Spiraling Motorboats

Rewriting the problem in more general terms: let $R$ be the radius of the circular path of motorboat 1, $S$ be the speed of the two motorboats, and $(r_{1}, \theta_{1})$ and $(r_{2}, \theta_{2})$ respectively be the positions of motorboats 1 and 2 (in polar coordinates, with the buoy at the origin). The velocity vector of a moving particle expressed in polar coordinates (when $\theta$ is in radians) is

$\hat{V}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}$

The speed of a moving object is the magnitude of this vector:

$S=\sqrt{\frac{dr}{dt}^{2}+r^{2}\frac{d\theta}{dt}^{2}}$

Since the speeds of the two motorboats are equal, we have the following equation:

$\sqrt{\frac{dr_{1}}{dt}^{2}+r_{1}^{2}\frac{d\theta_{1}}{dt}^{2}}=\sqrt{\frac{dr_{2}}{dt}^{2}+r_{2}^{2}\frac{d\theta_{2}}{dt}^{2}}$

Because the motorboats and the buoy are always in line, the angles $\theta_{1}$ and $\theta_{2}$ are always equal, so we can just use $\theta$ to describe them both. Because $\theta$ is measured in radians, $\theta=\frac{S}{R}t$ ; therefore $\frac{d\theta}{dt}=\frac{S}{R}$ . $\frac{dr_{1}}{dt}=0$ because $r_{1}=R$ , a constant. Using these facts, we can turn the above equation into the following differential equation:

$\frac{dr_{2}}{dt}=S\sqrt{1-(\frac{r_{2}}{R})^{2}}$

This is a separable DE, and can be turned into the following:

$\frac{dr_{2}}{S\sqrt{1-(\frac{r_{2}}{R})^{2}}}=dt$

Integrating both sides:

$\frac{R}{S}\sin^{-1}(\frac{r_{2}}{R})=t+C$

Because $r_{2}$ , the distance of the second motorboat from the buoy, is zero at time $t=0$ , we can say that the constant $C$ above is simply zero. Therefore $t=\frac{R}{S}\sin^{-1}(\frac{r_{2}}{R})$ . The two boats will meet when $r_{2}=R$ , which will mean that $t=\frac{R}{S}\sin^{-1}(1)$ . In this case, that would be equivalent to $\frac{2500\pi}{2\times75}$ or approximately $\boxed{52.359877}$ .