark: if for example the speed was ##v(t) =\frac{1}{t}##, it would still tend to 0 as ##t## grows to infinity, but the displacement would be infinite. It doesn't prove anything.

    The correct explanation would be that the displacement is :

    ##x(t) - x(0) = \int_0^t v(s) ds = - \frac{mu}{k}(e^{-\frac{k}{m}t} - 1)##.

    Since ##|x(t)-x(0)| \le \frac{mu}{k}##, the displacement is bounded independently of time, so ##a## cannot be equal to 1.
  11. Dec 29, 2014 #10

    Try ##a=2## ! In that case, your differential equation can be written more simply: ##\frac{k}{m} = \frac{d}{dt}(\frac{1}{v})##;
    and the displacement will grow infinitely large as time grows.
  12. Dec 29, 2014 #11


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    | am somewhat surprised by the sign of k. If there is mention of drag force, I would expect the force vector to oppose the velocity vector. It is fine to use magnitudes, but then I would still expect to see F = - kva o_O

    and then a=2 is no good either !
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