Riddler Classic
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?
Solution
Let the chaser be at \((0,0)\) and the runner be \((x_0,0)\) at the time \(t=0\) respectively, the instant the pursuit begins, with the runner running at constant speed \(V_r\) along the line \(x=x_0\). The chaser runs at a constant speed \(V_c\) along a curved path such that he is always moving directly toward the runner, that is, the velocity vector of the chaser points directly at the runner at every instant of time.
To find the curve of pursuit of the chaser, we assume that the chaser is at the location \((x,y)\) at time \(t \geq 0\). At time \(t\), the runner is at the point \((x_0,V_rt)\) and so, the slope of the tangent line to the pursuit curve (the value of \(dy/dx\) at \((x,y)\)) is given by
\[ \frac{dy}{dx} = \frac{V_rt - y}{x_0 - x} \]
We also know that the chaser would have ran a distance \(V_ct\) along it by the time \(t\). This arc-length is also given by the expression on the right below:
\[ V_ct = \int_0^x \sqrt{1 + \left(\frac{dy}{dz}\right)^2} dz \]
Eliminating \(t\) from the above two equations, we get
\[ \begin{aligned} \frac{1}{V_c} \int_0^x \sqrt{1 + \left(\frac{dy}{dz}\right)^2} dz &= \frac{y}{V_r} - \frac{(x-x_0)}{V_r} \cdot \frac{dy}{dx} \\\\ \implies \frac{1}{V_c} \int_0^x \sqrt{1 + p^2(z)} dz &= \frac{y}{V_r} - \frac{(x-x_0)}{V_r} \cdot p(x) \text{ , where $dy/dx= p(x)$}. \end{aligned} \]
Differentiating under the integral sign with respect to \(x\), we arrive at
\[ \begin{aligned} \frac{1}{V_c} \sqrt{1 + p^2(x)} &= \frac{p(x)}{V_r} - \frac{(x-x_0)}{V_r} \cdot \frac{dp(x)}{dx} - \frac{1}{V_r}p(x) \\\\ \implies (x-x_0)\frac{dp}{dx} &= -\frac{V_r}{V_c} \sqrt{1 + p^2(x)} \\\\ \implies \frac{dp}{\sqrt{1 + p^2(x)}} &= \frac{n dx}{(x_0-x)} \text{, where $n=V_r/V_c$} \end{aligned} \]
Integrating the above equation, we get
\[ \ln(p + \sqrt{1+p^2}) + c = -n\ln(x_0-x). \]
We see at \(t=0\), that \(p=dy/dx=0\) when \(x=0\) because at that instant, the runner as well as the chaser are on the \(x-axis\). It follows that \(c = -n\ln(x_0)\) and so
\[ \begin{aligned} \ln\left[ \left(p + \sqrt{1+p^2}\right)\left(1 - \frac{x}{x_0}\right)^n\right] &= 0 \\\\ \implies \left(p + \sqrt{1+p^2}\right)\left(1 - \frac{x}{x_0}\right)^n &= 1. \end{aligned} \]
From the above, we get
\[ p(x) = \frac{dy}{dx} = \frac{1}{2}\left[\left(1-\frac{x}{x_0}\right)^{-n} - \left(1 - \frac{x}{x_0} \right)^n \right] \]
Integrating the above equation, we get
\[ y(x) + c = \frac{1}{2}(x_0-x)\left[\frac{(1-x/x_0)^n}{1+n} - \frac{(1-x/x_0)^{-n}}{1-n}\right]. \]
Since \(y= 0\) when \(x=0\), pursuit curve equation is given by
\[ y(x) = \frac{n}{1-n^2}x_0 + \frac{1}{2}(x_0-x)\left[\frac{(1-x/x_0)^n}{1+n} - \frac{(1-x/x_0)^{-n}}{1-n}\right] \]
In the given problem, we have \(V_r=15\) miles/hr, \(x_0=50\) and \(y = 100\) when \(x=50\). Substituting these values in the above equation, we get
\[ \begin{aligned} 100 &= \frac{n}{1-n^2}\cdot50 \\\\ \implies 2n^2+n-2 &= 0 \end{aligned} \]
Solving the above quadratic and taking the positive value for \(n\), we see that the speed of the chaser \(V_c\) should be
\[ \frac{\sqrt{17}+1}{4}\cdot15 \text{ miles/hr} \]