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Thread: Circular Track (First Meeting and Second Meeting)

  1. #1
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    Circular Track (First Meeting and Second Meeting)

    Natsu and Gray both run around a circular track. They both started to run at the same time. Gray started at the westernmost point of the track and runs counterclockwise, and Gray takes about ninety-five seconds to run one lap of the track. Natsu runs clockwise and he passes Gray for the 1st time after eleven seconds. Natsu runs at 3 meters per sec (speed) and takes 80 seconds to run 1 lap of the track.

    What are Natsu's coords when he passes Gray for the 1st time?
    What I did:
    I knew that it took Gray about 95 seconds to run 1 full lap (which is equal to 2pi radians). So 2pi/95 rad/sec is his angular speed.
    While Natsu's angular speed is then 2pi/80 rad/sec
    Therefore speed = angular speed * radius
    I knew one of the speed was 3, so 3 = (2pi/80)*r, and solved for r to be 240/2pi = 120/pi
    Then I knew θ = angular speed * time, so I found the θ to be 11 * 2pi/95

    There I put everything into this: rcos(θ+pi), rsin(θ+pi)
    Where I got the coords: (-28.52, -25.40) for when Natsu meets Gray for the first time. (So I got the answers correct here, the second question is where I need help).

    What are Natsu's coords when he passes Gray for the second time?
    This is the part I got lost on:
    Ok, so I know Natsu's angular speed is 2pi/80
    While Gray's angular speed is 2pi/95

    What I tried doing was to find the time, through θ = angular speed * time
    So then θ = 2pi/80 * t (for Natsu)
    And 2pi - θ = 2pi/95 * t (for Gray)

    Though I intended to solve for t (time) and plug in t+11 to the coordinate equation: rcos(θ+pi), rsin(θ+pi)
    Though I got lost on how to solve for the angle this time... which then gave me troubles finding time.

    I knew that the angle had to be something like 2pi - θ at least... which just basically means the angle they swept through to meet the second time.
    Anyway, what should I do from here? Unless there's another easier way.

    Answers for the second part: (34.256, 16.898)

    Thanks
    Last edited by Chaim; Dec 2nd 2014 at 07:41 PM.
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  2. #2
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    Re: Circular Track (First Meeting and Second Meeting)

    Hey Chaim.

    Your total track length is 2*pi*r = 240 metres. If you can convert your co-ordinates to a distance on that line then you have two linear systems where:

    d(1) = start1 + rate1*time
    d(2) = start2 + rate2*time

    You are solving for time and start1 represents the proportion of the distance travelled with respect to the radius (i.e. the value of r*theta for a given theta) while the rate represents the speed rate in metres per second.

    If the two distances are the same (i.e. d(1) = d(2)) then it means that both meet and this will allow you to solve for the unknown time value.
    Thanks from Chaim
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  3. #3
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    Re: Circular Track (First Meeting and Second Meeting)

    Oh I got it now. Thanks
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