Velocity-time graphs
Determining acceleration
If an object moves along a straight line, its motion can be represented by a velocity-time graph. The gradient of the line is equal to the The rate of change in speed (or velocity) is measured in metres per second squared. Acceleration = change of velocity 梅 time taken. of the object.
The table shows what each section of the graph represents:
| Section of graph | Gradient | Velocity | Acceleration |
| A | Positive | Increasing | Positive |
| B | Zero | Constant | Zero |
| C | Negative | Decreasing | Negative |
| D (v = 0) | Zero | Stationary (at rest) | Zero |
| Section of graph | A |
|---|---|
| Gradient | Positive |
| Velocity | Increasing |
| Acceleration | Positive |
| Section of graph | B |
|---|---|
| Gradient | Zero |
| Velocity | Constant |
| Acceleration | Zero |
| Section of graph | C |
|---|---|
| Gradient | Negative |
| Velocity | Decreasing |
| Acceleration | Negative |
| Section of graph | D (v = 0) |
|---|---|
| Gradient | Zero |
| Velocity | Stationary (at rest) |
| Acceleration | Zero |
Calculating displacement - higher
The area under the graph can be calculated by:
- using geometry (if the lines are straight)
- counting the squares beneath the line (particularly if the lines are curved)
Example
Calculate the total displacement of the object, whose motion is represented by the velocity-time graph below.
Here, the displacement can be found by calculating the total area of the shaded sections below the line.
Find the area of the triangle:
\(\frac{1}{2} \times base \times height\)
\(\frac{1}{2} \times 4 \times 8 = 16\)
The area of the triangle is 16 m2
Find the area of the rectangle:
base 脳 height
(10 - 4) 脳 8 = 48
The area of the rectangle is 48 m2
Add the areas together to find the total displacement:
(16 + 48) = 64
Total displacement = 64 m