This week’s Riddler Classic is about catching

Hames Jarrison has just intercepted a pass at one end zone of a football field, and begins running — at a constant speed of 15 miles per hour — to the other end zone, 100 yards away.

At the moment he catches the ball, you are on the very same goal line, but on the other end of the field, 50 yards away from Jarrison. Caught up in the moment, you decide you will always run directly toward Jarrison’s current position, rather than plan ahead to meet him downfield along a more strategic course.

Assuming you run at a constant speed (i.e., don’t worry about any transient acceleration), how fast must you be in order to catch Jarrison before he scores a touchdown?

Here is the solution.

[Show Solution]

If the football field is $L$ long and $w$ wide, and Jarrison has a speed of $v_0$, then in order to catch him just in time, our speed $v$ should satisfy

\[

\frac{v}{v_0} \gt \frac{1+\sqrt{4\frac{L^2}{w^2}+1}}{2\frac{L}{w}}.

\]For the data given in the problem, we have $L=100$ and $w=50$, so $\frac{v}{v_0} \gt \frac{1+\sqrt{17}}{4}\approx 1.28$. Therefore, we would need to run at a speed of at least $19.21\,\text{mph}$ to catch Jarrison. Here is a plot of $\frac{v}{v_0}$ as a function of $\frac{L}{w}$.

In the limit $\frac{L}{w} \to \infty$, the field becomes infinitely long. Therefore, we spend most of the time running almost parallel to Jarrison. So we only need to be a tiny bit faster in order to eventually catch him, and we have $\frac{v}{v_0}\to 1$.

In the limit $\frac{L}{w} \to 0$, the field becomes infinitely wide. We must cover an increasing amount of lateral ground to catch Jarrison, so $\frac{v}{v_0} \to \infty$.

**Fun fact:** if the field is square ($L=w$), we must be $\varphi$ times faster than Jarrison to catch him, where $\varphi = \frac{1+\sqrt{5}}{2} \approx 1.618$ is the golden ratio!

We can also ask: what is the shape of the path that we will follow? (This is called a pursuit curve.) Using the coordinate system depicted below, our trajectory will satisfy:

\[

\frac{x}{w} = \frac{1}{2}\left[ \frac{\left(1-\frac{y}{w}\right)^{1+\frac{v_0}{v}}-1}{1+\frac{v_0}{v}}-\frac{\left(1-\frac{y}{w}\right)^{1-\frac{v_0}{v}}-1}{1-\frac{v_0}{v}}\right]

\]

For a detailed derivation (warning: calculus!) click below.

[Show Solution]

**Note:** This is the solution I came up with… Admittedly, it’s a bit tedious and not particularly intuitive. If you have a more direct or more elegant solution approach, I would love to hear about it!

Based on the diagram above, Jarrison starts at $(0,w)$ and runs toward $(L,w)$ at a speed $v_0$. Therefore, Jarrison’s position as a function of time is $(v_0t, w)$. We start at $(0,0)$ and we run at a speed $v$. Let’s also suppose our position at time $t$ is $(x(t),y(t))$.

We run in such a way that our velocity always points toward Jarrison. Therefore, we have:

\[

\frac{\dot y}{\dot x} = \frac{w-y}{v_0t-x},

\]where the dots denote time derivatives, i.e. $\dot x = \frac{\mathrm{d}}{\mathrm{d}t}x(t)$. To keep notation simple, we’ll omit the $(t)$ when writing $x$ or $y$. We also know that our speed is constant at $v$, therefore, we have:

\[

\dot x^2 + \dot y^2 = v^2

\]These two coupled differential equations, together with the initial conditions $x(0)=y(0)=0$, completely describe our motion. Our task is to solve these equations, and then find the value of $v$ such that the solution passes through the point $(L,w)$, i.e., we catch Jarrison right as he scores a touchdown.

To solve this problem, we’ll use the change of variables $u = \frac{\dot x}{\dot y}$. Substitute this into both equations. For the first equation, also isolate $t$ and differentiate so that $t$ no longer appears explicitly. Ultimately, we obtain:

\[

\dot u = \frac{v_0}{w-y}

\quad\text{and}\quad

\dot y = \frac{v}{\sqrt{u^2+1}}

\quad\text{with: }\begin{cases}y(0)=0 \\ u(0)=0\end{cases}

\]Combining these equations, we obtain the single ODE (in differential form)

\[

\frac{\mathrm{d}y}{w-y} = \frac{v}{v_0} \cdot \frac{\mathrm{d}u}{\sqrt{u^2+1}}

\]Integrating from $t=0$ ($y=u=0$) to an arbitrary later point, we obtain:

\[

\log\left( \frac{w}{w-y} \right) = \frac{v}{v_0} \log \left| \sqrt{u^2+1}+u \right|

\]Therefore,

\[

\frac{w}{w-y} = \left| \sqrt{u^2+1}+u \right|^{v/v_0}

\]Now note that if $\sqrt{u^2+1}+u = f$, we have $u = \frac{1}{2}\left( f-f^{-1} \right)$. Therefore, we can solve the above equation and obtain:

\[

u = \frac{1}{2}\left[ \left(1-\frac{y}{w}\right)^{-v_0/v}-\left(1-\frac{y}{w}\right)^{v_0/v} \right]

\]Now recall the definition $u = \frac{\dot x}{\dot y} = \frac{\mathrm{d}x}{\mathrm{d}y}$. Therefore,

\[

\mathrm{d}x = \frac{1}{2}\left[ \left(1-\frac{y}{w}\right)^{-v_0/v}-\left(1-\frac{y}{w}\right)^{v_0/v} \right]\mathrm{d}y

\]Integrating from $t=0$ ($y=u=0$) to an arbitrary later point, we obtain:

\[

\frac{x}{w} = \frac{1}{2}\left[ \frac{\left(1-\frac{y}{w}\right)^{1+v_0/v}-1}{1+\frac{v_0}{v}}-\frac{\left(1-\frac{y}{w}\right)^{1-v_0/v}-1}{1-\frac{v_0}{v}}\right]

\]This tells us $x$ as a function of $y$. The only thing we’re missing is $v$. To find $v$, we use the fact that the curve must pass through $(L,w)$, which leads to

\[

\frac{L}{w} = \frac{v_0 v}{v^2-v_0^2}.

\]Solving for $v$ yields:

\[

\frac{v}{v_0} = \frac{1+\sqrt{4\frac{L^2}{w^2}+1}}{2\frac{L}{w}}.

\]Note that the right-hand side is always greater than $1$, regardless of the value of $L$ or $w$. This makes sense; we need to be going faster than Jarrison if we ever hope to catch him.