Running track and stagger · 3 marks

What is the difference in radius between the first and second lanes? Use the figure to find the stagger needed by the runner in the second lane. Will an equal stagger be needed between the third and second lanes?

Answer: The radius goes up by one lane width, 1.22 m. Over one lap (two semicircles = one full circle) lane 2 is longer by 2π × 1.22 ≈ 7.67 m, so that is the stagger. Yes, lane 3 needs the same 7.67 m stagger ahead of lane 2, because its radius is again 1.22 m more.

Step-by-step solution

Given: Lane width = 1.22 m; Each runner keeps the same distance (0.3 m) from her inner line; One lap has two semicircular bends

Idea: Straights are equal for all lanes, so only the bends differ. Two semicircles make a full circle, and a circle whose radius is bigger by w is longer by 2πw.

Each lane is 1.22 m wide, so the runner in lane 2 runs on a circle whose radius is 1.22 m more than the lane 1 runner’s: 36.8 m and 38.02 m.1 mark
Extra length on the bends = 2π × 38.02 − 2π × 36.8 = 2π × 1.22 = 2 × 3.1416 × 1.22 ≈ 7.67 m. So the lane 2 runner starts 7.67 m ahead.1 mark
Lane 3’s radius is another 1.22 m more than lane 2’s, so it is longer by the same 2π × 1.22 ≈ 7.67 m. Yes, an equal stagger is needed between every pair of neighbouring lanes.1 mark
Note: actual competition rules measure lane 1 at 0.30 m from its inner line but the other lanes at 0.20 m from theirs. Then lane 2’s radius is only 1.12 m more than lane 1’s, and the real stagger is 2π × 1.12 ≈ 7.04 m. From lane 2 outwards the difference is again a full 1.22 m.

The radius increases by 1.22 m from lane 1 to lane 2. For one lap the stagger is 2π × 1.22 ≈ 7.67 m. The same stagger is needed between lanes 2 and 3 (and between any two neighbouring lanes), since the radius increases by 1.22 m each time.

Check: 2π × 36.8 = 231.22 m and 2π × 38.02 = 238.89 m; difference 7.67 m ✓.

Answer to write in the exam

Difference in radius = lane width = 1.22 m

Stagger = 2π(r + 1.22) − 2πr = 2π × 1.22

= 2 × 3.1416 × 1.22 ≈ 7.67 m

Lane 3 − lane 2: radius again differs by 1.22 m ⇒ stagger = 7.67 m

∴ Yes, an equal stagger is needed.

Common mistakes that cost marks

Using π × 1.22 ≈ 3.83 m. That is for one bend only; one lap has two bends.
Thinking outer lanes need bigger and bigger staggers. Each new lane adds the same 1.22 m to the radius, so each stagger is the same.
Taking the radius difference as 2 × 1.22 m.

How this can come in the exam

MCQ (1 mark)

Lanes are 1.25 m wide. The stagger between neighbouring lanes for a race of one full lap (two semicircular bends) is about (use π = 3.14)

3.93 m
7.85 m
1.25 m
15.7 m

Show answer

(B) 7.85 m
2π × 1.25 = 2 × 3.14 × 1.25 = 7.85 m.

Try one yourself

For a race of half a lap (one semicircular bend) on the same track (lanes 1.22 m wide), what stagger is needed between lanes 1 and 2? (Use π = 3.1416.)