I was modeling a situation I've come across before, and derived an equation for the solution. WolframAlpha calculated the solution, but gave no explanation as to how it did it. I'll give the equation below, explain the issue, and then explain how it was derived.
$t=\frac{\sqrt{(r*\cos(θ_0 +ωt)-x_0)^2+(r*\sin(θ_0 +ωt)-y_0)^2}}{v}$.
I was thinking to myself, "If I was running around a track, and decided to meet up with someone else running around that track, to where would I have to run to meet up with them?"
I reasoned that, at the time at which we make contact, they would have traversed $ω*t$ radians (with $ω$ being the angular velocity and $t$ being the time elapsed until the collision occurs). If the radius of the circular track is $r$, then the person running on the track would be at the position $(r*\cos(θ_0 +ωt),r*\sin(θ_0 +ωt))$ at time $t$. Since I am running in a straight line towards where we'll meet, and $v=\frac{d}{t}$, $t=\frac{d}{v}$. Since this line is at an angle, it can be expressed as the hypotenuse of the triangle formed by the distance between where I begin and where we meet—that is, $\sqrt{(r*\cos(θ_0 +ωt)-x_0)^2+(r*\sin(θ_0 +ωt)-y_0)^2}$. Combining this with the previous equation for time, I obtain $t=\frac{\sqrt{(r*\cos(θ_0 +ωt)-x_0)^2+(r*\sin(θ_0 +ωt)-y_0)^2}}{v}$.
Now, WolframAlpha gave me a value for $t$: 1.51244. However, it gave no explanation as to how it was derived. How would I even begin to find what $t$ is?
Edit: I forgot to mention that $t=1.51244$ only in the particular situation I modeled, where: $r=5m$
$θ_0=0$
$ω=\frac{π}{4}$radians/second.
$x_0=0$
$y_0=-1$
$v=\frac{5π}{4}$
P.S.—I'm asking this on the Mathematics StackExchange as opposed to the Physics StackExchange because my concern is solving for $t$; I've already done the heavy lifting for mathematically modeling the physical situation, which would have been the purview of the Physics StackExchange.