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Math Help - distance question

  1. #1
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    distance question

    I'm havin a problem with a word problem that read
    "a runner sprints around a circular track with a radius 100m at the constant speed of 7m/s. The runner'sf riend is standing at a distance 200m from the center of the track. How fast is the distance between them changing when the distane is 200m?"
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  2. #2
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    Draw a diagram first.

    Setup a coordiante system. Make the (0,0) the circle's center. Make (200,0) the position of the friend.

    Can you write the equations for the runner's position on the circle over time? Something like x(t) = 100cos(At), y(t) = 100sin(At), but you have to find A so that the speed is 7m/s
    Now can you write the equation for the distance between the runner and the friend over time? Call this function dist(t)
    When is the distance 200? dist(t) = 200
    How fast is the distance changing at this time?
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  3. #3
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    okay that makes sense but how do you find A and what would t stand for??
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  4. #4
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    t stands for time.

    A is your angular speed. Remember linear speed = angular speed * radius.

    You can also take derivative of x and y w.r.t t
    speed is sqrt(x'^2 + y'^2). Which you will find speed = 100A.
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  5. #5
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    Hello, skyhawk714!

    \text{A runner sprints around a circular track with a radius 100 m}
    \text{at a constant speed of 7 m/s. }\:\text{The runner's friend is standing}
    \text{at a distance 200m from the center of the track.}

    \text{How fast is the distance between them changing}
    \text{when the distance is 200 m?}
    Code:
                  * * *
              *           *  R
            *               o
           *          100 *  * *   x
                        *         *
          *           * @     *      * 
          *         o - - - - * - - - - o F
          *         O   100   *P  100
    
           *                 *
            *               *
              *           *
                  * * *

    The track is a circle with center \,O and radius 199 m.

    The runner is at \,R\!:\;OR = 50..
    His friend is at \,F\!:\;OF = 200
    Let \theta \,=\,\angle ROF.
    Let x\,=\,RF.

    \text{The runner sprints along arc }\,s \,=\,\overline{PR}\,\text{ at }\,\dfrac{ds}{dt} = 7\text{ m/s.}

    \text{Since }s \:=\:r\theta,\,\text{ then: }\:\dfrac{ds}{dt} \:=\:r\dfrac{d\theta}{dt} \quad\Rightarrow\quad \dfrac{d\theta}{dt} \:=\:\dfrac{1}{r}\dfrac{ds}{dt}

    . . Hence: . \dfrac{d\theta}{dt} \:=\:\dfrac{7}{100} \:=\:0.07 .[1]


    Law of Cosines: . x^2 \;=\;100^2 + 200^2 - 2(100)(200)\cos\theta

    . . . . . . . . . . . . x^2 \;=\;50,\!000 - 40,\!000\cos\theta


    Differentiate with respect to time:

    . . 2x\dfrac{dx}{dt} \;=\;40,\!000\sin\theta\dfrac{d\theta}{dt} \quad\Rightarrow\quad \dfrac{dx}{dt} \;=\;\dfrac{20,\!000\sin\theta}{x}\dfrac{d\theta}{  dt} .[2]



    When x = 200, we have isosceles triaagle F\!RO with sides 100, 200, 200.
    . . And we find that: . \sin\theta \,=\,\frac{\sqrt{15}}{4} .[3]


    Substitute [1] and [3] into [2]:

    . . \dfrac{dx}{dt} \;=\;\dfrac{20,\!000\left(\frac{\sqrt{15}}{4}\righ  t)}{200}(0.07) \;\approx 6.78\text{ m/sec}

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