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Free falling with air resistance question

  1. Jul 10, 2016 #1
    1. The problem statement, all variables and given/known data
    An object falls under gravity and air resistance force bv. It is known there is a characteristic time τ=m/b such that the object comes close to the terminal velocity for t>>τ. Derive an exact expression for the distance d travelled at time t.

    2. Relevant equations
    not sure, dv/dt=g-γv....(1) ,where γ=b/m, or v=v∞(1-e^(-γt))....(2) , v∞=g/γ...(3)
    3. The attempt at a solution
    the answer is d=mgτ/eb.
    First, I don't understand why cant I integrate equation (1) this way.....
    dv/dt=g-γv
    v=gt-γd
    d=(gt-v)/γ =gτ^2-gτ^2 =0 which is wrong

    Then, I tried to integrate equation (2),
    d=v∞(t+γe^(-γt))+C
    put t=0,v=0,d=0
    C=-v∞γ=-g
    so,
    d=v∞(t+γe^(-γt))-g
    putting the information inside
    d=gτ(τ+(1/τe))-g
    =gτ^2 + g/e -g
    =(gτ^2e+g-ge) /e

    PLEASE HELP , THX
     
  2. jcsd
  3. Jul 10, 2016 #2

    cnh1995

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    dv/dt=g-γv is a differential equation which describes velocity as a function of time i.e. v(t). Velocity is not constant w.r.t. time but is a function of time .So, you can't integrate the RHS as ∫(g-γv)dt. You'll have to separate the appropriate variables.
     
  4. Jul 10, 2016 #3

    cnh1995

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    The answer given for d is true for t=τ and not t>>τ.
     
  5. Jul 10, 2016 #4
    You can integrate that way (taking definite integrals in both sides with time points t=0 and ##t=\tau##) but ##v(\tau)## is not equal to ##g\tau## because the object doesn't do free fall exactly.

     
  6. Jul 10, 2016 #5
    Thanks. But at time t>>τ, I set dv/dt = 0, then dv/dt = g-γv , so itsn't v=gτ (1/γ = τ)?
     
  7. Jul 10, 2016 #6
    yes that's correct. But this doesn't mean that d=0, it is still ##d=(gt-v)/\gamma## where ##t>>\tau## not ##t=\tau##
     
  8. Jul 10, 2016 #7

    cnh1995

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    t>>τ implies that the velocity at time t is g/γ. At time τ, the velocity is still changing.
     
  9. Jul 10, 2016 #8
    Oh I got it.:) But how to solve the question with equation (2) now?
     
  10. Jul 10, 2016 #9
    You can continue with the initial approach and plug in ##v(\tau)## from the equation ##v(t)=v_{\infty}(1-e^{-\gamma t})##.

    Or as you go by integrating (2) I see a mistake in the integration, it is ##\int -e^{-\gamma t}=\frac{e^{-\gamma t}}{\gamma}+C##

    But you got me confused, you want d at time ##\tau## or d at time ##t>>\tau##?
     
  11. Jul 10, 2016 #10
    Oh my god.... thanks for reminding me, now I've solved it
     
  12. Jul 10, 2016 #11
    I think the question need me to find t tends to τ, as v at τ is the same as t>>τ ,so my dv/dt can't set to be 0 as the curve hasn't level off. Not sure about it
     
  13. Jul 10, 2016 #12
    No sorry velocity at time ##t=\tau## is quite different from velocity at time ##t>>\tau##.

    The velocity at time ##t=10\tau## is probably close enough to the final velocity ##v_{\infty}##
     
  14. Jul 10, 2016 #13

    cnh1995

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    No. Terminal velocity is attained at t>>τ. As Delta2 said, it could be 10τ. In fact, the output is approximately equal to the steady state value after 5τ (99%). So practically, for any value of t after t=5τ, the output can be said to have reached steady state.
     
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