Chapter 7 · Circular-track travel

Lesson 7.4 — Stops, Corners, and Piecewise Motion

第7讲 · 环形路上的行程问题

A runner can be moving, waiting at a corner, or changing speed. Learn to separate running time from elapsed time, build event schedules, and verify a catch instead of trusting one relative-speed equation.

Speeds describe movement. Schedules determine when travelers actually meet.

Grade 5 enrichment45–60 minutes10 missionsNo this lesson required

Mission 1

Keep two clocks: running time and elapsed time

In this lesson, stated running speeds apply while moving; speed is zero during a rest. All travelers begin moving at time 0, with no initial rest. Use each constant speed with its own running interval. A stopwatch for the whole race keeps ticking during every stop.

Not completed

The time ledger

elapsed time = running time + stopping timedistance = speed × running time

Never write distance = speed × elapsed time when the traveler has stopped during that interval.

A quick story

A runner moves for 60 seconds, rests for 10 seconds, and then moves for 20 more seconds.

80 srunning time

10 sstopping time

90 selapsed time

At 5 m/s, the distance is 5×80=400 m—not 5×90.

Check the two clocks

A child runs 75 s and rests 15 s. Elapsed time:

Elapsed time is 100 s; stopping time is 20 s. Running time:

Which time belongs in distance = speed × time?

A runner can be caught while:

Mission 2

Build a corner-arrival schedule

The square track ABCD has side length 100 m and perimeter 400 m. Runner A starts at vertex A, follows A→B→C→D→A at 7 m/s, and rests 5 seconds on every corner arrival, including later returns to A.

Not completed

Choose an arrival number

Corner arrival4th

400distance, m

57.14running time, s

72.14arrival time, s

77.14leave time, s

arrival = 400÷7 + 3×5 ≈ 72.14 s

Before the kth arrival, the runner has already completed k−1 stops.

Square track and current arrivalInteractive schedule

Arrival time does not include the new stop that begins at the highlighted corner; leave time does.

Use the schedule

Running time for one 100-m side at 7 m/s, to 2 decimals:

Arrival at A after 800 m, to 2 decimals:

Leave A after that stop, to 2 decimals:

Why are there 7 completed stops before the 800-m arrival?

Mission 3

A relative-motion equation gives a candidate—not always the first catch

For Missions 3–5, runner A starts at vertex A and runner B at opposite vertex C on Mission 2’s square. Both start together in the direction A→B→C→D→A. A runs at 7 m/s and B at 5 m/s; each rests 5 seconds whenever reaching a corner. Thus A initially trails B by 200 m along the route.

Not completed

If nobody stopped

relative speed = 7−5=2 m/scatch time = 200÷2=100 s

This calculation is useful as a first estimate. But the real runners stop at corners, and a catch can happen while one or both are waiting.

Why the estimate can failState comparison

The graph follows the actual stop schedule. Relative gain falls while only the faster runner rests, rises while gaining, and stays level while both rest.

Interpret the estimate

No-stop catch estimate:

Relative speed while both are running:

Can the estimate alone prove the first real catch?

What can happen at a corner?

Mission 4

Account for the extra corner stops

Try a candidate catch while both runners are moving, assuming A has completed two more corner stops than B. Under that assumption A has spent 10 more seconds stopped. This does not yet prove when the first catch occurs.

Not completed

Compare pure running times

Let A's pure running time be y seconds. Because A has spent two extra 5-second stops, B has been running for:

y+10 seconds

A must gain the initial 200 m:

7y−5(y+10)=2002y=250 → y=125

The equation predicts that A would run:

7×125=875 m

Why this is still a checkpoint

The equation predicts a catch while comparing running distances. Before A reaches 875 m, however, both runners may occupy the same corner during overlapping stops.

Next step: inspect every relevant arrival and departure time before accepting 875 m as the first catch.

Read the model correctly

A's pure running time y:

B's pure running time y+10:

Predicted A running distance:

How should 875 m be treated?

Mission 5

Verify the first catch by overlapping stop windows

Both runners reach corner A before the predicted 875-m catch. Their waiting intervals overlap.

Not completed

Runner A's A-corner window

arrival = 800÷7 + 7×5 ≈149.29 sleave ≈149.29+5=154.29 s

Runner B's A-corner window

arrival = 600÷5 + 5×5=145 sleave =145+5=150 s

The overlap proves the catch

A: 149.29–154.29

B: 145–150

A stopB stopoverlap

From about 149.29 s to 150 s, both are at A. The first catch occurs when A arrives, and B has run 600 m.

Audit the stop windows

A arrives at A, to 2 decimals:

B leaves A at:

Where does the first catch occur?

How far has B run?

Mission 6

Guided practice: a 100 m by 60 m rectangle

On rectangle ABCD, AB = CD = 100 m and BC = DA = 60 m. Runner A starts at vertex A at 6.25 m/s; runner B starts at vertex C at 5 m/s. Both begin together, follow A→B→C→D→A counterclockwise, and rest 5 seconds on each later vertex arrival.

Not completed

Rectangle trackGuided practice problem

Perimeter: 320 m. A's first catch occurs at vertex A during an overlapping stop.

Build the two stop windows at A

A: 960÷6.25 + 11×5 = 208.6 sA waits from 208.6 to 213.6 sB: 800÷5 + 9×5 = 205 sB waits from 205 to 210 s

The intervals overlap from 208.6 to 210 seconds.

Result: A has run 960 m, B has run 800 m, and the catch occurs at A at 208.6 s.

Complete the guided practice

Rectangle perimeter:

A's running distance at the catch:

B's running distance at the catch:

Catch time, in seconds:

Mission 7

Piecewise speed: split the trip into time blocks

A child completes one 360-metre lap. During the first half of the total time the speed is 5 m/s; during the second half it is 4 m/s.

Not completed

Let each half last t seconds

Half-time t40 s

200first-half distance

160second-half distance

360total distance

80total time, s

5t+4t=360 → 9t=360 → t=40 s

Two equal time blocksPiecewise-motion model

The time blocks are equal, but the distances are not.

Solve the lap

Use the completed 360 m lap, regardless of the trial slider. Give times in seconds and distance in metres.

Length of each half-time block:

Total elapsed time:

Distance during the faster half:

Why can we add 5t+4t?

Time for the last half of the distance (180 m), in seconds

Mission 8

Periodic rest cycles: build an event table

When each traveler follows a repeating run–rest rule, list the event intervals in time order. Do not replace the pattern with one unverified average speed.

Not completed

Opposite directions around a 24-km lake

They start together from the same point in opposite directions. Runner A repeatedly runs for 50 minutes at 6 km/h, then rests for 10 minutes. Runner B repeatedly runs for 60 minutes at 4 km/h, then rests for 5 minutes.

A runs 50

rests 10

B runs 60

rests 5

Time	A state	B state	Combined distance so far
0–50	run	run	8⅓ km
50–60	rest	run	9 km
60–65	run	rest	9.5 km
65–110	run	run	17 km
110–120	rest	run	17⅔ km
120–125	run	run	18½ km
125–130	run	rest	19 km
130–160	run	run	24 km → meet

First meeting: 160 minutes.

Same direction; rest after every 100 m

On a 400-m track, A and B begin together in the same direction, with B initially 100 m ahead along the route. A runs 5 m/s; B runs 4 m/s. Each repeats 100 m of running followed by 10 seconds of rest, measuring the 100 m from their own last departure.

30 sA's 100-m run–rest cycle

35 sB's 100-m run–rest cycle

500 mA's distance at 140 s

400 mB's distance at 140 s

500−400=100 m

A has gained the full initial gap, so the first catch occurs at 140 seconds.

Check both repeating schedules

Lake problem first meeting, minutes:

A's 100-m run–rest cycle in the track problem:

B's 100-m run–rest cycle:

Track problem first catch, seconds:

Mission 9

Stops and piecewise-motion workshop

Correct at least 6 of 8. Questions 2–4 use the square scenario in Missions 3–5; Question 5 uses Mission 6; Question 6 uses Mission 7; Questions 7–8 use the two Mission 8 schedules. Give distances in metres and times in seconds, except Question 7 asks for minutes.

Not completed

1Two clocks

Run 70 s, rest 20 s. Elapsed time?

2No-stop estimate

Gap 200 m; speeds 7 and 5 m/s. No-stop catch time?

3Mission 4 candidate

From 7y−5(y+10)=200, how far would A run?

4Mission 5 first catch

How far has B run when caught at A?

5Rectangle practice

How far has A run at the first catch?

6Piecewise lap

How long does the last half of the 360 m distance take?

7Lake rest cycles

First meeting time, in minutes?

8Rest every 100 m

First catch time, in seconds?

Mission 10

Exit ticket

Use the fixed scenarios in Missions 3–8. Give distances in metres and times in seconds. Complete all five. The reflection is optional and is not automatically graded.

Not completed

1. Elapsed time equals:

2. In the square problem from Missions 3–5, B's distance at the first catch:

3. In the rectangle practice, A's distance at the first catch:

4. In the 360-m piecewise-speed lap, time for the last 180 m:

5. In the 100-m run–rest cycle problem, first catch time:

Optional reflection — not automatically graded

⏱

Piecewise-Motion Schedule Engineer

This records completed checkpoints on running and elapsed time, corner stops, piecewise speeds and rest schedules.

Lesson 7.4 completed

Teaching notes

The rectangle practice solution is derived from the worked example data: A reaches vertex A after 960 m at 208.6 s and waits until 213.6 s; B reaches A after 800 m at 205 s and waits until 210 s. Their first overlap begins at 208.6 s.