Secondary Curriculum Linked
High Jumping
How high can a high jumper jump? How can a high jumper jump higher without jumping higher? Read on...
This is our collection of favourite mathematics and sport materials.
Speed-time Problems at the Olympics
Have you ever wondered what it would be like to race against Usain Bolt?
Track Design
Well done to Daniel from Kings School, New Zealand, Josh from Chatham Grammar School for Boys, and Anna from the Sandon School, who all correctly worked out the dimensions of the running track. Here is Josh's solution:
Radius of the curved sections:
As the straight sections of the track are both 85m long, 170m of the track is straight. That means that the remaining 230m of the track is comprised of the two semicircular sections. As there are two semicircular sections, each of equal radii, the sum of their inside edges (of the inside lanes) is equal to the circumference of a circle that would fit neatly into the inside curves of the track.
The circumference of a circle, $c$, is given by the equation: $c = 2 \times \pi \times r$, where $r$ is the radius.
Therefore, as we have deduced that the circumference of this imaginary circle is equal to the sum of the inside curves of the track, we can write: $230 = 2 \times \pi \times r$, so $r = \frac{230}{2\pi} = 36.606$ metres (to 3 d.p.)
This measurement is only the measurement of the radii of the semicircles which comprise the inside edge of the inside lane on the curved section. We want to find a measurement for the radii of the semicircles which comprise the whole of both curved sections; there are 8 lanes that we must consider in order to do this. As each lane is a constant 1.25 metres, the sum of the thickness of the lanes is equal to 8 x 1.25 = 10 metres. Therefore, the radius of the curved sections, in order to produce a scale drawing, is equal to: 36.606 + 10 = 46.606 metres (to 3 d.p.)
The 200m Staggered Start:
As it is mentioned in the problem for the 400m start that the measurement for the length of a given lane is that of its inside edge, I will assume that it is the inside edges of each lane that must be made constant in the 200m race in order to make the race fair. The runner in lane 1 starts at the curved section on the bottom right of the track, so I will use that lane for comparison with the others. The straights do not need to be compensated for in the staggered start, as the runners would all run the same distance here. Therefore, we only need to consider the 115m of track that the runners will run on before they hit the straight; in this time, all runners will be running on the curved section of track.
The runner in lane 2 is running on a track with an inside edge that is 1.25m further outwards than that of lane 1. Therefore, the radius of runner 2's track is equal to $36.606 + 1.25 = 37.856$ metres. Thus, the circumference of the semicircle that makes up the curved section is equal to: $\frac{2 \pi r}{2} = \pi r = 118.927$ metres This is 3.927 metres longer than the inside edge of lane 1, so the rune in lane 2 will start 3.927 metres in front of the runner in lane 1.
We can also generate an nth term sequence for the lanes. The nth term for this sequence is: $(36.606 + 1.25(n-1)) \times \pi$ I arrived at this conclusion because the formula for the arc of a semicircle is pi x r, hence the reason for multiplying the sequence by pi, and the section in brackets is the way to determine the radius of the inside edge of any given track. Each runner, as we move from lane 1 to lane 8, will start 3.927 metres (3 d.p.) in front of the previous runner.
The obvious conclusion for this problem would be to say that each runner starts twice as far behind the runner in the next inside lane as they did in the 200m race, but I will investigate this mathematically.
Each runner is running 400 metres, but 170 metres of the track they will run (85 x 2) is made up of straight sections, where the runners will run the same distance regardless of the lane they are in, so this does not need to be compensated for. Thus, 230m of the track they will run on is comprised of a curved section, which must be compensated for. When the runner in lane 2 reaches a curved section, the radius of the inner edge of his/her lane is equal to $36.606 + 1.25 = 37.856$ metres. Thus, the arc lengths of both semicircles together is equal to: $2 \times \pi \times 37.856 = 237.854$ metres (3 d.p.) This is 7.854 metres longer than the length of curved track in lane 1, so the runner in lane 2 will start 7.854 metres in front of the runner in lane 2.
We can also generate an nth term sequence for the lanes. The nth term for this sequence is: $2 \times ((36.606 + 1.25(n-1)) \times pi)$. I arrived at this conclusion using the nth term formula I generated before to calculate the arc length of one of the semicircular pieces of track, then multiplied it by 2 (for the two semicircles involved in the 400m. As we move from lane 1 to lane 8, each runner starts 7.854 metres in front of the runner on the next inside lane to them.
Rajeev, from Haberdashers' Aske's Boys' School, sent us his calculations for the staggers presented clearly in a table. You can see it here.