Problem of the Week
Problem E and Solution
There and Back

Problem

Silvan and Nash are training for a race. They run $600$ m along a straight path from their school to a bridge. Once they reach the bridge they turn around and run along the same path back to their school. Silvan and Nash run at different, but constant speeds for each part of the run. For each of them, their constant speed from the bridge back to the school was twice their constant speed on the way to the bridge.

Silvan reached the bridge first, turned around, and immediately started running back to school. Silvan then passed Nash running in the opposite direction when they were $50$ m away from the bridge.

When Silvan reaches the school, how far behind him will Nash be?

Solution

Let $s$ represent Silvan’s speed, in metres per second, when running to the bridge, and $2s$ represent his speed when running from the bridge back to school. Let $n$ represent Nash’s speed, in metres per second, when running to the bridge, and $2n$ represent his speed when running from the bridge back to school.

When Silvan and Nash passed each other $50$ m away from the bridge, they had been running for the same amount of time. In this time, Silvan had run $600$ m to the bridge at a speed of $s$ m/s and $50$ m back to school at a speed of $2s$ m/s, and Nash had run $550$ m to the bridge at a speed of $n$ m/s. Thus, $$\begin{array}{rl}\frac{600}{s}+\frac{50}{2s}& =\frac{550}{n}\\ \frac{600}{s}+\frac{25}{s}& =\frac{550}{n}\\ \frac{625}{s}& =\frac{550}{n}\\ \frac{n}{s}& =\frac{550}{625}=\frac{22}{25}\end{array}$$ Let $x$ represent the distance, in metres, that Nash is behind Silvan when Silvan reaches the school. At this time, Silvan will have run $600$ m to the bridge at a speed of $s$ m/s and $600$ m back to school at a speed of $2s$ m/s, and Nash will have run $600$ m to the bridge at a speed of $n$ m/s and $(600-x)$ m back to school at a speed of $2n$ m/s. Thus, $$\begin{array}{rl}\frac{600}{s}+\frac{600}{2s}& =\frac{600}{n}+\frac{600-x}{2n}\\ \frac{600}{s}+\frac{300}{s}& =\frac{1200}{2n}+\frac{600-x}{2n}\\ \frac{900}{s}& =\frac{1800-x}{2n}\\ \frac{2n}{s}& =\frac{1800-x}{900}\\ \frac{n}{s}& =\frac{1800-x}{1800}\end{array}$$ From earlier, we know that ${\displaystyle \frac{n}{s}}={\displaystyle \frac{22}{25}}$. Thus, $$\begin{array}{rl}\frac{22}{25}& =\frac{1800-x}{1800}\\ 22(1800)& =25(1800-x)\\ 39600& =45000-25x\\ 25x& =5400\\ x& =216\end{array}$$ Therefore, when Silvan reaches the school, Nash will be $216$ m behind him.