What you'll learn
This revision guide covers everything you need to know about distance–time and velocity–time graphs for AQA GCSE Physics. You'll learn how to interpret motion graphs, calculate gradients to find speed and acceleration, and determine distance travelled from velocity–time graphs. These skills are essential for both Paper 1 and Paper 2 of your AQA GCSE Physics exams.
Key terms and definitions
Distance — how far an object has travelled from its starting point, measured in metres (m) or kilometres (km); it is a scalar quantity with magnitude only.
Displacement — the distance travelled in a particular direction from the starting point; it is a vector quantity with both magnitude and direction.
Speed — the rate of change of distance, calculated as distance ÷ time, measured in metres per second (m/s) or kilometres per hour (km/h); it is a scalar quantity.
Velocity — the rate of change of displacement in a particular direction, calculated as displacement ÷ time, measured in metres per second (m/s); it is a vector quantity.
Acceleration — the rate of change of velocity, calculated as change in velocity ÷ time, measured in metres per second squared (m/s²).
Gradient — the slope of a line on a graph, calculated as change in y-axis ÷ change in x-axis.
Uniform motion — motion at a constant speed or velocity, represented by a straight line on a distance–time graph.
Deceleration — negative acceleration, where an object slows down; also called retardation.
Core concepts
Distance–time graphs
Distance–time graphs show how the distance travelled by an object changes over time. Distance is always plotted on the y-axis and time on the x-axis.
Interpreting the shape of the line:
- A horizontal line means the object is stationary (not moving). The distance remains constant, so speed = 0 m/s.
- A straight diagonal line indicates constant speed. The steeper the line, the faster the object is travelling.
- A curved line shows changing speed (acceleration or deceleration). If the curve is getting steeper, the object is speeding up. If the curve is getting less steep, the object is slowing down.
Calculating speed from a distance–time graph:
The gradient of a distance–time graph equals the speed of the object.
For a straight line:
- Speed = gradient = rise ÷ run = change in distance ÷ change in time
For a curved line:
- Draw a tangent to the curve at the point of interest
- Calculate the gradient of the tangent to find the instantaneous speed at that moment
Distance–time graphs show total distance travelled, not displacement, so they cannot show direction of motion.
Velocity–time graphs
Velocity–time graphs show how the velocity of an object changes over time. Velocity is plotted on the y-axis and time on the x-axis.
Interpreting the shape of the line:
- A horizontal line means constant velocity (uniform motion). The acceleration is zero.
- A straight diagonal line indicates constant acceleration. The steeper the line, the greater the acceleration.
- A line with positive gradient shows positive acceleration (speeding up).
- A line with negative gradient shows negative acceleration or deceleration (slowing down).
- A curved line indicates changing acceleration (non-uniform acceleration).
- A line below the time axis (negative velocity) means the object is moving in the opposite direction to the original direction of travel.
Calculating acceleration from a velocity–time graph:
The gradient of a velocity–time graph equals the acceleration of the object.
Acceleration = gradient = change in velocity ÷ change in time
For example, if velocity increases from 5 m/s to 15 m/s over 4 seconds:
- Acceleration = (15 − 5) ÷ 4 = 2.5 m/s²
A negative gradient indicates deceleration.
Calculating distance travelled from a velocity–time graph:
The area under a velocity–time graph equals the distance travelled (or displacement if direction is considered).
For regular shapes:
- Rectangle: area = base × height = time × velocity
- Triangle: area = ½ × base × height = ½ × time × velocity
- Trapezium: area = ½ × (sum of parallel sides) × height = ½ × (u + v) × t
For irregular shapes:
- Count squares under the graph
- Each square represents a certain distance based on the scale
- Add up all complete and partial squares
Comparing distance–time and velocity–time graphs
| Feature | Distance–time graph | Velocity–time graph |
|---|---|---|
| Gradient represents | Speed | Acceleration |
| Horizontal line means | Stationary (zero speed) | Constant velocity |
| Straight diagonal line means | Constant speed | Constant acceleration |
| Curved line means | Changing speed | Changing acceleration |
| Area under graph represents | No physical meaning | Distance travelled |
Uniform and non-uniform motion
Uniform motion occurs when an object travels at constant velocity (constant speed in a straight line). On graphs:
- Distance–time: straight diagonal line
- Velocity–time: horizontal line
Non-uniform motion occurs when velocity changes (acceleration or deceleration). On graphs:
- Distance–time: curved line
- Velocity–time: diagonal or curved line
For non-uniform motion, you can calculate:
- Average speed = total distance ÷ total time
- Instantaneous speed = gradient of tangent at a specific point on a distance–time graph
Interpreting real-world motion scenarios
You must be able to sketch and interpret graphs for everyday situations:
A car journey with traffic lights:
- Accelerating from rest: increasing gradient (distance–time) or positive gradient (velocity–time)
- Constant speed: straight diagonal line (distance–time) or horizontal line (velocity–time)
- Braking to stop: decreasing gradient (distance–time) or negative gradient (velocity–time)
- Stopped at lights: horizontal line (distance–time) or line on time axis (velocity–time)
An object thrown upwards:
- Initial upward velocity decreases due to gravity (deceleration of approximately 9.8 m/s²)
- Velocity reaches zero at maximum height
- Object falls back down with increasing downward velocity (acceleration)
- Velocity–time graph shows straight line with negative gradient going through zero
Equations of motion and graphs
The suvat equations (studied at Higher Tier) relate to graph features:
- v = u + at (where v = final velocity, u = initial velocity, a = acceleration, t = time)
- s = ½(u + v)t (where s = displacement)
- v² = u² + 2as
- s = ut + ½at²
From velocity–time graphs:
- u is the initial velocity (y-intercept)
- v is the final velocity (y-value at time t)
- a is the acceleration (gradient)
- s is the displacement (area under graph)
Worked examples
Example 1: Distance–time graph analysis
Question: The graph shows the journey of a cyclist.
[Imagine a graph with: 0-10s: straight line from origin to (10, 40); 10-20s: straight line from (10, 40) to (20, 100); 20-30s: horizontal line at 100m]
(a) Calculate the speed of the cyclist between 10 and 20 seconds. [3 marks]
(b) Describe the motion of the cyclist between 20 and 30 seconds. [1 mark]
Solution:
(a)
- Change in distance = 100 − 40 = 60 m ✓
- Change in time = 20 − 10 = 10 s ✓
- Speed = 60 ÷ 10 = 6 m/s ✓
(b) The cyclist is stationary / not moving / stopped ✓
Mark scheme notes:
- Part (a) requires showing working for all 3 marks
- Units must be correct (m/s)
- Part (b) accepts any clear statement indicating zero motion
Example 2: Velocity–time graph — calculating distance
Question: A car accelerates uniformly from rest to 20 m/s in 8 seconds, then travels at constant velocity for 12 seconds before decelerating uniformly to rest in 5 seconds.
(a) Sketch a velocity–time graph for this journey. [3 marks]
(b) Calculate the total distance travelled. [4 marks]
Solution:
(a) Graph should show:
- Straight line from (0, 0) to (8, 20) ✓
- Horizontal line from (8, 20) to (20, 20) ✓
- Straight line from (20, 20) to (25, 0) ✓
(b) Distance = area under graph
Split into three sections:
- Triangle (acceleration): ½ × 8 × 20 = 80 m ✓
- Rectangle (constant velocity): 12 × 20 = 240 m ✓
- Triangle (deceleration): ½ × 5 × 20 = 50 m ✓
- Total distance = 80 + 240 + 50 = 370 m ✓
Example 3: Finding instantaneous speed
Question: The distance–time graph shows the motion of a bus. Calculate the speed of the bus at exactly 15 seconds. [4 marks]
[Imagine a curved line showing acceleration]
Solution:
- Draw a tangent to the curve at t = 15 s ✓
- Read values from tangent, e.g., at t = 10 s, distance = 30 m and at t = 20 s, distance = 130 m ✓
- Change in distance = 130 − 30 = 100 m
- Change in time = 20 − 10 = 10 s ✓
- Speed = 100 ÷ 10 = 10 m/s ✓
Mark scheme notes:
- Tangent must be drawn accurately with a ruler
- Clear working showing which points were used
- The exact answer depends on where the tangent touches; any value within an acceptable range (typically ±10%) scores full marks if working is correct
Common mistakes and how to avoid them
Confusing speed and velocity: Remember that speed is a scalar (magnitude only) while velocity is a vector (magnitude and direction). Distance–time graphs show speed, not velocity, unless direction is specified.
Calculating area instead of gradient on distance–time graphs: On distance–time graphs, gradient gives speed; area has no physical meaning. On velocity–time graphs, gradient gives acceleration and area gives distance travelled. Learn which is which.
Drawing curves when motion is uniform: Constant speed produces a straight line on distance–time graphs, not a curve. Only draw curves when speed is changing.
Forgetting units: Always include units in your final answer. Speed is m/s (or km/h), acceleration is m/s², distance is m (or km), and time is s (or minutes/hours depending on context).
Misreading scales: Always check what each square on the graph represents. Don't assume each square is 1 unit — read the axis labels carefully.
Negative gradients on distance–time graphs: Distance cannot decrease (it's the total path length), so distance–time graphs cannot have negative gradients. Velocity–time graphs can have negative sections showing opposite direction or deceleration.
Exam technique for "Distance–time and velocity–time graphs"
"Calculate" questions: Show all working clearly. Write the formula, substitute values, and give the answer with units. Marks are often awarded for method even if the final answer is incorrect.
"Describe" or "Explain" motion: Use precise physics terminology. State whether the object is stationary, moving at constant speed, accelerating, or decelerating. Give evidence from the graph (e.g., "horizontal line shows zero velocity").
Drawing tangents: Use a ruler and make the tangent line long enough to read coordinates accurately. Show the coordinates you've used in your calculation.
Area calculations: When finding distance from velocity–time graphs, split complex shapes into triangles, rectangles, and trapeziums. Label each section and show the area calculation for each part separately. Most marks come from method, not just the final answer.
Quick revision summary
Distance–time graphs plot distance (y-axis) against time (x-axis). The gradient represents speed; horizontal lines mean stationary, straight diagonals mean constant speed, and curves mean changing speed. Velocity–time graphs plot velocity (y-axis) against time (x-axis). The gradient represents acceleration and the area under the graph represents distance travelled. Horizontal lines indicate constant velocity, positive gradients show acceleration, and negative gradients show deceleration. Always show full working in calculations, include units, and use accurate tangents for curved lines.