Around and Back
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns around and heads back to the starting point where he meets the runner who is just finishing his first circuit. Find the ratio of their speeds.
Age
14 to 16
Challenge level
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns around and heads back to the starting point where he meets the runner who is just finishing his first circuit. Find the ratio of their speeds.
This is an unusual question in that you are given no numerical information but have to find a numerical answer.
This solution was sent in by Harry from Riccarton High School, Christchurch, New Zealand.
A cyclist and a runner are pracising on a race track going round at constant speeds, $V_c$ for the cyclist and $V_r$ for the runner. Let the fraction of the circuit covered by the runner when they first meet be $x$ and then the cyclist will have covered $1 + x$ circuits. Equating the time taken gives the first equation:
$${x \over V_r} = {1 + x \over V_c}.$$
Similarly the time taken between the cyclist first passing the runner and the finish gives the second equation: \begin{equation}{1 - x\over V_r} = {x\over V_c}. \end{equation} The ratio of $V_c$ to $V_r$ from the two equations gives: \begin{equation*}{V_c\over V_r} = {1 + x\over x} = {x \over 1 - x} \end{equation*} Hence \begin{equation*}x^2 = 1 - x^2. \end{equation*} From this we get $x = \sqrt{1\over 2}$ and this gives the ratio of the speeds as \begin{equation*}{V_c\over V_r} = {{1 + 1/\sqrt 2}\over {1/\sqrt 2}} = \sqrt 2 + 1. \end{equation*}