Distance-Time and Speed-Time Graphs

Motion graphs appear on almost every 0625 paper, and they separate students who memorise from students who understand. Examiners use them because one graph tests definitions, calculation and interpretation in a single question.

What does the gradient of each graph tell you?

On a distance-time graph, the gradient gives speed. On a speed-time graph, the gradient gives acceleration. Mixing these two readings is the single most common graph error in 0625.

Graph	Gradient means	Area under graph means	Horizontal line means
Distance-time	speed	nothing useful	object at rest
Speed-time	acceleration	distance travelled	constant speed

In words: speed equals the change in distance divided by the change in time, read from the slope. Acceleration equals the change in speed divided by the change in time, read from the slope of a speed-time graph. In symbols: $v = \dfrac{\Delta s}{\Delta t}$ and $a = \dfrac{\Delta v}{\Delta t}$.

How do you find distance from a speed-time graph?

Calculate the area under the line. For constant speed, that area is a rectangle: $\text{distance} = \text{speed} \times \text{time}$. For uniform acceleration from rest, it is a triangle: $\text{distance} = \dfrac{1}{2} \times \text{time} \times \text{final speed}$. For mixed journeys, split the shape into rectangles and triangles and add the areas. Extended students must also recognise curves: an increasing gradient on a speed-time graph means increasing acceleration, and a curve flattening towards horizontal means acceleration decreasing to zero, the terminal velocity shape.

Worked Exam Question

A cyclist starts from rest and accelerates uniformly to 6.0 m/s in 8.0 s. She then travels at a constant 6.0 m/s for 12 s. (a) Calculate her acceleration during the first 8.0 s. [2] (b) Calculate the total distance travelled in the 20 s journey. [3]

Solution (a). Equation: $a = \dfrac{\Delta v}{\Delta t} = 6.0 \div 8.0 = 0.75\ \text{m/s}^2$.

Solution (b). Distance = area under the speed-time graph. Triangle: $\dfrac{1}{2} \times 8.0 \times 6.0 = 24\ \text{m}$. Rectangle: $6.0 \times 12 = 72\ \text{m}$. Total: $24 + 72 = 96\ \text{m}$.

Mark scheme

• M1 (a): $a = \dfrac{\Delta v}{\Delta t}$ or $6.0/8.0$ seen. A1: $0.75\ \text{m/s}^2$ with unit.
• M1 (b): $\dfrac{1}{2} \times 8.0 \times 6.0 = 24\ \text{m}$ (triangle area).
• M1 (b): $6.0 \times 12 = 72\ \text{m}$ (rectangle area).
• A1 (b): $96\ \text{m}$.

Common Mistakes

• Reading a distance-time gradient as acceleration. Fix: check the y-axis label before you write anything.
• Forgetting the $\dfrac{1}{2}$ in the triangle area. That doubles the distance. Fix: sketch the shape and label it triangle or rectangle first.
• Saying a horizontal line on a speed-time graph means “stopped”. It means constant speed; stopped is the line along $v = 0$. Fix: learn both horizontal-line meanings from the table above.
• Using one point instead of a gradient triangle. Fix: take the change in y over the change in x across a large section of the line.
• Misreading scales. A small square is not always one unit. Fix: write the value of one square on the axis before calculating.

Exam Technique Tip

Before answering any graph question, annotate the graph itself: label each section “accelerating”, “constant speed” or “at rest”, and write the gradient or area you need next to it. Examiners mark working on the graph, and this 20-second habit prevents the axis mix-up that costs the whole question.

How This Is Examined

Graphs span Core and Extended. Papers 1 and 2 give a graph and ask which statement describes the motion, worth one mark with a high error rate. Papers 3 and 4 set structured questions like the worked example; Paper 4 (Extended) adds curved graphs, tangent gradients and terminal-velocity descriptions worth up to 6 marks. Paper 6 asks you to plot motion data, draw a best-fit line and find a gradient, so bring a sharp pencil and a 30 cm ruler. Core candidates stick to straight-line graphs; Extended candidates must handle changing acceleration too.