Hello, skyhawk714!

$\displaystyle \text{A runner sprints around a circular track with a radius 100 m}$

$\displaystyle \text{at a constant speed of 7 m/s. }\:\text{The runner's friend is standing}$

$\displaystyle \text{at a distance 200m from the center of the track.}$

$\displaystyle \text{How fast is the distance between them changing}$

$\displaystyle \text{when the distance is 200 m?}$

Code:

* * *
* * R
* o
* 100 * * * x
* *
* * @ * *
* o - - - - * - - - - o F
* O 100 *P 100
* *
* *
* *
* * *

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

The runner is at $\displaystyle \,R\!:\;OR = 50.$.

His friend is at $\displaystyle \,F\!:\;OF = 200 $

Let $\displaystyle \theta \,=\,\angle ROF.$

Let $\displaystyle x\,=\,RF.$

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

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

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

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

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

Differentiate with respect to time:

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

When $\displaystyle x = 200$, we have isosceles triaagle $\displaystyle F\!RO$ with sides 100, 200, 200.

. . And we find that: .$\displaystyle \sin\theta \,=\,\frac{\sqrt{15}}{4}$ .[3]

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

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