Describing motion along a line distance–time graphs

SPEC 4.5.6.1 • TOPIC 5 FORCES — PAPER 2

Distance-Time Graphs

Read the gradient — not the y-value. That single rule separates students who get 4/4 from students who get 1/4 on every d-t question.

What You'll Learn

The motion topic starts here. One equation to recall, one graph type to read, one HT trick. The maths is small — the marks are big.

Lesson Goals

Distinguish distance from displacement, and speed from velocity (scalar vs vector)

Recall typical speeds for walking, running, cycling, urban driving, and sound in air

Apply s = v t with confident unit conversion (km/h → m/s; hours → seconds)

Read a distance-time graph: speed = gradient, NOT y-value

(HT) Find instantaneous speed on a curved d-t graph by drawing a tangent

Key Definitions

Four words — two pairs. Match the right word to what the question asks.

Distance

Distance is how far an object has travelled along its path. Scalar — magnitude only, no direction. Measured in metres (m).

Displacement

Displacement is the straight-line distance from start to finish, with direction. Vector. A round trip has distance > 0 but displacement = 0.

Speed

Speed is the rate at which distance is covered. Scalar — no direction. Measured in metres per second (m/s).

Velocity

Velocity is speed in a given direction. Vector. A car going round a circle at constant speed has changing velocity (because the direction changes).

HT only — circular motion

An object moving in a circle at constant speed has a continuously changing velocity. Speed (scalar) is constant — but velocity (vector) depends on direction, and the direction is different at every point around the circle. Example: a satellite orbiting Earth at steady speed has changing velocity throughout its orbit.

Scalar vs Vector — Applied to Motion

The two columns aren't just labels. Mark schemes use the precise term — if a question asks for "displacement", an answer in "distance" loses marks.

PairScalar (no direction)Vector (has direction)

How far?Distance — total path lengthDisplacement — straight line + direction from start to end

How fast?Speed — rate of distance coveredVelocity — speed + direction

Quick check: a 400 m running track

A runner completes one full lap. Distance covered = 400 m. Displacement = 0 m (they're back where they started). Same physical journey, different scalar/vector descriptions.

Typical Speeds — Spec-Required Values

The spec specifically requires you to recall these values. They appear in "estimate" questions and in problems where you need to sanity-check your answer.

~1.5

m/s — walking

~3

m/s — running

~6

m/s — cycling

~13

m/s — urban driving (≈30 mph)

~50

m/s — inter-city train

~250

m/s — passenger aircraft

Other speeds

Speed of sound in air: ~330 m/s. Wind speed varies — there's no fixed "typical" value.

Sanity-check rule

If a "person walking" problem gives you an answer of 50 m/s, you've slipped a decimal. Walking speed is ~1.5 m/s. The typical values are your last line of defence against an arithmetic mistake.

The Equation — s = v t

Recall equation. Not on the equation sheet. For an object moving at constant speed, the distance covered in a given time. For non-uniform motion, this gives the average speed.

Distance = speed × times = v × ts — distance, mv — speed, m/st — time, s

Rearranged forms

v = s ÷ t | t = s ÷ v

Unit conversion — the killer step

Substitute in SI units: distance in m, time in s, speed in m/s. Common conversions:

1 km = 1,000 m | 1 hour = 3,600 s | km/h ÷ 3.6 = m/s

ESUA Worked Example

A train travels 240 km in a time of 2.0 hours.

Calculate the average speed of the train in m/s. 4 MARKS

E

Equation: v = s ÷ t

S

Convert at substitution:

240 km = 240,000 m

2.0 hours = 2.0 × 3,600 = 7,200 s

v = 240,000 ÷ 7,200

U

Unit: metres per second (m/s)

A

Answer: v ≈ 33 m/s (2 s.f.)

Note — average speed: 33 m/s is the train's average speed over the 2.0 hours. Its instantaneous speed varied (slower when accelerating from stations, faster at cruise). The spec specifically requires you to be able to "calculate average speed for non-uniform motion" — this is exactly that calculation.

The Distance-Time Graph

A d-t graph shows how an object's distance travelled changes over time. The convention is fixed: time on the x-axis, distance on the y-axis.

What it tells you

The shape of the line tells you the type of motion (stationary, constant speed, accelerating)

The gradient (steepness) tells you the speed at that moment

The y-value tells you the distance covered so far — NOT the speed

Plotting a d-t graph from data (spec skill)

The spec also requires you to draw a d-t graph from measurements. Method:

Put time on the x-axis, distance on the y-axis — with a scale and a unit on each

Plot each (time, distance) measurement as a small cross or dot

Draw a straight line or smooth curve of best fit through the points — not a zigzag connecting them

Translating the graph into a story

A good description of motion from a d-t graph names the time interval and what's happening: "From 0 to 5 s, the object moves at constant speed. From 5 to 10 s, it's stationary. From 10 to 20 s, it moves at constant speed again, but faster."

The Four Basic Shapes

Every d-t graph you'll see at GCSE is built from these four shapes. Recognise them on sight.

Stationary
flat line — Stationary flat line

Constant Speed
straight, sloped — Constant Speed straight, sloped

Faster constant speed
steeper line — Faster constant speed steeper line

Accelerating curving steeper

Decelerating?

A decelerating object on a d-t graph would show a curve that flattens out — the gradient gets smaller over time, eventually levelling off. Curve gets steeper → accelerating. Curve gets flatter → decelerating.

Speed = Gradient (NOT y-value)

The single most important rule on this topic. The speed of an object at any moment is the gradient — the steepness of the line — at that point, calculated as rise ÷ run: change in distance divided by change in time.

NEVER read the y-value as the speed

At t = 8 s, the line is at distance = 64 m. That is not the speed at t = 8 s. The speed is the gradient of the line — calculated by picking two points and dividing the rise by the run. Reading the y-value as speed is the most common error on every d-t question.

Worked Example — Speed From Gradient

On a distance-time graph, an object's line passes through two points:

(2 s, 10 m) and (8 s, 40 m).

The line is straight between these two points.

Calculate the speed of the object. 3 MARKS

E

Method: speed = gradient = Δs ÷ Δt (rise over run)

S

Substitute:

Δs = 40 − 10 = 30 m

Δt = 8 − 2 = 6 s

speed = 30 ÷ 6

U

Unit: metres per second (m/s)

A

Answer: speed = 5 m/s

Instantaneous Speed on a Curve HT

For accelerating motion the line is curved, so the gradient changes from moment to moment. To find the speed at a specific instant, draw a tangent — a straight line touching the curve at that point — and measure its gradient.

3-step method

Find the point on the curve at the time you want

Draw a straight line through that point that just touches the curve (don't cross it) — this is the tangent

Pick two points on the tangent (not the curve!) and calculate its gradient = rise ÷ run

Examiner Alerts

Trap 1 — Reading the y-value as speed

The most-examined error on d-t graphs. The y-value at any point is the distance covered, not the speed. Speed = gradient. Always pick two points and compute rise ÷ run.

Trap 2 — Forgetting unit conversion (km/h or hours)

"Calculate the speed in m/s" usually involves a km-to-m or hours-to-seconds conversion. Quick shortcut: km/h ÷ 3.6 = m/s. Convert at substitution, not after.

Trap 3 — Drawing the tangent on the curve, not measuring its gradient

(HT) Students often draw a beautiful tangent and then read the y-value of the curve underneath it. Wrong. Pick two points on the tangent — well separated — and use them for rise ÷ run.

Trap 4 — "Describe" vs "Calculate"

"Describe the motion" = words ("constant speed, then stationary, then constant speed"). "Calculate the speed" = numbers with units. The wrong type of answer scores zero, even if the physics is correct.

Exam-Style Multi-Part Question

(a)

1 MARK

(b)

3 MARKS

(c)

3 MARKS

(a)

The drone is stationary (the line is horizontal — distance doesn't change). (1)

(b)

Method: speed = gradient = Δs ÷ Δt (M1)

Substitute: Δs = 80 − 0 = 80 m; Δt = 10 − 0 = 10 s; v = 80 ÷ 10 (M1)

Answer: v = 8 m/s (A1)

(c)

Speed between 20 and 40 s: Δs = 200 − 80 = 120 m; Δt = 20 s; v = 120 ÷ 20 = 6 m/s (M1)

6 m/s < 8 m/s (M1) — the speed is less than in part (b) (A1).

Alternative reasoning: the second segment is less steep than the first → smaller gradient → lower speed.

Weak vs Strong Answers

Two answers to: "From the d-t graph, calculate the speed of the object at t = 4 s. The graph is a straight line through the points (0, 0) and (10, 50). [3 marks]"

✗ Weak

"At t = 4 s, distance = 20 m, so speed = 20 m/s."Read the y-value as the speed. That's the distance travelled, not the speed. Scores 0/3.

✓ Strong

"speed = gradient = Δs ÷ Δt = 50 ÷ 10 = 5 m/s"Method named, substituted, answer with unit. 3/3.

Two answers to: "Describe the motion of the object between t = 5 s and t = 15 s. [2 marks]"

✗ Weak

"It moves quickly then stops."No time intervals named. Vague. Usually 0/2.

✓ Strong

"From t = 5 s to t = 10 s the object moves at a constant speed. From t = 10 s to t = 15 s the object is stationary."Time intervals named; motion type identified for each. 2/2.

Common Errors

Students often writeCorrect versionWhy it matters

Read y-value of d-t graph as "speed"Speed = gradient = Δs ÷ ΔtThe y-value is distance covered, not speed

"Distance" used where displacement is meantUse displacement for straight-line + direction; distance for total pathDifferent scalar/vector — mark scheme picks this up

Substitute km or hours directly into s = v tConvert to m and s at the substitution stepAnswer wrong by factor of 1000 or 3600

(HT) Tangent drawn, then y-value of curve read as speedPick two points on the tangent , compute its gradientThe tangent gives speed; the curve doesn't directly

"Describe" question answered with numbersFor "describe": words — name time intervals and motion type ("constant speed", "stationary", "accelerating")Wrong format scores 0 even with correct physics

"5" as answer (no unit)"5 m/s""Calculate" answers always require a unit

Checkpoint

You can now

Distinguish distance / displacement and speed / velocity using scalar vs vector

Recall typical speeds (walking 1.5, running 3, cycling 6, urban 13, sound 330 m/s)

Apply s = v t with confident unit conversion (km → m and hours → seconds)

Read a d-t graph: speed = gradient (not y-value); identify stationary, constant speed, and accelerating motion

(HT) Find instantaneous speed on a curved d-t graph by drawing a tangent and measuring its gradient

One rule to take away

Speed is the gradient of a d-t graph. The y-value is just where the object has reached. Pick two points, do rise ÷ run, state the unit. Every d-t question becomes the same routine.

Next lesson

5.09 — Velocity-time graphs and uniform acceleration. A new equation (a = Δv ÷ t, recall), the v-t graph's twin readings (gradient = acceleration, area = distance), and the third equation on the equation sheet for uniform acceleration: v² − u² = 2 a s.