Circuit training
Mike and Monisha meet at the race track, which is 400m round. Just to make a point, Mike runs anticlockwise whilst Monisha runs clockwise. Where will they meet on their way around and will they ever meet at the start again? If so, after how many circuits?
Problem
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Why do runners always run around the track in an anticlockwise direction?
Do they develop a clinical lean?
Mike and Monisha meet at the race track, which is $400 \; \text{m}$ round. Just to make a point, Mike runs anticlockwise whilst Monisha runs clockwise, normally Mike can complete a circuit of the track in two thirds of the time Monisha takes.
They agree to stop when they both meet at the start again. Where will they meet on their way around and will they ever meet at the start again? If so, after how many circuits?
Getting Started
Try some different times and see what distances each runner will have covered.
How will you know when they meet on the track?
How will you know when they meet on the track?
Student Solutions
A neat solution by Nazia, and similar ones from a number of other students from Nottingham Girls' High School, showed that an organised approach could be fruitful; but where else did they meet?
When Mike has finished the first circuit, Monisha will be $\frac{2}{3}$ of the way round.
When Mike has finished the second circuit, Monisha will be $1\frac{1}{3}$ of the way round.
When Mike has finished the third cicuit, Monisha will have just finished her second circuit.
This is where they will cross for the first time, and when they both will meet at the start again.
The answer is that they will meet at the start again after Mike has run $3$ circuits and Monisha has run $2$ circuits.
Fred and Matt of Albion Heights took a slightly different approach and were able to identify an intermediate passing place:
The first question asks if they will ever meet at the start if Mike runs a lap in $2/3$ the time it takes Monisha. Since speed is directly related to time, Mike's speed is $3/2$ greater than Monisha's speed.
They will meet on Mike's third lap and Monisha's second lap.
The second question asks where will they meet on the first lap.
Well, if Mike can travel $3$ units in one minute and and Monisha can travel $2$ units in one minute, then the total distance travelled will be $5$ units.
Therefore, if they were on opposite ends of a $400$ m track, Mike will be $\frac{3}{5}$ of the way round the track and Monisha $\frac{2}{5}$ of the way around when they meet. If you make the denominator of $\frac{3}{5}$ $400$, you get $\frac{240}{400}$.
Therefore, they will meet $240 \; \text{m}$ down the track starting from Mikes end of the track.
But will they meet again part of the way
around this circular track?
Last but not least Andrei of the Tudor Vianu
National College suggests the following:, which starts from exactly
the same argument as the one used by Fred and Matt:
Let Mike be runner No. $1$, and Monisha runner number $2$. All physical quantities (period: time for each to run around the circuit, velocity, distance travelled) for Mike will be indexed with $1$, while for Monisha with $2$.
From the data of the problem, I see that the relation between their periods is: $$ T_1 = \frac{2}{3}T_2$$ This implies that their velocities are in an inverse relation: $$v_1 = \frac{3}{2}v_2$$ These relations could be read as: in the same time, runner $1$ completes $3$ units, runner $2$ only completes $2$ circuits.
So, if the circuit is divided into $5$ parts, runner $1$ completes $3$ and runner $2$ only $2$ parts. They meet first at $\frac{3}{5}$ of the circuit in the direction Mike runs, and at $\frac{2}{5}$ of circuit in the direction Monisha runs.
For the first meeting they must complete a circuit ($400 \; \text{m}$).
The second meeting corresponds to completing $2$ circuits, with $\frac{6}{5}\times 400 = 480$ m travelled by Mike, and $\frac{4}{5} \times 400 = 320 \; \text{m}$ by Monisha. This point is at $80 \; \text{m}$ from the start in the direction Mike runs.
The third meeting is at a multiple of three of the first (as distances travelled) so at $320 \; \text{m}$ from the start in the direction Mike runs.
The fourth is at $160 \; \text{m}$ from the start, in the same direction.
The fifth is at the start, because the circuit was divided into $5$ parts, as dictated by the relation between the periods.
More generally, if the relation between the periods would be $$\frac{T_1}{T_2} = \frac{p}{q}$$ with $p$ and $q$ coprime, then the runners meet again at the start after completing in total $(p+q)$ circuits, this means they meet the $(p+q)$th time.
Teachers' Resources
It is worth spending some time discussing the strategies pupils have adopted.
Another question might be:
What is the ratio of the speeds of the two runners if they are to meet at the start after $1$ lap, $2$ laps, $3$ laps... $n$ laps?.
What is the relationship betweent he ratio of the speeds of the runners and where they meet?