# 2440.63 – Rashon and Santiago

Rashon and Santiago walk at constant speeds along Main and Elm streets respectively. When Rashon is at the intersection---they are perpendicular streets---Santiago is still 500 yards away. Two minutes later, they are the same distance from the intersection. And in another eight minutes, they are again equidistant from the intersection. Got that?

OK! Suppose Rashon walks at $r$ yards/minute and Santiago at $s$ yards/minute. What is the ratio $r/s$?

Solution

Let $R_t$ and $S_t$ be the positions of Rashon and Santiago along their respective streets at time $t$, so that, for example, $R_2$ represents the position of Rashon along Main Street after two minutes. Note that we take Main Street to be vertical while Elm Street is horizontal. The figure below sketches the action. $R$ gives the vertical or $y$ coordinate; $S$ gives the horizontal or $x$ coordinate. In particular, note that $R_2$ and $R_10$ are negative.

In terms of $r$ and $s$, the positions of $R$ and $S$ are given in this table:

\begin{aligned} t &|& R &|& S \\ 0 &|& 0 &|& -500 \\ 2 &|& -2r &|& 2s - 500 \\ 10 &|& -10r &|& 10s - 500 \end{aligned}

The relative locations of Rashon and Santiago after 2 and 10 minutes, then boil down to the two equations:

\begin{aligned} S_2 &=& R_2 \\ S_{10} &=& -R_{10}. \end{aligned}

Using the table, this becomes the set of equations

\begin{aligned} 2s - 500 &=& -2r \\ 10s - 500 &=& 10r. \end{aligned}

or

\begin{aligned} r + s &=& 250 \\ r - s &=& 50. \end{aligned}

It turns out that $r = 150$ and $s = 100$. Thus $r/s = 3/2$.