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This lesson covers distance-time graphs and velocity-time graphs — as required by the Edexcel GCSE Physics specification (1PH0), Topic 2: Motion and Forces. You need to be able to interpret, draw and extract information from both types of graph, including calculating speed, acceleration and distance.
A distance-time graph plots the distance an object has travelled (y-axis) against the time taken (x-axis).
| Shape of Graph | Meaning |
|---|---|
| Horizontal line | The object is stationary (not moving) |
| Straight line going up | The object is moving at constant speed |
| Steep straight line | Moving at a higher constant speed |
| Shallow straight line | Moving at a lower constant speed |
| Curve getting steeper | The object is accelerating (speeding up) |
| Curve getting shallower | The object is decelerating (slowing down) |
The gradient (slope) of a distance-time graph gives the speed:
Speed = Distance ÷ Time = Gradient of the line
For a straight line: choose two points, calculate rise ÷ run.
For a curved line: draw a tangent at the required point and calculate the gradient of the tangent.
A distance-time graph shows an object travels 120 m in 30 s in a straight line.
Speed = 120 / 30 = 4 m/s
On a distance-time graph, an object is at 50 m at t = 10 s and at 200 m at t = 40 s.
Speed = (200 − 50) / (40 − 10) = 150 / 30 = 5 m/s
Exam Tip: On a curved distance-time graph, you must draw a tangent (a straight line that just touches the curve) at the specific time you are interested in. Then calculate the gradient of the tangent. The Edexcel mark scheme awards separate marks for drawing the tangent correctly and for calculating the gradient.
A velocity-time graph plots the velocity of an object (y-axis) against time (x-axis). These graphs give much more information than distance-time graphs.
| Shape of Graph | Meaning |
|---|---|
| Horizontal line | Object moves at constant velocity (no acceleration) |
| Straight line going up | Constant acceleration (uniform acceleration) |
| Straight line going down (towards zero) | Constant deceleration (uniform deceleration) |
| Steep line | Greater acceleration (or deceleration) |
| Line at v = 0 | Object is stationary |
| Curve | Changing acceleration (non-uniform) |
1. Gradient = Acceleration
Acceleration = Change in velocity ÷ Time taken = Gradient
a = (v − u) / t
2. Area Under the Graph = Distance Travelled
The total area between the line and the time axis gives the total distance (or displacement) covered.
3. The y-value at any point gives the velocity at that time.
A car accelerates from 0 m/s to 20 m/s in 8 seconds.
a = (v − u) / t = (20 − 0) / 8 = 2.5 m/s²
The same car (0 to 20 m/s in 8 s) — what distance does it cover?
The graph forms a triangle:
Distance = ½ × base × height = ½ × 8 × 20 = 80 m
A cyclist:
Total distance:
flowchart TD
A["Distance-Time Graph"] -- "Gradient gives" --> B["Speed / Velocity"]
B -- "Plot speed values<br/>against time" --> C["Velocity-Time Graph"]
C -- "Area under graph gives" --> D["Distance"]
D -- "Plot cumulative distance<br/>against time" --> A
C -- "Gradient gives" --> E["Acceleration"]
style A fill:#2980b9,color:#fff
style B fill:#27ae60,color:#fff
style C fill:#e67e22,color:#fff
style D fill:#8e44ad,color:#fff
style E fill:#c0392b,color:#fff
| From this graph... | You can find... | By calculating... |
|---|---|---|
| Distance-time | Speed | Gradient |
| Velocity-time | Acceleration | Gradient |
| Velocity-time | Distance | Area under the graph |
A car leaves home, drives to the shops, parks for a while, then drives back.
On a distance-time graph:
On a velocity-time graph:
On a velocity-time graph (taking upward as positive):
Exam Tip: When calculating area under a velocity-time graph, split the shape into simple rectangles and triangles. Show your working clearly — the Edexcel mark scheme awards marks for method even if the final answer is wrong.
Confusing distance-time and velocity-time graphs. A straight line going up on a distance-time graph means constant speed, but on a velocity-time graph it means constant acceleration.
Forgetting that area under a distance-time graph has no useful physical meaning — only the area under a velocity-time graph gives distance.
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