Real-Life Graphs Distance-Time, Velocity-Time & Financial Contexts

SPEC A14 • F + H • ⭐ HIGH PRIORITY

Real-Life Graphs

Distance-time, velocity-time, and financial graphs — where gradient and area have concrete meanings.

Distance-Time Graphs

A distance-time graph shows how far something is from a starting point at each moment in time.

The two axes

Horizontal: time (seconds, minutes, hours).

Vertical: distance from a fixed starting point (metres, km, etc.).

Gradient = speed. Steeper line = faster. In the example: gradient = 10 m ÷ 5 s = 2 m/s.

Reading a Journey

Real journeys have different segments. Each segment tells a story.

How to read the shape

Sloping up → moving away from the start.

Horizontal → stationary (same distance as time passes — NOT constant speed).

Sloping down → returning towards the start.

Steeper → faster. The walker-out segment (3 m/s) is steeper than the return (1.5 m/s).

Finding Speed from a d-t Graph

Speed is the gradient of the segment you're interested in. Pick two points on that segment and apply the standard gradient formula.

speed = distance ÷ time = ^Δd⁄_Δt

Worked calculation (from the journey above)

First segment (0 → 10 s): distance changes from 0 to 30 m. Speed = 30 ÷ 10 = 3 m/s.

Middle segment (10 → 20 s): distance stays at 30 m. Speed = 0 ÷ 10 = 0 m/s (stationary).

Return segment (20 → 40 s): distance changes from 30 to 0 m. Speed = 30 ÷ 20 = 1.5 m/s (magnitude only — d-t doesn't show direction).

Always include units

Mark schemes require units on your speed answers — "3" scores M1 only. "3 m/s" scores the A1. The same applies for m/s², km/h, etc. on v-t graphs.

Velocity-Time Graphs

A velocity-time graph looks similar to d-t, but the vertical axis shows velocity (or speed) — not distance. Everything means something different.

What the features mean on v-t

Sloping up → accelerating (speeding up).

Horizontal → constant velocity (zero acceleration — but still moving!).

Sloping down → decelerating (slowing down).

Gradient = acceleration (m/s²).

The Key Rule: Area Under v-t = Distance

On a velocity-time graph, the area under the curve tells you how far the object has travelled. This is the single most important fact in the lesson.

area under v-t graph = distance travelled

Total distance here

25 + 200 + 25 = 250 metres covered over the 30-second journey. On a v-t graph, the gradient would give acceleration — but it's the area that gives distance.

Shape Formulas You'll Use

Break the area into straight-line shapes and add them up. These three formulas cover nearly every GCSE v-t question.

ShapeWhen to use itArea formula

RectangleConstant velocity (horizontal segment)area = velocity × time

TriangleStarts or ends at 0 m/s (accel from rest, or decel to rest)area = ½ × base × height

TrapeziumBetween two non-zero velocities v₁ and v₂ (both vertical edges present)area = ½(v₁ + v₂) × t

Shortcut: the whole trapezoidal journey

If the journey is a single trapezium (accelerate up, cruise, decelerate down), you don't need to split it — just use: area = ½(parallel + parallel)(height) directly.

area = ½(short parallel side + long parallel side) × peak velocity

01Worked Example — Total Distance from v-t

A car accelerates uniformly from rest to 20 m/s over 8 seconds, travels at a steady 20 m/s for 12 seconds, then decelerates uniformly to rest over 5 seconds. Find the total distance travelled. [4 marks]

M

Method: area under a v-t graph is the distance travelled. Split the trapezoidal journey into triangle + rectangle + triangle and sum the areas.

C

Triangle 1 (acceleration, 0–8 s): ½ × 8 × 20 = 80 m.

Rectangle (constant, 8–20 s, width 12 s): 12 × 20 = 240 m.

Triangle 2 (deceleration, 20–25 s): ½ × 5 × 20 = 50 m.

Total = 80 + 240 + 50 = 370 m.

E

Evaluate: Total distance = 370 m. M1 each shape's area (3 marks), A1 correct total with units. Always include "m" — mark schemes reject bare numbers.

02Worked Example — Trapezium-Shape v-t

A scooter's speed-time graph for a 30-second journey is shown below. The scooter accelerates uniformly from rest to 20 m/s over 10 seconds, travels at a steady 20 m/s for 15 seconds, then decelerates uniformly to rest over 5 seconds.

(a) Find the acceleration at t = 20 seconds. [1 mark]

(b) Work out the total distance travelled. [3 marks]

M

Method (a): at t = 20 s, the scooter is in the constant-speed section (between t = 10 s and t = 25 s). Acceleration = gradient of that segment = 0.

C

(a) Horizontal line → gradient = 0 → acceleration = 0 m/s².

(b) Two methods give the same answer:

Trapezium shortcut: parallel sides 15 s (top) and 30 s (bottom), height 20 m/s.

Area = ½(15 + 30)(20) = ½(45)(20) = 450 m.

Split method: ½(10)(20) + 15(20) + ½(5)(20) = 100 + 300 + 50 = 450 m.

E

Evaluate: (a) 0 m/s², (b) 450 m. B1 in (a); M1 for any complete method (trapezium or split), M1 for correct substitution, A1 450 m with units.

Estimating Area Under a Curve HIGHER

Real v-t graphs aren't always made of straight lines — sometimes they're curves. You can't compute the exact area, but you can estimate it by splitting the region into vertical strips of equal width and treating each strip as a trapezium. (Estimating areas under non-linear graphs is spec A15 — Higher only.)

area ≈ ^h⁄₂ (y₀ + 2y₁ + 2y₂ + … + 2yₙ₋₁ + yₙ)

Trapezium rule steps

Divide the time interval into n equal strips, each of width h.

Read off the velocity y₀, y₁, y₂, … yₙ at the edges of each strip.

Multiply the first and last (y₀ and yₙ) by 1. Multiply all the middle values by 2.

Add them all up and multiply by h ÷ 2.

Why it works

Each strip is the area of a trapezium with parallel sides yᵢ and yᵢ₊₁: ½(yᵢ + yᵢ₊₁)h. When you add up all the strips, each middle reading is shared between two strips — hence the "2" coefficient.

Worked Example — Trapezium Rule Estimation

HIGHER

The velocity-time graph below shows a runner over 40 seconds. By dividing the area under the graph into four sections of equal widths, estimate the distance travelled during the 40 seconds. [3 marks]

M

Method: split into 4 strips of width h = 40 ÷ 4 = 10 s. Read off the velocity at t = 0, 10, 20, 30, 40. Use the trapezium rule.

C

Readings: y₀ = 0, y₁ = 4, y₂ = 7, y₃ = 8, y₄ = 6 (all m/s).

Area ≈ ¹⁰⁄₂ (0 + 2(4) + 2(7) + 2(8) + 6)

= 5 × (0 + 8 + 14 + 16 + 6)

= 5 × 44

= 220 m.

E

Evaluate: Estimated distance ≈ 220 m. M1 correct strip-edge readings, M1 correct trapezium-rule substitution, A1 final answer in an accepted range (examiners allow a tolerance of roughly ±5 m on trapezium-rule estimates because graph readings involve reading precision).

Financial Graphs

Graphs also appear in money contexts. Two patterns crop up repeatedly: simple interest (linear, F + H) and compound interest (exponential, Higher only at A14).

Simple interest (linear)

Same amount of interest every year. Graph is a straight line. Balance grows by a constant amount each year.

balance = P + Prt

Compound interest (exponential) H

Interest earns more interest. Graph curves upwards. Growth accelerates year on year.

balance = P × (1 + r)^t

Examiner Alert: d-t vs v-t Confusion

⚠️ EXAMINER ALERT: The biggest trap in this topic. d-t and v-t graphs look similar but their features mean completely different things. Label the axes before doing anything else.

FeatureOn d-t graphOn v-t graph

GradientSpeed (m/s)Acceleration (m/s²)

Horizontal segmentStationary (not moving)Constant velocity (moving at steady speed)

Area under graph(Not meaningful)Distance travelled (m)

Typical units on ym, kmm/s

⚠️ EXAMINER ALERT — Units matter: The mark scheme wording is "M1 correct gradient from specified interval; A1 speed value with units". Writing "5" instead of "5 m/s" loses the A1. Same for "800" instead of "800 m" on distance answers.

Examiner Alert: Reading the Question Carefully

⚠️ EXAMINER ALERT — "Halfway through the journey": A recurring AQA phrasing. It means halfway in time — not halfway across the graph visually, and not halfway along the distance axis. For a 30-second journey, halfway is at t = 15 s. For a 50-second journey, it's at t = 25 s.

Why this matters — a typical trap

A journey has three phases: accelerating, constant speed, decelerating.

If you're asked the acceleration "halfway through the journey", first compute the total time, then halve it to locate the time point.

If that time lies in the constant-speed section, the acceleration is 0 m/s² (flat line = gradient 0). Common distractors include the acceleration DURING the ramp phase — which is NOT what the question asks.

⚠️ EXAMINER ALERT — Back-solving from a total distance: Harder AQA questions give you the total distance travelled and only a partial speed-time graph (say, the first two phases). You're then asked for a missing final velocity. Strategy: compute the areas you can from the graph, subtract from the given total to find the area of the missing final section, and solve a trapezium equation ½(v₁ + V)(t) = remaining area for the unknown V.

Practice Question 1

The distance-time graph below shows Amira's walk to a shop and back home.

(a) How far is the shop from home? [1 mark]

(b) For how long was she at the shop? [1 mark]

(c) Calculate her speed on the return journey, in m/s. [2 marks]

M

Method: read distance and time values off the graph at the corner points; use speed = distance ÷ time for the return segment's gradient.

C

(a) Maximum distance reached = 40 m.

(b) Stationary segment runs from t = 15 s to t = 30 s: 15 seconds.

(c) Return segment: from (30 s, 40 m) to (50 s, 0 m). Distance covered = 40 m over 20 s. Speed = 40 ÷ 20 = 2 m/s.

E

Evaluate: (a) 40 m, (b) 15 s, (c) 2 m/s. Always include units. B1 each for (a) and (b); M1 correct gradient method + A1 answer with units for (c).

Practice Question 2

The velocity-time graph shows a lorry's motion over 20 seconds.

(a) Find the acceleration during the first 4 seconds. [2 marks]

(b) Calculate the total distance travelled. [3 marks]

M

Method: (a) acceleration = gradient of v-t over first 4 s. (b) distance = area of the trapezium.

C

(a) Gradient = Δv ÷ Δt = 15 ÷ 4 = 3.75 m/s².

(b) Trapezium: parallel sides 12 s (top: 16 − 4) and 20 s (bottom), height 15 m/s.

Area = ½(12 + 20)(15) = ½(32)(15) = 240 m.

E

Evaluate: (a) 3.75 m/s², (b) 240 m. M1 each calculation, A1 each answer with units. Note: you can also split into triangle + rectangle + triangle: ½(4)(15) + 12(15) + ½(4)(15) = 30 + 180 + 30 = 240 m. Same answer either way.

Practice Question 3 HIGHER

The velocity-time graph below shows a runner over 60 seconds. By dividing the area into four equal-width sections, estimate the total distance travelled. [3 marks]

M

Method: strip width h = 60 ÷ 4 = 15 s. Read velocity at t = 0, 15, 30, 45, 60. Apply trapezium rule.

C

Readings: y₀ = 0, y₁ = 5, y₂ = 9, y₃ = 11, y₄ = 9 (m/s).

Area ≈ ¹⁵⁄₂ × (0 + 2(5) + 2(9) + 2(11) + 9)

= 7.5 × (10 + 18 + 22 + 9)

= 7.5 × 59

= 442.5 m.

E

Evaluate: Estimated distance ≈ 442.5 m. M1 strip-edge readings, M1 correct trapezium-rule substitution, A1 value in range (accept 440–445). Mark schemes on estimate questions always allow a tolerance.

Summary

Distance-time graphs

Gradient = speed . Steeper = faster.

Horizontal = stationary. Sloping down = returning.

Area under graph has no special meaning on d-t.

Velocity-time graphs

Gradient = acceleration (m/s²).

Area under graph = distance travelled (m). This is the big one.

Horizontal = constant velocity (still moving).

Methods for area

Straight-line v-t: split into rectangles, triangles, or use trapezium area formula ½(v₁ + v₂)t.

Curved v-t: trapezium rule — area ≈ ^h⁄₂(y₀ + 2y₁ + … + 2yₙ₋₁ + yₙ).

Financial graphs

Simple interest: linear growth (straight line).

Compound interest: exponential growth (curve that accelerates). H

Final checklist

Before answering any real-life graph question: (1) read the axis labels to decide if it's d-t or v-t. (2) Decide whether the question asks for a gradient (speed or acceleration) or an area (distance). (3) Always give the answer with units.