Graphs in Practical Situations: Revision Notes

Condensed for the final weeks. For worked examples, use the Graphs in Practical Situations study guide.

The two travel graphs — never confuse them

	Distance–time	Speed–time
Gradient =	Speed	Acceleration
Horizontal line =	Stationary	Constant speed
Area under =	Nothing	Distance travelled
Straight slope =	Constant speed	Constant acceleration
Negative gradient =	Returning to start	Deceleration

The two most valuable facts: gradient of a speed–time graph is acceleration, and area under a speed–time graph is distance. Always check which quantity sits on the vertical axis before interpreting a graph — the single most common source of lost marks in this topic is applying a distance–time rule to a speed–time graph, or vice versa.

Finding gradient

gradient = change in y / change in x

Use a large triangle spanning most of the line for accuracy, and read coordinates from the line, not from plotted data points.

For a curve, draw a tangent at the point and find the gradient of the tangent — that gives the instantaneous rate of change.

Worked example. A distance–time graph shows a curve of increasing gradient. What does this describe, and how do you find the speed at a particular instant? The object is speeding up, because the gradient — and therefore the speed — is increasing along the curve. To find the speed at a particular instant: draw a tangent to the curve at that point, then construct a large right-angled triangle on the tangent, and the speed is the change in distance divided by the change in time for that triangle.

Area under a speed–time graph

Split into rectangles, triangles and trapezia.

rectangle  = base x height
triangle   = 1/2 x base x height
trapezium  = 1/2 (a + b) x h

Worked: a car accelerates from rest to 20 m/s in 8 s, holds 20 m/s for 12 s, then stops in 5 s.

Triangle:   1/2 x 8 x 20   = 80 m
Rectangle:  12 x 20        = 240 m
Triangle:   1/2 x 5 x 20   = 50 m
                             ------
Total distance             = 370 m

Displacement vs distance travelled

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

• Displacement = the difference between the two areas (net position change).
• Distance travelled = the sum of their magnitudes (total ground covered).

Worked example. A velocity–time graph shows a line starting at +6 m/s and falling steadily to −4 m/s over 10 s. Find the acceleration, and explain the difference between the displacement and the distance travelled.

acceleration = (-4 - 6) / 10 = -1.0 m/s^2

Where the line crosses the axis, the object is momentarily at rest, then moves in the opposite direction. The area above the axis is a triangle of ½ × 6 × 6 = 18 m; the area below is a triangle of ½ × 4 × 4 = 8 m. The displacement is 18 − 8 = 10 m, but the distance travelled is 18 + 8 = 26 m — always add the areas for distance, and subtract for displacement.

Conversion graphs

A straight line converting between two units (currency, miles/km, °C/°F). Read across and down; state the units in your answer.

The line passes through the origin only when the two quantities are directly proportional — miles and kilometres, for instance, since 0 miles is 0 kilometres. Celsius and Fahrenheit is the standard counter-example: the line crosses the Fahrenheit axis at 32, not at the origin, because 0 °C is 32 °F — the relationship is linear but not proportional.

The gradient is the conversion factor only when the relationship is directly proportional (a line through the origin). For a non-proportional conversion such as Celsius to Fahrenheit, the gradient (1.8) is the rate of change between the scales, but no single multiplying factor converts one reading directly to the other, since a fixed offset (the 32) must also be applied.

Rates of change

Any straight-line graph’s gradient is a rate: cost per item, litres per minute, wages per hour. Interpret the gradient in context with units — that phrasing is what earns the mark.

Exam traps

• Do not calculate the area under a distance–time graph — it has no meaning.
• Read the axis labels before deciding what the gradient represents.
• Check units: km/h vs m/s. Divide km/h by 3.6 for m/s.
• Use a large triangle; small ones magnify reading errors.
• A horizontal line on a speed–time graph means constant speed, not stopped.
• When a velocity–time graph dips below the axis, add the areas for distance but subtract them for displacement.
• Using a small triangle when drawing a tangent — it magnifies any reading error.

Self-test

1. What does the gradient of a distance–time graph represent?
2. What does the area under a speed–time graph represent?
3. A horizontal line on a speed–time graph — what is happening?
4. A car goes 0 to 15 m/s in 6 s. Find the acceleration and the distance covered.
5. Convert 72 km/h to m/s.
6. A distance–time graph is a curve with increasing gradient. What does this describe?
7. A velocity–time graph’s area below the axis represents what?

Answers: 1. Speed. 2. Distance travelled. 3. The object is moving at constant speed (zero acceleration). 4. a = 15/6 = 2.5 m/s²; distance = ½ × 6 × 15 = 45 m. 5. 72 ÷ 3.6 = 20 m/s. 6. The object is speeding up, since the gradient — and so the speed — is increasing. 7. Negative displacement — motion in the opposite direction, still counted positively when totalling distance travelled.

For exam-style questions with full mark schemes on this topic, see the Travel Graphs practice questions.