Primary & Middle School Physics · Grades 1–9

65Motion: Uniform and Varied

Two years ago you diagnosed motion by eye: evenly spaced dots, uniform; spreading dots, speeding up. This year the dots get millimetre rulings and the diagnosis gets numbers — interval by interval, speed by speed. And at the chapter’s end stands a quiet sentence about force and motion that will one day carry half of physics on its back.

65.1 The record of motion

Definition 65.1 (Timed position records)

A timed position record of a motion marks the moving object’s positions at equal ticks of time — video frames stepped through, a blinking strobe photograph, or the laboratory’s dot-timer stamping a paper tape pulled by the object, fifty dots a second. Between any two neighboring marks lies one tick’s worth of travel: the record turns motion into measurable segments.

Method 65.2 (Reading a record with numbers)

For a record with tick length $\tau$ (say $0.02\,\mathrm{s}$ for fifty a second, or a comfortable $0.1\,\mathrm{s}$ from video):

measure each gap between neighboring marks with the millimetre ruler;
divide each gap by the tick: $v = d/\tau$ gives the (average) speed over that little interval — in $\mathrm{m}/\mathrm{s}$ if the gaps are in metres;
lay the speeds side by side and read the story: steady numbers, uniform motion; climbing, accelerated; sinking, decelerated;
quote any interval’s speed as the speed “at” that moment: over so short a tick, average and instantaneous shake hands.

Three tapes from the dot-timer, one dot per tick. This year the gaps are measured, and each one becomes a speed.

Example 65.3 (A tape, worked)

A trolley’s tape, ticks of $0.1\,\mathrm{s}$ , gaps in centimetres: $2.0$ , $3.0$ , $4.0$ , $5.0$ , $6.0$ . Speeds, gap by gap: $0.02 \div 0.1 = 0.2\,\mathrm{m}/\mathrm{s}$ , then $0.3$ , $0.4$ , $0.5$ , $0.6\,\mathrm{m}/\mathrm{s}$ . Verdict: accelerated — and beautifully regularly so: the speed grows by exactly $0.1\,\mathrm{m}/\mathrm{s}$ each tick. Motions whose speed climbs by equal steps in equal times have a special name and a great future — uniformly accelerated, the signature of steady causes.

Example 65.4 (The most famous accelerated motion)

Drop a stone and film it: the gaps stretch tick by tick — free fall is accelerated. Measured, its speed grows by about $9.8\,\mathrm{m}/\mathrm{s}$ in each second of falling: after one second, $9.8\,\mathrm{m}/\mathrm{s}$ ; after two, nearly $20\,\mathrm{m}/\mathrm{s}$ . That the growth rate equals the place’s $g$ is no coincidence — it is the deepest rhyme in this year’s physics, and the High School volume builds its mechanics upon it. Note it, underline it, and let it wait.

65.2 Instantaneous speed, honestly

Definition 65.5 (Instantaneous speed)

The instantaneous speed of a motion at some moment is the average speed over a very short interval around that moment — so short that the motion has no room to change its pace within it. It is the speedometer’s number and the radar’s: both, in fact, measure exactly as Method 65.2 does, over ticks of a wink. (What “very short” means with full rigor is the great question the last year of high school mathematics answers — and the answer created modern science.)

Example 65.6 (Average and instantaneous, side by side)

A metro run between stations: average speed — total distance over total time — $35\,\mathrm{km}/\mathrm{h}$ . Instantaneous speed: $0$ at both platforms, $70\,\mathrm{km}/\mathrm{h}$ mid-tunnel. In uniform motion, and only there, the two notions merge into one steady number: uniform motion is precisely the motion that makes speed a single, honest, interval-free quantity.

65.3 The quiet sentence

Proposition 65.7 (Balanced forces leave motion unchanged)

When the forces on a body cancel each other out — or when none act at all — the body’s motion does not change: at rest it remains at rest, and moving it continues in a straight line at constant speed. Change of motion — speeding, slowing, turning — occurs only while some unbalanced force acts. This sentence, stated here and honestly deferred, is the opening law of all mechanics; the High School volume begins with it and never lets go.

Example 65.8 (Reading the world with the sentence)

The puck on smooth ice glides straight and steady — forces balanced (weight down, ice up), motion unchanged: no push needed to keep going. The braking car slows: one unbalanced force, the road’s grip on the tires, acting backward. The turning cyclist: unbalanced force sideways, supplied by the leaning tires. And the cruising airliner at steady $900\,\mathrm{km}/\mathrm{h}$ : thrust balancing drag, lift balancing weight — four forces, perfect tie, uniform motion. Stillness and cruising are, to physics, the same peaceful state.

Remark 65.9 (Why the sentence surprises)

Daily life seems to protest: stop pedaling and the bicycle slows — surely motion needs force? But the slowing bicycle is not force-free: friction and air push backward, unbalanced, and that is exactly why the pace decays. Remove them — the ice rink, the void of space — and motion coasts forever: the space probes launched before your grandparents met are still coasting now, engines cold. Humanity needed two thousand years to see through friction’s disguise; you have the advantage of ice rinks and space probes.

65.4 Exercises

Exercise 65.1 ★

What is a timed position record? Name two ways of making one.

Solution

Solution of Exercise 65.1.

Marks of the object’s positions at equal ticks of time. Two makers: stepping through video frames; the dot-timer stamping a pulled paper tape (or a strobe photograph).

Exercise 65.2 ★

A tape with $0.1\,\mathrm{s}$ ticks shows gaps of $4.0\,\mathrm{cm}$ , $4.0$ , $4.0$ , $4.0$ . Diagnose the motion and give its speed in $\mathrm{m}/\mathrm{s}$ .

Solution

Solution of Exercise 65.2.

Equal gaps: uniform. $v = 0.04 \div 0.1 = 0.4\,\mathrm{m}/\mathrm{s}$ .

Exercise 65.3 ★

Another tape: gaps $1.0$ , $2.5$ , $4.5$ , $7.0\,\mathrm{cm}$ . Diagnose — and compute the first and last interval speeds.

Solution

Solution of Exercise 65.3.

Growing gaps: accelerated. First interval: $0.01 \div 0.1 = 0.1\,\mathrm{m}/\mathrm{s}$ ; last: $0.07 \div 0.1 = 0.7\,\mathrm{m}/\mathrm{s}$ .

Exercise 65.4 ★

Distinguish average from instantaneous speed on a school run: which does the radar by the gate read, and which does the family’s “twenty minutes door to door” compute?

Solution

Solution of Exercise 65.4.

The radar reads instantaneous speed — the pace at the gate’s instant. “Twenty minutes door to door” computes the average — total distance over total time, stops and sprints blended.

Exercise 65.5 ★

In which motions do average and instantaneous speed coincide? Why there and only there?

Solution

Solution of Exercise 65.5.

In uniform motions: the pace never changes, so every interval — long or wink-short — returns the same number, and the two notions collapse into one.

Exercise 65.6 ★

By how much does a freely falling stone’s speed grow each second, and what local number does that growth equal?

Solution

Solution of Exercise 65.6.

By about $9.8\,\mathrm{m}/\mathrm{s}$ each second — numerically the local $g$ , $9.8\,\mathrm{N}/\mathrm{kg}$ : the year’s deepest rhyme.

Exercise 65.7 ★

State the quiet sentence. What does it say about a body with no forces at all — and about the cruising airliner’s four-way tie?

Solution

Solution of Exercise 65.7.

Balanced (or absent) forces leave motion unchanged: rest stays rest, and steady straight motion continues. The force-free body coasts forever; the airliner’s four-way tie — thrust against drag, lift against weight — is exactly why its $900\,\mathrm{km}/\mathrm{h}$ holds steady.

Exercise 65.8 ★

“Motion needs force to continue.” Convict this ancient error with the coasting space probe and friction’s disguise.

Solution

Solution of Exercise 65.8.

The probe, force-free in the void, has coasted for decades with cold engines — motion needing no force to continue. The bicycle that “proves” otherwise slows under friction and air, unbalanced backward forces in disguise; on ice, with the disguise thinned, it coasts far.

Exercise 65.9 ★★

A tape ( $0.1\,\mathrm{s}$ ticks) reads, in centimetres: $6.0$ , $5.0$ , $4.0$ , $3.0$ , $2.0$ . Diagnose; compute each interval speed; and predict the tape’s future if the pattern holds.

Solution

Solution of Exercise 65.9.

Shrinking gaps: decelerated. Speeds: $0.6$ , $0.5$ , $0.4$ , $0.3$ , $0.2\,\mathrm{m}/\mathrm{s}$ — falling by $0.1\,\mathrm{m}/\mathrm{s}$ per tick. If the pattern holds, two more ticks ( $1.0\,\mathrm{cm}$ , then nothing) bring the tape to rest.

Exercise 65.10 ★★

The worked trolley gained $0.1\,\mathrm{m}/\mathrm{s}$ per tick of $0.1\,\mathrm{s}$ . Express that growth per second — and compare the trolley’s gain with free fall’s.

Solution

Solution of Exercise 65.10.

$0.1\,\mathrm{m}/\mathrm{s}$ per $0.1\,\mathrm{s}$ is $1\,\mathrm{m}/\mathrm{s}$ gained per second — about a tenth of free fall’s $9.8\,\mathrm{m}/\mathrm{s}$ per second: a gentle push beside the Earth’s.

Exercise 65.11 ★★

A parachutist falls faster and faster — then, canopy open, at a steady $5\,\mathrm{m}/\mathrm{s}$ . Read both phases with the quiet sentence: what is unbalanced at first, and what tie has formed by the end?

Solution

Solution of Exercise 65.11.

First phase: weight outpulls the young air resistance — unbalanced downward force, speed grows. Canopy open: the blossoming air drag rises until it ties the weight exactly — balanced forces, and the fall settles to the steady $5\,\mathrm{m}/\mathrm{s}$ the sentence promises.

Exercise 65.12 ★★★

Design the full laboratory verdict on a toy car released down a ramp onto a carpet: the record to take, the numbers to compute, the expected two-act diagnosis (which act on the ramp, which on the carpet), and the force-sentence reading of each act.

Solution

Solution of Exercise 65.12.

Record: film the release and step frames at $0.1\,\mathrm{s}$ (or tape the car to the dot-timer); measure gap after gap down ramp and across carpet. Numbers: interval speeds, listed in order. Expected verdict: act one, growing gaps — accelerated on the ramp (weight’s unbalanced share along the slope); act two, shrinking gaps — decelerated on the carpet (friction’s unbalanced backward force). One toy, both faces of the quiet sentence.

65.5 Problem: The Tape Bureau

Problem 65.1

Weekend problem — an afternoon at the motion-analysis bureau; four tapes, four verdicts; the inspector’s rhyme

The bureau analyzes timed records for schools, sports clubs and courts. Ticks are $0.1\,\mathrm{s}$ throughout; gaps arrive in centimetres. You hold the ruler.

Part I — Routine verdicts.

Tape A (a corridor walker): $8.0$ , $8.0$ , $8.0$ , $8.0$ , $8.0$ . Verdict and speed — in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ .
Tape B (a sprinter’s start): $2.0$ , $4.0$ , $6.0$ , $8.0$ , $10.0$ . Verdict, first and last speeds, and the per-second growth of speed.
Tape C (a rolling ball meeting sand): $10.0$ , $7.0$ , $4.5$ , $2.5$ , $1.0$ . Verdict — and the force-sentence’s account of the sand.
Tape D (a mystery): $5.0$ , $5.0$ , $5.0$ , $7.5$ , $10.0$ . Tell D’s two-act story, and mark the tick where something happened.

Part II — The bureau’s physics desk.

A client insists their delivery scooter “was doing $30\,\mathrm{km}/\mathrm{h}$ , officer, on average”. The radar logged $55\,\mathrm{km}/\mathrm{h}$ at the crossing. Explain to the client why both numbers can be true and which one the law reads.
Tape B’s sprinter gains speed by equal steps. What single word does the bureau stamp on such motion, and what does the steadiness of the growth suggest about the pushing force?
The corridor walker of tape A, says the force-sentence, walks under balanced forces. Name the balance for a steady walker (what pushes, what resists).
A stone’s fall-tape shows interval speeds $0.98$ , $1.96$ , $2.94\,\mathrm{m}/\mathrm{s}$ . Verify the per-second growth and name the number it reproduces.

Part III — The court case. A skateboard rolled from a ramp across a schoolyard into a flowerbed; the caretaker blames “reckless speed”, the skater claims “I was slowing all along”. The yard camera yields gaps: ramp exit $12.0$ , then $11.5$ , $11.0$ , $10.5$ , $10.0$ into the flowerbed.

Verdict on the skater’s claim — with the interval speeds.
Exit speed and flowerbed-entry speed in $\mathrm{km}/\mathrm{h}$ : was either “reckless” beside a brisk cyclist’s $15\,\mathrm{km}/\mathrm{h}$ ?
The caretaker asks why the board did not simply stop on flat ground “as things naturally do”. The bureau’s answer, in one sentence of force-language.
Close the file with the inspector’s rhyme — two lines: what equal gaps mean, and what changing gaps demand (a cause, by the quiet sentence).

Solution

Solution of Problem 65.1.

1. Uniform; $v = 0.08 \div 0.1 = 0.8\,\mathrm{m}/\mathrm{s} \approx 2.9\,\mathrm{km}/\mathrm{h}$ — a stroll. 2. Accelerated: from $0.02 \div 0.1 = 0.2\,\mathrm{m}/\mathrm{s}$ to $1.0\,\mathrm{m}/\mathrm{s}$ ; gaining $0.2\,\mathrm{m}/\mathrm{s}$ per tick — $2\,\mathrm{m}/\mathrm{s}$ per second. 3. Decelerated, steeply. The sand supplies a large unbalanced backward force — deep friction — and the speeds tumble: $1.0$ , $0.7$ , $0.45$ , $0.25$ , $0.1\,\mathrm{m}/\mathrm{s}$ . 4. Three ticks of uniform gliding at $0.5\,\mathrm{m}/\mathrm{s}$ , then acceleration — gaps leaping to $7.5$ and $10.0\,\mathrm{cm}$ : between ticks three and four something began to push (a slope, a shove). The event lives at the third gap’s end. 5. Both can be true: the average blends waiting and riding over the whole trip; the radar reads the crossing’s instant. Traffic law reads the radar — limits govern instantaneous speed, and no gentle average excuses a fast instant. 6. Uniformly accelerated. Equal speed-steps in equal times point to a steady, unchanging net push — constant cause, constant growth. 7. The walker’s forward push from the ground on the shoes ties the backward drags of air and ground on the body: balanced, hence the steady $0.8\,\mathrm{m}/\mathrm{s}$ . 8. Growth: $0.98\,\mathrm{m}/\mathrm{s}$ per tick of $0.1\,\mathrm{s}$ — $9.8\,\mathrm{m}/\mathrm{s}$ per second: the local $g$ , reproduced by a falling stone’s tape. 9. The claim holds: speeds $1.2$ , $1.15$ , $1.10$ , $1.05$ , $1.0\,\mathrm{m}/\mathrm{s}$ — slowing, gently, all along. 10. Exit $1.2 \times 3.6 = 4.3\,\mathrm{km}/\mathrm{h}$ ; entry $3.6\,\mathrm{km}/\mathrm{h}$ — both far below the cyclist’s benchmark: recklessness acquitted; steering, perhaps, another matter. 11. “On flat ground nothing stops a rolling board except unbalanced backward friction — smooth yards supply little, so the board keeps most of its motion, exactly as the law of balanced forces says it must.” 12. For example: “Equal gaps, a pace at peace — no net force disturbs its lease. Changing gaps demand a cause: some unbalanced push, by the quiet clause.”

Terms defined in this chapter

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