Chapter 7 · Circular-track travel

Lesson 7.3 — Reconstruct a Track from Meeting Points

第7讲 · 环形路上的行程问题

A point on a loop can represent several traveled distances. Learn to connect first and second encounters, unwrap directed arcs, test lap cases, and rebuild the track length from where the travelers meet.

Name the direction. Count the laps. Turn each point into a traveled distance.

Grade 5 enrichment40–55 minutesDirected arcsCase analysis

Mission 1

One point on a loop can have many traveled distances

Throughout this lesson, travelers start simultaneously, maintain constant speeds, and do not stop or reverse. Encounters are counted after time 0. “Before” and “after” a point mean along the named traveler’s direction around the track, not along a straight chord. Give distances in metres and times in minutes unless stated otherwise. On a straight road, position and distance traveled often match. On a loop, the same point is reached again after every full lap.

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Choose a side of the starting point

Extra full laps1 lap

Target point is 50 m…

400circumference, m

50first-lap distance, m

450total distance, m

50 afterphysical location

1 × 400 + 50 = 450

Position is a remainder. The number of full laps disappears when we wrap the route back onto the circle.

Closed track and unwrapped routeConcept model

Distances 50, 450, 850, … all reach the same point 50 m after the start on a 400 m track.

Check the idea

Use the 400 m track and the fixed distances in these questions, regardless of the model controls.

Two full laps plus 50 m equals:

A point 50 m before the start is how far into the first lap?

One extra lap, then the point 50 m before the start, gives total distance:

The same physical point may represent:

Mission 2

Why the second meeting happens at three times the first-meeting time

Two travelers begin at opposite ends of a diameter and move in opposite directions. Their first combined gap is half a lap.

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Combined-distance ladder

First meeting: together they close ½C.
From the first meeting to the second, they add one full relative lap, C.
From the original start to the second meeting, they close ½C + C = 1½C.

1½C ÷ ½C = 3

Since speeds stay constant, three times the combined distance takes three times the time. Each traveler also travels three times as far.

First, second, and third meetingsoriginal structure

From opposite ends: the combined distances are ½C, 1½C, 2½C, …

Build the scale factor

First-meeting combined distance, as a fraction of one lap:

Second-meeting combined distance from the original start:

Second-meeting time ÷ first-meeting time:

If one runner traveled 50 m by the first meeting, by the second meeting that runner has traveled:

Mission 3

Reconstruct the track from 50 m and 30 m

Two travelers start at opposite ends A and B of a diameter and move in opposite directions. The runner from A travels 50 m to their first meeting at C, before reaching B. By their second meeting at D, that runner has reached B for the first time and traveled another 30 m in the same direction. These are distances along the track.

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Meeting-point mapWorked original example

The same runner travels three times the first-meeting distance by the second meeting.

Reconstruction chain

First-to-second time scale: 3.
Runner from A has traveled 50 × 3 = 150 m by the second meeting.
That 150 m consists of a half-circumference plus the 30 m arc from B to D.
Half-circumference: 150 − 30 = 120 m.
Full circumference: 120 × 2 = 240 m.

C = 2(3 × 50 − 30) = 240 m

Complete the reconstruction

Distance traveled by the A-runner by the second meeting:

Half-circumference:

Full circumference:

Why subtract 30?

Mission 4

Try the companion problem: 100 m and 80 m

Use Mission 3’s opposite-diameter starts and opposite directions. The runner from A travels 100 m before the first meeting; by the second meeting that runner has reached B for the first time and traveled another 80 m along the track.

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Fill the four boxes

3 × 100 = ? ? − 80 = half a lap 2 × half a lap = circumference

Do not add 100 and 80 directly. The 100 m belongs to the first-meeting distance. The 80 m is removed from the three-times distance at the second meeting.

Companion meeting mapGuided original practice

The diagram is schematic. Use the stated arc lengths, not ruler measurements.

Calculate without copying the previous numbers

Three times the first-meeting distance:

Half-circumference:

Full circumference:

The correct structure is:

Mission 5

Use a track-reconstruction machine

For Mission 3’s directed-arc pattern, let x be the A-to-first-meeting arc and y be the B-to-second-meeting arc shown after B in the A-runner's direction.

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Change the two arc clues

First arc, x50 m

Second arc, y30 m

1503x, m

120half-lap, m

240circumference, m

3time scale

C = 2(3 × 50 − 30) = 240 m

The formula is not a magic rule for every circular problem. It belongs to this exact starting arrangement and directed diagram.

Live directed-arc modelInteractive generalization

The arc positions move with the values. Labels show directed route distances rather than straight-line chords.

Test the machine at x=70, y=40

3x:

Half-circumference:

Full circumference:

The formula may be used when:

Mission 6

First and second catches: the slower runner's distance doubles

Two runners start together in the same direction at constant speeds. The faster runner catches the slower at equal time intervals.

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The time scale controls every traveled distance

second catch time = 2 × first catch time slower distance at second catch = 2 × slower distance at first catch

If the slower runner has traveled 250 m at the first catch, then by the second catch:

250 × 2 = 500 m

Important: 500 m is a traveled distance. It is not automatically the track's circumference.

Unwrapped catch timelineWorked original structure

The circle must be reconstructed by asking where 500 m lands relative to the starting point.

Check the doubling step

Slower distance at the second catch:

Why does the distance double?

Is 500 m automatically one lap?

To reconstruct the track, we must also know:

Mission 7

Test the possible lap cases

For this investigation, the slower runner has traveled 250 m at the first catch. At the second catch, the runner is 50 m before or 50 m after the first return to the start. This lap restriction is part of the problem: later returns are excluded.

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50 m before the first return

500 = C − 50C = 550 m

The runner is 50 m short of completing the first lap.

50 m after one return

500 = C + 50C = 450 m

The runner has completed one lap and moved another 50 m.

After two complete laps

500 = 2C + 50C = 225 m

This is excluded because it uses the second return, while our problem specifies the first return. A traveled distance may exceed a circumference; that fact alone would not reject a candidate. The stated first-return condition is what matters here.

Case-analysis habit: Write one distance equation for each plausible lap count and side of the start. Then test every candidate against all earlier clues.

Classify the cases under the first-return restriction

50 m before completing the first lap gives circumference:

50 m after completing one lap gives circumference:

The two answers under this first-return restriction are:

A candidate circumference must be checked against:

Mission 8

Reconstruct a lap time from three event times

Two travelers start at distinct points A and B and move toward each other along the approaching arc, in opposite directions. They first meet after 8 minutes. Six minutes later, traveler A reaches B’s starting point for the first time. Ten minutes after that, they meet again.

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Turn the timeline into speed units

0start

81st meet

14A reaches B

242nd meet

The initial approaching A-to-B arc equals the A-runner’s 14-minute distance. In the equations below, A and B stand for the runners’ speeds, not the points. At the first meeting:

8(A + B) = 14A 8B = 6A ⇒ B : A = 3 : 4

AAAABBB

Use the interval between meetings

From 8 minutes to 24 minutes:

24 − 8 = 16 minutes

In 16 minutes, the combined relative motion completes one full circumference. Using A = 4 units/min and B = 3 units/min:

C = 16 × (4 + 3) = 112 distance units

A's lap time is:

112 ÷ 4 = 28 minutes

Rebuild the timeline result

Time between the first and second meetings:

Speed ratio B:A:

Combined speed in ratio units:

A's lap time:

Mission 9

Track-reconstruction workshop

Correct at least 6 of 8. Questions 1–4 and 8 use Mission 3’s directed-arc pattern: opposite diameter starts, opposite directions, and the A-runner’s second distance equals half a lap plus the given arc beyond B. Question 7 uses Mission 8’s approaching-arc setup, and A reaches B for the first time at the stated time. Give circumferences in metres and the lap time in minutes.

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1Diameter-start pattern

First meeting is 60 m from A; second is 20 m from B. Find the circumference.

2Another pair

First meeting is 80 m from A; second is 50 m from B. Find the circumference.

3Use the machine

In the same pattern, x=70 m and y=40 m. Find C.

4original companion

First arc 100 m, second arc 80 m. Find the circumference.

540 m after the first return

The slower runner travels 180 m by the first catch and 360 m by the second. The second catch is 40 m after one completed lap. Find C.

640 m before the start

The slower runner travels 360 m by the second catch and is 40 m before completing the first lap. Find C.

7New event timeline

First meeting at 6 min; A reaches B at 10 min; second meeting at 18 min. Find A's lap time.

8Shorter meeting arcs

First meeting is 40 m from A; second is 20 m from B. Find the circumference.

Mission 10

Exit ticket

Use the named patterns from this lesson: opposite-direction diameter starts for Questions 1–3, same-start same-direction catches for Question 4, and Mission 8’s first-arrival timeline for Question 5. Try the questions on your own first. You can check an attempt before opening worked review. The reflection is optional and is not automatically graded.

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1. From opposite diameter endpoints, second-meeting time ÷ first-meeting time:

2. First arc 50 m, second arc 30 m: circumference:

3. First arc 100 m, second arc 80 m: circumference:

4. Slower runner travels 250 m at first catch. Distance at second catch:

5. First meet at 8 min, A reaches B at 14 min, second meet at 24 min. A's lap time:

Optional reflection — not automatically graded

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Track Reconstruction Navigator

This records completed checkpoints on directed arcs, meeting-time scaling, lap cases and event timelines.

Lesson 7.3 completed

Teaching notes

Interpretation note for Example 5: This lesson explicitly restricts the second catch to 50 m before or after the first return. Without that restriction, 500=2C−50 also gives 275 m; the original positional wording alone is insufficient. The worked example gives 450 m and 550 m as the intended answers and rejects a 225 m two-lap case. The phrase “50 m from the starting point” does not specify a directed side, so a fully modular interpretation can be ambiguous unless the before/after position and allowed lap count are stated. The student lesson follows the worked example's intended case structure while making “50 m before” and “50 m after” explicit.