Graphs in Practical Situations

This guide covers subtopic 2.9 Graphs in practical situations, from Topic 2, Algebra and graphs, for Cambridge O Level Mathematics (Syllabus D) 4024, 2025–2027 series.

Where this fits in 4024

This subtopic applies gradient and area ideas to real, physical situations — the same distance-time and speed-time graph skills also appear in Kinematics and Motion Graphs for Cambridge O Level Physics, approached here from the mathematics side.

Syllabus coverage

CAMBRIDGE O LEVEL MATHEMATICS (SYLLABUS D) 4024

• Use and interpret graphs in practical situations, including travel graphs and conversion graphs (2.9)
• Draw graphs from given data (2.9)
• Apply the idea of rate of change to simple kinematics involving distance–time and speed–time graphs, acceleration and deceleration, including estimation and interpretation of the gradient of a tangent at a point (2.9)
• Calculate distance travelled as the area under a speed–time graph, for linear sections only (2.9)

4024 is not tiered — every candidate covers all of the above.

Travel graphs and conversion graphs

A travel graph (distance against time) shows a journey directly: horizontal sections mean stationary periods, and the steepness of a sloping section shows how fast the journey is progressing at that point.

A conversion graph shows the relationship between two different units or quantities (currency conversion, temperature scales) as a straight line, letting a value be read off in one unit once it’s known in the other.

Drawing graphs from given data means plotting the data points accurately on suitable axes and joining them appropriately — with a straight line where the relationship is linear, or a smooth curve where it isn’t.

Distance–time graphs

On a distance–time graph, the gradient at any point gives the speed at that instant:

speed = gradient = change in distance / change in time

For a straight-line section, this is a single calculation between two points. For a curved section, the gradient at a point is found by drawing a tangent to the curve at that point and calculating the tangent line’s gradient — this is the “estimation and interpretation of the gradient of a tangent” the syllabus specifically names.

Speed–time graphs

On a speed–time graph, the gradient gives acceleration, and a negative gradient indicates deceleration:

acceleration = gradient = change in speed / change in time

The area under the graph gives distance travelled — and 4024 restricts this calculation to linear sections only, meaning the areas involved are always rectangles, triangles or trapezia, found with standard area formulas rather than any calculus technique.

Worked example. An object accelerates uniformly from rest to 12 m/s over 6 s, then travels at a constant 12 m/s for a further 10 s. Find the total distance travelled.

Stage 1 (triangle): distance = ½ × base × height = ½ × 6 × 12 = 36 m
Stage 2 (rectangle): distance = base × height = 10 × 12 = 120 m

total distance = 36 + 120 = 156 m

Displacement versus distance travelled (enrichment, beyond this syllabus)

This syllabus sub-topic is restricted to speed-time graphs, and speed cannot be negative, so a graph of the kind described below cannot arise in this course. The section is included as enrichment for students who may meet velocity-time graphs elsewhere, not as examinable content here.

If a speed–time graph — strictly, a velocity–time graph — dips below the time axis, the object is moving in the opposite direction. The area above the axis represents positive displacement, and the area below represents negative displacement.

• Displacement = the difference between the two areas (net change in position).
• Distance travelled = the sum of their magnitudes (total ground covered) — always the larger, or equal, value.

Worked example. A velocity–time graph shows a line starting at +9 m/s and falling steadily to −3 m/s over 12 s. Find the displacement and the distance travelled.

gradient = (-3 - 9) / 12 = -1 m/s^2
time to reach zero velocity: 9 / 1 = 9 s

area above axis (0 to 9 s): 1/2 x 9 x 9 = 40.5 m
area below axis (9 to 12 s): 1/2 x 3 x 3 = 4.5 m

displacement = 40.5 - 4.5 = 36 m
distance travelled = 40.5 + 4.5 = 45 m

Always add the two areas for distance travelled, and subtract for displacement.

Rates of change beyond travel

The same gradient-as-rate idea applies beyond distance and speed: on a graph of cost against quantity bought, the gradient gives the price per item; on a graph of volume against time for a tap filling a container, the gradient gives the flow rate, in litres per minute. Always state the units when interpreting a gradient in context — a bare number rarely earns the mark.

Common mistakes

• Reading a distance–time graph as if it were a speed–time graph, or vice versa. A straight sloping line means constant speed on the former, but constant acceleration on the latter — always check which quantity is on the vertical axis first.
• Drawing a poor tangent to a curve when estimating a gradient at a point — a tangent should touch the curve at exactly one point locally, without crossing through it there.
• Treating deceleration as needing a separate method. It’s simply a negative gradient on a speed–time graph — the same calculation, with a negative result.
• Trying to find the area under a curved section. This syllabus restricts the area-under-the-graph calculation to linear sections, which can always be split into rectangles, triangles and trapezia.
• Adding the areas above and below the axis when displacement is asked for, instead of subtracting — this gives distance travelled, not displacement.
• Interpreting a gradient without stating its units in context — a correct number with no units, or no stated meaning, rarely earns the mark.

Quick revision checklist

• Reading travel graphs and conversion graphs
• Drawing an accurate graph from a table of data
• Gradient of a distance–time graph = speed (including via a tangent, for curved sections)
• Gradient of a speed–time graph = acceleration (negative = deceleration)
• Area under a speed–time graph = distance, for linear sections only
• Displacement versus distance travelled for a velocity-time graph that crosses the axis (subtract vs. add the areas) – enrichment beyond this syllabus, which only requires speed-time graphs
• Interpreting a gradient as a rate in context, with correct units

Related resources

• Sequences and Proportion — the previous subtopic
• Graphs of Functions and Sketching Curves — the next subtopics
• Cambridge O Level Mathematics subject hub

Written against Cambridge O Level Mathematics (Syllabus D) 4024, 2025–2027 series. Always check the current syllabus for your examination year.