Hello Reid,
Let's work this problem in general terms, derive a formula, and then plug in the given data.
Let $S$ be the distance of the spectator from the center of the track, $r$ be the radius of the circular track, $D$ be the distance between the runner and the spectator and $v$ be the speed of the runner.
Please refer to the following diagram:
View attachment 893
We see, using the law of cosines, that we may state:
(1) $$D^2=r^2+S^2-2rS\cos(\theta)$$
Differentiating with respect to time, we find:
$$2D\frac{dD}{dt}=2rS\sin(\theta)\frac{d\theta}{dt}$$
$$\frac{dD}{dt}=\frac{rS}{D}\sin(\theta)\frac{d \theta}{dt}$$
Now, since the distance the runner has covered is $$d=r\theta$$ and the relationship between distance, constant velocity and time is $$d=vt$$ we may state:
$$vt=r\theta$$
Differentiating with respect to time $t$, we have:
$$v=r\frac{d\theta}{dt}$$
Hence:
$$\frac{d\theta}{dt}=\frac{v}{r}$$
And so we may state:
$$\frac{dD}{dt}=\frac{Sv}{D}\sin(\theta)$$
We also have via Pythagoras and (1):
$$\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}=\pm\frac{\sqrt{\left((r+S)^2-D^2 \right)\left(D^2-(r-S)^2 \right)}}{2rS}$$
And so we may state:
$$\left|\frac{dD}{dt} \right|=\frac{v\sqrt{\left((r+S)^2-D^2 \right)\left(D^2-(r-S)^2 \right)}}{2Dr}$$
Plugging in the given data, we find in meters per second:
$$v=4,\,r=130,\,S=210,\,D=210$$
$$\left|\frac{dD}{dt} \right|=\frac{4\sqrt{\left((130+210)^2-210^2 \right)\left(210^2-(130-210)^2 \right)}}{2\cdot210\cdot130}=\frac{2\sqrt{1595}}{21}\approx3.803566770993497$$
Hence, the correct answer is d), rounded to the nearest thousandth.
