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Velocity of person on zip line

  1. Dec 18, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find out how I can find the velocity function and position function (and the like) of a person on a zip line using calculus (this is for a calc 1 class).

    Here is what I have:

    height at start of zip line: 10m
    height at end of zip line: 8m
    Length of zip line: 25m
    Height of person: 1.8m
    Mass of person: 65 kg
    gravity: -9.8 ms/s^2

    Thanks!

    2. Relevant equations



    3. The attempt at a solution

    I know that the equation of the zip line is y = -2/25x +10
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Dec 18, 2012
  2. jcsd
  3. Dec 18, 2012 #2

    haruspex

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    According to your equation for y, after 25m it will be at 12m height.
    Do the usual free body diagram and figure out the net force to get the acceleration. Write down the resulting differential equation and post your working.
     
  4. Dec 18, 2012 #3
    thanks

    It's actually 8 ( I mistakenly forgot to add the negative sign). Not sure what a free body diagram is. It's for a calc class
     
  5. Dec 18, 2012 #4

    haruspex

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    A free body diagram considers the forces, torques and accelerations on a rigid body. The body may be part of a larger system. So for the person on the zipline at some distance x from the start, what are the forces acting and what is the acceleration?
    (The much easier way to solve the question is by conservation of energy, but I assume you are supposed to use calculus here.)
     
  6. Dec 18, 2012 #5

    I like Serena

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    Welcome to PF, trest! :smile:

    What formulas are you supposed to use?
    ##v=at## with ##x=\frac 1 2 at^2##?
    Or ##mgh=\frac 1 2 mv^2##?
    Or something else?
     
  7. Dec 18, 2012 #6
    thank you both but honestly we were given nothing. I estimated all those from a video we were given and this is for a calc 1 class so I suppose we have to use antiderivatives to derive the formulas or something.

    anyways thanks
     
  8. Dec 18, 2012 #7

    I like Serena

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    Well, you have a gravity constant.
    And you are mentioning velocity and position functions.

    Sounds like you're supposed to use for instance that the velocity function is the time derivative of the position function.
    Aren't you?

    And it suggests that you are supposed to use a time derivative of the velocity in a way that it relates to the gravity constant.
    But before you can do that, you need a little more about how gravity and an object on a zip line behaves....
     
  9. Dec 18, 2012 #8
    I think i figured it out using a = sin(theta)*g
    then took the antiderivative giving me the velocity function. However, according to this function the velocity increases as time goes on indefinitely. does this make sense physically?
     
  10. Dec 18, 2012 #9

    I like Serena

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    Yes.
    In ideal circumstances (that is in the absence of friction and air resistance) the velocity increases linearly and indefinitely.
     
  11. Dec 18, 2012 #10
    Alright thank you! Do you happen to know of a simple way to calculate air resistance of a person on a zip line?
     
  12. Dec 18, 2012 #11

    I like Serena

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    Nope.
    That is out of scope for calc 1.
    If you want to know more about it, read http://en.wikipedia.org/wiki/Drag_(physics)

    In short, air resistance is a constant times speed squared.
    But the value of the constant is hard to predict, and the resulting equation can't be properly integrated.
     
  13. Dec 18, 2012 #12
    okay I think I have everything I need and thanks for your time :)
     
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