1MA1 · A14–A15, R14–R15AlgebraFoundation & HigherGrades 4–7

Real-Life Graphs & Rates of Change

A graph of a journey tells a story: when the car sped up, when it stopped, how far it went. Two measurements unlock that story — the $\textbf{gradient}$ , which gives a rate such as speed or acceleration, and, for velocity–time graphs, the $\textbf{area}$ under the graph, which gives the distance travelled.

55 min Graphs

The big picture

Gradient always means "how fast the vertical quantity changes per unit of the horizontal one". On a distance–time graph that is speed; on a velocity–time graph it is acceleration; on a graph of water depth against time it is the rate of filling. For curves, the gradient changes, so it is estimated by drawing a tangent.

Area under a velocity–time graph is distance because velocity multiplied by time is distance. Edexcel Higher questions ask you to estimate the area under a curve using trapezia and to say whether the estimate is too big or too small.

What you'll learn

Interpret distance–time graphs and calculate speeds
Interpret velocity–time graphs, finding acceleration from gradients
Find distance travelled from the area under a velocity–time graph
Estimate gradients of curves using tangents and areas using trapezia (Higher)

Distance–time graphs

The gradient of a distance–time graph is the $\textbf{speed}$ . Steeper means faster; a horizontal section means the object is stationary.

A line sloping back down to the time axis means the object is returning to its starting point.

Average speed for the whole journey is total distance divided by total time — not the average of the individual speeds.

Example 1A cyclist rides 12 km in 40 minutes, rests for 20 minutes, then rides 18 km in 60 minutes. Find her speed on the first section and her average speed overall, in km/h.

1First section: 12 km in

\tfrac23

h, so speed

= 18

km/h.

2Overall: 30 km in 2 hours.

AnswerFirst section 18 km/h; average speed 15 km/h.

Your turnSame method, new numbers — try it before moving on.

A car travels 90 km in 1 hour 30 minutes. What is its average speed in km/h?

Your answer

Show worked solution

Solution

11 hour 30 minutes $= 1.5$ hours.
2 $90 \div 1.5 = 60$ km/h.

Answer: $60$ km/h

Tip — Convert minutes to hours before calculating km/h: 40 minutes is $\tfrac{40}{60} = \tfrac23$ hour.

Velocity–time graphs

The gradient of a velocity–time graph is the $\textbf{acceleration}$ . A negative gradient is deceleration; a horizontal line means constant velocity.

The $\textbf{area}$ between the graph and the time axis is the distance travelled. Split it into rectangles, triangles and trapezia.

\text{acceleration} = \dfrac{\text{change in velocity}}{\text{time}}

Gradient of a velocity–time graph.

\text{distance} = \text{area under the graph}

Area of a velocity–time graph.

Example 1A train accelerates uniformly from rest to 12 m/s in 4 s, travels at 12 m/s for 10 s, then decelerates uniformly to rest in 6 s. Find its initial acceleration and the total distance.

1Acceleration:

\dfrac{12}{4} = 3

m/s².

2Distance: triangle

\tfrac12 \times 4 \times 12 = 24

, rectangle

10 \times 12 = 120

, triangle

\tfrac12 \times 6 \times 12 = 36

AnswerAcceleration 3 m/s²; distance 180 m.

Your turnSame method, new numbers — try it before moving on.

A cyclist accelerates uniformly from rest to 15 m/s in 5 seconds. What is the acceleration, in m/s²?

Your answer

Show worked solution

Solution

1Acceleration is the gradient of the velocity–time graph.
2 $15 \div 5 = 3$ m/s².

Answer: $3$ m/s²

Gradients of curves

When a graph is curved, the rate of change is different at every point. To estimate it at a point, draw a $\textbf{tangent}$ — a straight line touching the curve at that point — and find the tangent’s gradient.

The $\textbf{average}$ rate of change between two points is the gradient of the chord joining them.

Tangent gradients are estimates, so a range of answers is accepted if the tangent is drawn carefully.

Why this works

The difference between the chord and the tangent is the difference between average speed over a journey and the speedometer reading at an instant. At A-Level, differentiation calculates the tangent gradient exactly.

Areas under curves

To estimate the area under a curve, split it into vertical strips of equal width and treat each as a trapezium.

If the curve bends so that the straight tops of the trapezia lie below it, the estimate is an underestimate; if they lie above, it is an overestimate.

\text{trapezium area} = \tfrac12(a + b)h

Parallel sides a and b, width h.

Example 1A car’s velocity is 0, 5, 8 and 9 m/s at

t = 0, 2, 4, 6

s, and the graph is a smooth curve that flattens out. Estimate the distance travelled in 6 s and say whether it is an over- or underestimate.

1Three trapezia of width 2:

\tfrac12(0 + 5)(2) + \tfrac12(5 + 8)(2) + \tfrac12(8 + 9)(2)

= 5 + 13 + 17 = 35

3The curve bulges above the straight chords between points.

AnswerAbout 35 m — an underestimate.

Your turnSame method, new numbers — try it before moving on.

A velocity–time graph is a straight line from $(0, 0)$ to $(10, 20)$ , with time in seconds and velocity in m/s. How far does the object travel?

Your answer

Show worked solution

Solution

1Distance is the area under the graph: a triangle.
2 $\frac{1}{2} \times 10 \times 20 = 100$ m.

Answer: $100$ m

Tip — Sketch the trapezia on the curve to decide over- or underestimate: if the curve is above the trapezium tops, the trapezia miss some area.

Think like an examiner

Show gradient calculations with the values read from the graph.
Split areas into clear shapes and show each area.
Include units in every answer: km/h, m/s, m/s², m.
Justify over- or underestimates by referring to the shape of the curve.

Rates and areas

\text{speed} = \tfrac{\text{distance}}{\text{time}}

Gradient of distance–time.

\text{acceleration} = \tfrac{\Delta v}{t}

Gradient of velocity–time.

\text{distance} = \text{area under } v\text{–}t

Area of velocity–time.

Watch out for these

Reading distance from the gradient of a velocity–time graph.

Gradient is acceleration; area is distance.

Averaging the speeds of each section.

Average speed is total distance over total time.

Using minutes directly in a km/h calculation.

Convert time to hours first.

Drawing a tangent that crosses the curve near the point.

A tangent touches the curve at the point and follows its direction there.

Stretch yourself

A lift starts from rest, accelerates uniformly for 3 s to 2 m/s, travels at 2 m/s for $T$ seconds, then decelerates uniformly to rest in 2 s. It rises 25 m in total. Find $T$ .

Hint — Write the area under the velocity–time graph in terms of $T$ and set it equal to 25.

Reveal solution

SolutionArea

= \tfrac12 \times 3 \times 2 + 2T + \tfrac12 \times 2 \times 2 = 3 + 2T + 2 = 2T + 5

. Setting

2T + 5 = 25

gives

T = 10

s. Check:

3 + 20 + 2 = 25

m ✓.

\blacksquare

Questions students ask

What does a horizontal line mean on a velocity–time graph?

Constant velocity — zero acceleration. On a distance–time graph, a horizontal line means stationary.

Why is area under a velocity–time graph a distance?

Each small strip has area velocity × time, which is distance; adding the strips gives the total.

How accurate must a tangent gradient be?

It is an estimate. Draw a long, careful tangent and use a large triangle; mark schemes accept a range.

Key takeaways

Gradient of distance–time is speed; horizontal means stationary.
Gradient of velocity–time is acceleration; area underneath is distance.
Average speed is total distance over total time.
Tangents estimate instantaneous rates; chords give average rates.
Trapezia estimate areas under curves; decide over- or underestimate from the curve’s shape.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

A graph of taxi cost against distance is a straight line from $(0, 3)$ to $(10, 18)$ , with distance in km and cost in pounds.

(a)Write down the fixed charge.[1]
Show mark scheme (a)
- B1£3
(b)Work out the cost per kilometre.[2]
Show mark scheme (b)
- M1 $\frac{18 - 3}{10}$
- A1£1.50
Answer: £1.50

Question 2Grade 63 marks

A runner accelerates uniformly from rest to 8 m/s in 4 seconds, then runs at 8 m/s for 6 seconds. Work out the total distance run.[3]
Show mark scheme
- M1Area of triangle $\frac{1}{2} \times 4 \times 8 = 16$
- M1Area of rectangle $6 \times 8 = 48$
- A1 $64$ m
Answer: $64$ m

Question 3Grade 72 marks

On a distance–time curve (distance in metres, time in seconds), the tangent at $t = 5$ passes through $(2, 10)$ and $(8, 40)$ . Estimate the speed at $t = 5$ .[2]
Show mark scheme
- M1 $\frac{40 - 10}{8 - 2}$
- A1 $5$ m/s
Answer: $5$ m/s

Unlimited practice

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