1.2.2Dynamics
Determining Displacement from V-T Graphs
5 min read 5 parts Lesson 7 of 22 Virtual lab
Part 1 of 5Area Equals Displacement
You cannot read distance directly from a velocity–time graph, but you can calculate it by finding an area.
Key idea
- The area enclosed between the line of a velocity–time graph and the time axis equals the displacement of the object.
Why it works
- Area is height multiplied by width, which is velocity multiplied by time, and that product is displacement.
Units
- With velocity in m/s and time in s, the area comes out in metres, which is the unit of displacement.
Distance
- For motion in one direction the area is also the distance travelled, so the two quantities have the same size.
Part 2 of 5Calculating Areas
Calculating Areas
Rectangle
- A section where the velocity is constant forms a rectangle with area = velocity × time.
Triangle
- A section starting from rest or ending at rest forms a triangle with area = ½ × base × height.
Trapezium
- A section joining two different velocities is a trapezium with area = ½ × (sum of parallel sides) × width.
Combining
- Split a complicated graph into simple shapes, find each area separately and add them together.
| Shape | Area formula |
|---|---|
| Rectangle | base × height |
| Triangle | ½ × base × height |
| Trapezium | ½ × (a + b) × h |
Part 3 of 5Areas Below the Axis
1
When the line is below the time axis the velocity is negative, so that area represents displacement in the opposite direction.
2
To find the final displacement, subtract the area below the axis from the area above it.
3
To find the total distance travelled, add the sizes of all the areas as positive numbers.
4
For example a ball thrown upwards has an area above the axis on the way up and an equal area below it on the way down, so its displacement is zero when it returns.
5
Average speed is then total distance divided by total time, and average velocity is displacement divided by total time.
Part 4 of 5Worked Example
Question
- A graph rises from 0 to 12 m/s in 4.0 s, stays at 12 m/s for 6.0 s, then falls to 0 in 2.0 s.
- Find the total displacement and the average velocity.
- Triangle (0 to 4 s): ½ × 4.0 × 12 = 24 m.
- Rectangle (4 to 10 s): 12 × 6.0 = 72 m.
- Triangle (10 to 12 s): ½ × 2.0 × 12 = 12 m; total = 24 + 72 + 12 = 108 m.
- Total time = 12 s, so average velocity = 108 ÷ 12 = 9.0 m/s.
Answer
- Displacement 108 m;
- average velocity 9.0 m/s in the direction of travel.
Part 5 of 5Exam Checklist & Tips
1.2.2 – Key Points for the Exam
- Area under graph
- The area between the line and the time axis;
- equals displacement on a v–t graph.
- The area between the line and the time axis;
- Trapezium A four-sided shape with one pair of parallel sides.
- Total distance The sum of the sizes of every area on the graph.
- Average velocity Total displacement divided by total time.
Common Mistakes
- Using the gradient instead of the area when asked for displacement.
- Using base × height for a triangle without multiplying by one half.
- Subtracting areas below the axis when the question asks for total distance, which needs them added.
- Forgetting that the time on the x-axis must be in seconds when the velocity is in m/s.