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Change in motion of a track-runner

  1. Mar 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A runner sprints around a circular track of radius 100m at a constant speed of 7m/s. The runner's friend stands 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m?


    2. Relevant equations



    3. The attempt at a solution

    I'm having a hard time setting this problem up. I drew a picture but honestly that wasn't of much help to me, I'm stuck :/.
     
  2. jcsd
  3. Mar 28, 2013 #2

    mfb

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    How does the picture look like?
    You already know the distance runner<->friend. The track itself gives you another distance, and there is a known third distance. This allows to find the position of the runner.
     
  4. Mar 28, 2013 #3
    Untitled.png

    Not a right triangle so I can't use right-triangle geometry to relate the sides. I was thinking law of cosines for a while but idk.
     
  5. Mar 28, 2013 #4

    HallsofIvy

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    Set up a coordinate system with origin at the center of the track and positive x-axis through the friend.

    At 7 meters per second the runner will run around the track, which has length [itex]100\pi[/itex], in [itex]\frac{100\pi}{7}[/itex] seconds so will run the [itex]2\pi[/itex] radians in [itex]\frac{100\pi}{7}[/itex] seconds. That is, his angular speed is [itex]\frac{2\pi}{\frac{100\pi}{7}}= \frac{7}{5}[/itex] radians per second. Assuming that the runner started at the point on the track closest to his friend, we can write his position at time t seconds after starting as (100 cos(5t/7), 100 sin(5t/7)). His friends position is the constant (200, 0) so the distance between them at time t is [itex]\sqrt{(200- 100cos(5t/7))^2+ 10000sin^2(5t/7)}[/itex] and the rate of change of that distance is the deriative.
     
  6. Mar 28, 2013 #5
    How did you get the length to be 100pi? Shouldn't it be 200pi?
     
  7. Mar 28, 2013 #6

    mfb

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    @HallsofIvy: Setting this equal to 200, solving it for t and putting it into the derivative looks quite messy.

    You can split it in two symmetric, right triangles.
     
  8. Mar 28, 2013 #7
    Yeah I noticed that but I don't understand how I'm supposed to use that information. My problem is I can't figure out a way to set the problem up.
     
  9. Mar 28, 2013 #8

    haruspex

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    I think it will help if you draw the diagram a bit more accurately. If it makes this easier, dispense with the circle and just draw the isosceles triangle with its axis of symmetry, plus a tangent to the circle where the runner is. What do you notice about the axis and the tangent?
    In terms of angles and lengths in the diagram, what is the relationship between the runner's speed and his speed relative to the observer?
     
  10. Mar 29, 2013 #9
    I give up. I can't solve this problem :/. I only got as far as solving for the axis of symmetry. I labeled the tangent line x and figured out that I want dx/dt but I can't figure out what to do with the rest of the information.
     
  11. Mar 29, 2013 #10

    haruspex

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    You need no calculus here - it's mostly geometry.
    Call the centre of the circle O, the runner's position R, the observer's position P. Let A be the midpoint of OR.
    The axis of symmetry is AP, right? What you're after is the component of the runner's velocity in the PR direction.
    The axis of symmetry makes what angle to the radius OR? The tangent at R makes what angle to OR? What does that tell you about the relationship between the tangent and the line AP? Can you work out the angle RP makes to the tangent?
     
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