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Drag Force Physics Problem

  1. Mar 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Assuming that the drag force magnitude is given by the equation D= bv, where b is the drag parameter and v is the instantaneous velocity magnitude of the object.

    (a) Show that the vertical displacement through which a dropped object must fall from to reach X% of its terminal velocity is given by the equation:
    Δy= (v^2ty/g)*[(X/100)-ln((100-x)/100))
    where vty is the object's terminal velocity.

    2. Relevant equations
    v=(-mg/b)*(1-e^(-bt/m))

    3. The attempt at a solution
    I honestly ran around in circles with this problem trying to integrate the given equation desperately trying to figure out what this problem was asking for exactly. Then I integrated v=(-mg/b)*(1-e^(-bt/m)). No luck there attempting to recreate the given equation for about three hours now.
     
  2. jcsd
  3. Mar 14, 2016 #2

    TSny

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    Welcome to PF!

    If you had an expression for dy/dv, maybe you could integrate it to get a relation between y and v.

    Recall the chain rule: dy/dt = dy/dv ⋅ dv/dt
     
  4. Mar 14, 2016 #3
    In your relevant equations you have the equation for velocity as it depends on time. Note that as t approaches infinity limit, [itex] v = -\frac{mg}{b} [/itex]. This is terminal velocity. You want to find how far the distance must be to reach [itex] -\frac{x}{100}\frac{mg}{b} [/itex]. Using this expression, you can solve for time needed to reach that speed. Finding distance is straightforward from there.
     
  5. Mar 14, 2016 #4
    But how would I lead it all back to proving this Δy= (v^2ty/g)*[(X/100)-ln((100-x)/100))
     
  6. Mar 14, 2016 #5
    You are looking for the distance required to reach a specific velocity, yes? If you know the time required and you have the function of the velocity, how would you then go from velocity to distance?
     
  7. Mar 14, 2016 #6

    haruspex

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    Yes, but TSny's method avoids the need to find the time.
     
  8. Mar 14, 2016 #7
    Since v=(-x/100)*(mg/b), I used kinematics equation d=vit+1/2 at^2, assuming a=0, I found t to be -100d*b/mgx
     
  9. Mar 14, 2016 #8

    haruspex

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    That is a SUVAT formula. Those are only valid for constant acceleration.
     
  10. Mar 14, 2016 #9
    What equation should be used then to isolate t?
     
  11. Mar 14, 2016 #10

    haruspex

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    There are several ways open to you.
    You can start with the differential equation for the acceleration, then use TSny's method to make it a diffeential equation only involving velocity, distance, and a derivative of one with respect to the other (so no time in the equation). Solve that.
    Since you are given the solution for velocity as a function of time, you can integrate that. You posted that you tried that but have not posted your working. If you post it we can lead you through it.
     
  12. Mar 14, 2016 #11
    how would you go from t=(-m/b) ln((100-x)/(100)) to the vertical displacement?
     
  13. Mar 14, 2016 #12

    haruspex

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    Which post is that a response to? (Please use the Reply/Quote buttons)
     
  14. Mar 14, 2016 #13
    The original post i guess. I'm working on the same problem.
    The time it takes to reach X% of its terminal velocity is t=(-m/b) ln((100-x)/(100)).
    Vy=(mg/b)(e^(-bt/m)-1)
    If i plug in t to this equation i get
    Vy=(mg/b)(e^(ln((100-x)/(100)) -1)
    But i'm not sure what to do next? i know i need to integrate velocity to get displacement but I'm not sure how
     
  15. Mar 14, 2016 #14

    haruspex

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    You don't know how to integrate ##\frac{mg}b(1-e^{-\frac{bt}m})## with respect to t?
     
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