On a sunny summer day, you are taking the scenic boat ride from Stein am Rhein, Switzerland, to Schaffhausen, down the Rhein River. This nonstop trip takes 40 minutes, but the return trip to Stein, upstream, will take a full hour. Back in Stein, you decide to stay on the boat and continue on to Constance, Germany,
now traveling on the still waters of Lake Constance. How long will this nonstop trip from Stein to Constance take? You may assume that the boat is traveling at a
constant speed relative to the water throughout and that the Rhein River flows at a constant speed between Stein and Schaffhausen. The traveling distance from Stein to
Schaffhausen is the same as from Stein to Constance.
I'm not sure how to set up this problem, any help would be great!
Best Answer
There are three legs of the journey: (1) Stein to Schaffhausen, (2) Schaffhausen to Stein, and (3) Stein to Constance.
Let $$\newcommand{\sft}[1]{\mathsf{\text{#1}}} \begin{align} s&=\sft{speed of boat in still water (in meters/minute)}\\\\\\ r&=\sft{speed of river (in meters/minute)}\\\\\\ d&=\sft{distance from Stein to Schaffhausen (in meters)}\\ &=\sft{distance from Schaffhausen to Stein (in meters)}\\ &=\sft{distance from Stein to Constance (in meters)}\\\\\\ t&=\sft{time it takes to travel from Stein to Constance (in minutes)} \end{align}$$ For each leg of the journey, we have that $$\sft{distance}=\sft{speed}\times\sft{time}$$ When the boat is traveling downstream, the river speeds up the boat, and likewise when the boat is traveling upstrem, the river slows down the boat. In still water the boat just has speed $s$.
For leg (1), $$d=(s+r)\times (40\sft{ min})$$ For leg (2), $$d=(s-r)\times (60\sft{ min})$$ For leg (3), $$d=s\times t.$$ Use the first two equations to solve for $s$ in terms of $d$, then plug into the third equation to find $t$.