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# Running around a track I

## Task

An Olympic $400$ meter track is made up of two straight sides, each measuring $84.39$ meters in length, and two semi-circular curves with a radius of $36.5$ meters as pictured below:

The picture is drawn to scale with one centimeter in the picture representing $20$ meters on an actual olympic track. The one of the eight lanes which is closest to the center of the track is called the first lane.

- What is the perimeter of the track, measured on the innermost part of the first lane?
- Each lane on the track is $1.22$ meters wide. What is the perimeter of the track measured on the outermost part of the first lane?
- In order to run the intended $400$ meters in a lap, how far away from the inside of the first lane would a runner need to be?

Below is an enlarged picture of one of the straight sections of the track with the blue line representing the line around the track with perimeter exactly $400$ meters:

## IM Commentary

In this problem geometry is applied to a $400$ meter track. The specifications for building a track for the Olympics are very precise and are laid out on pages $14$ and $15$ of

This task uses geometry to find the perimeter of the track. Students may be surprised when their calculation does not give $400$ meters but rather a smaller number. The teacher may wish to stop students at this point for a discussion of how this can be. Hopefully students will have the idea that if a runner's body were centered on the border of the track, the runner would only be half on the track and hence considered out of bounds. The next two parts of the problem give an idea where the runner needs to be in order to be running $400$ meters in one lap. If the students have a track at school they may want to measure out $30$ centimeters and see if it is realistic to run around the track with their body centered on this line.

This task is further developed in ''Running around a track II'' where the staggered starts for different lanes are considered. This task is ideal for instruction but could be use for assessment as well. If used for instruction purposes, the teacher may wish to ask students if the lane lines of the track are similar geometric shapes. This is a subtle question as the curved sections and the straight sections are similar but the entire shapes are not because there is a scale factor in the curves (with the outer lanes being radii of larger circles) while the straight sections are the same length for all lanes.

A simplified version of this task, without part (c) or with a different solution method, could be used in Grade 7 for 7.G.4, "Know the formulas for the area and circumference of a circle and use them to solve problems ...." And a more challenging modeling task for high school could remove some of the scaffolding and expect students to research the dimensions themselves.

## Solution

- The two straightaway sections of the track are each $84.39$ meters. The two semi-circular sections can be joined to form a circle whose radius is $36.5$ meters and so the diameter of this circle is $2 \times 36.5 = 73$ meters. The circumference of this circle will be $\pi \times 73$ meters and so the total perimeter of the track is $$ 2 \times 84.39 + \pi \times 73 \approx 398.12. $$ So the perimeter of the track is less than $400$ meters.
- For the first lane on the track, the straightaway sections are each $84.39$ meters long. However, the curved sections form a circle whose radius is now $36.5 + 1.22 = 37.72$ meters. The diameter of the circle will be $2 \times 37.72 = 75.44$ meters. So the perimeter of lane 1 is $$ 2 \times 84.39 + \pi \times 75.44 \approx 405.78. $$ So the perimeter of lane 1 on the track is more than $400$ meters and is almost $8$ meters more than the perimeter of the inside of the track.
- Suppose we let $x$ denote the distance from the inside of lane 1 which gives a perimeter of $400$ meters. This perimeter will be consist of the two straight sections which contribute $2 \times 84.39$ meters to the perimeter. In addition, there will be two semi-circular sections of radius $36.5 + x$ meters. Combining these gives a circle whose diameter is $ 2 \times (36.5 + x)$ meters. So we want $$ 2 \times 84.39 + \pi \times 2 \times (36.5 + x) = 400. $$ Rewriting this we find $$ 2 \pi \times x = 400 - 2 \times 84.39 - 2\pi \times 36.5 $$ Solving for $x$ we find $$ x \approx 0.30. $$ Note that this value for $x$ is not exact but approximate. It is accurate to within about two ten thousandths of a meter or a fraction of a millimeter. So approximately $30$ centimeters from the inside of lane 1 the perimeter of the track is $400$ meters.

## Running around a track I

An Olympic $400$ meter track is made up of two straight sides, each measuring $84.39$ meters in length, and two semi-circular curves with a radius of $36.5$ meters as pictured below:

The picture is drawn to scale with one centimeter in the picture representing $20$ meters on an actual olympic track. The one of the eight lanes which is closest to the center of the track is called the first lane.

- What is the perimeter of the track, measured on the innermost part of the first lane?
- Each lane on the track is $1.22$ meters wide. What is the perimeter of the track measured on the outermost part of the first lane?
- In order to run the intended $400$ meters in a lap, how far away from the inside of the first lane would a runner need to be?

Below is an enlarged picture of one of the straight sections of the track with the blue line representing the line around the track with perimeter exactly $400$ meters:

## Comments

Log in to comment## Jennifer says:

over 3 yearsI blogged about using this task with my students: http://easingthehurrysyndrome.wordpress.com/2014/03/09/running-around-a-track-i/

## wij says:

over 4 yearsI don't see why this is provided as a representative example for high school.

Certainly this does "model" a situation using "geometric shapes, their measures, and their properties" but I would expect that at high school students would be using shapes and properties they've been emphasizing in high school. Knowing circumference of circles is a Grade 7 topic.

I think middle school standard 7.G.B.4 would be a better place to house this activity: Know the formula[s] for the ... circumference of a circle and use [it] to solve problems ....

I suppose we can presume that students are responsible for having mastered middle school content and that any of the topics from middle school could be "fair game" for a high school assessment item. But I wonder if this is the best exemplar to give for G.MG.A.1.

I do acknowledge that the reasoning in part c. involves some equation solving work that is possibly out of bounds for Grade 7 the way it's presented here, which would likely be something one could expect of students who have finished Algebra 1. So in that respect, the item (as a whole) fits better in high school than in middle school.

I just wonder if by providing this exemplar you are conveying (intentionally) that content from prior years (in this case Grade 7) is likely to be present in high school assessment tasks.

## Bill says:

about 4 yearsThanks for these comments. I agree that with a different solution to part (c) (maybe trial and error), and possibly some other simplifications, this could be a Grade 7 task.

On the question of whether high school tasks should include content from prior years, it depends on the complexity of the modeling aspect. It is often the case that the difficulty of a modeling task lies not only in the mathematical ingredients but also in the complexity of the modeling demands: how complicated is the situation, what choices and decisions must the student make, what are the issues involved in interpreting the result, and so on. A true modeling task should depend on securely held content, often from a previous grade level.

Now, this particular task is fairly scaffolded, so your concerns are reasonable. Perhaps one could consider both a simplified version of it for Grade 7 and a less scaffolded version as a true example of modeling in high school (for example, by requiring students to do the research on Olympic track dimensions themselves).

That said, this task also sits somewhere in the continuum between those extremes and, as you point out, the solution to (c) puts it in high school. I'll make some changes in the commentary to reflect this discussion.

## carydc says:

over 4 yearsCheck out your solution b. Maybe being picky, or is it attending to precision, but the diameter is 2 x 37.72 = 75.44m. Therefore, the perimeter of lane 1 is 2 × 84.39 + π x 75.44 = 405.8m.

## Michael Nakamaye says:

over 4 yearsthank you!!! there is a typo in the line before finding the diameter of the circle which created the problem. You are to be applauded for attending to precision! thanks again, I will change it in a moment...