Velocity-time graphs
Acceleration
The rate of change in speed (or velocity) is measured in metres per second squared. Acceleration = change of velocity ÷ time taken. is the rate of change of velocity. It is the amount that velocity changes per unit time.
The change in velocity can be calculated using the equation:
change in velocity = final velocity – initial velocity
\(\Delta \text{v} = \text{v - u}\)
The average acceleration of an object can be calculated using the equation:
\(\text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}}\)
\(\text{a} = \frac{\Delta \text{v}}{\text{t}}\)
This is when:
- acceleration (\(\text{a}\)) is measured in metres per second squared (m/s2 )
- change in velocity (\(\Delta \text{v}\)) is measured in metres per second (m/s)
- time taken (\(\text{t}\)) is measured in seconds (s)
If an object is slowing down, it is decelerating (and its acceleration has a negative value).
Example
A car takes 8.0 s to accelerate from rest to 28 m/s. Calculate the average acceleration of the car.
The speed, in a particular direction, of a body after it accelerates (after it changes speed or direction). , \(\text{v}\) = 28 m/s
The speed, in a particular direction, of a body before it accelerates (before it changes speed or direction)., \(\text{u}\) = 0 m/s (because it was at rest – not moving)
change in velocity, \(\Delta \text{v}\) = (28 – 0) = 28 m/s
\(\text{a} = \frac{\Delta \text{v}}{\text{t}}\)
= 28 ÷ 8
= 3.5 m/s2
Question
A car takes 25 s to accelerate from 20 m/s to 30 m/s. Calculate the acceleration of the car.
final velocity, \(\text{v}\) = 30 m/s
initial velocity, \(\text{u}\) = 20 m/s
change in velocity, \(\Delta \text{v}\) = (30 – 20) = 10 m/s
\(\text{a} = \frac{\Delta \text{v}}{\text{t}}\)
= 10 ÷ 25
= 0.4 m/s2
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 acceleration 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
Scientists draw graphs of data to help analyse a situation. A velocity-time graph of a journey can give information about acceleration (the gradient) and distance travelled (displacement).
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:
- ½ × base × height
- ½ × 4 × 8 = 16 m2
- Find the area of the rectangle:
- base × height
- (10 – 4) × 8 = 48 m2
- Add the areas together to find the total displacement:
- (16 + 48) = 64 m
More guides on this topic
- Newton's laws - AQA Synergy
- Circuits - AQA Synergy
- Mains electricity - AQA Synergy
- Acids and alkalis - AQA Synergy
- Rates of reaction - AQA Synergy
- Energy, rates and reactions - AQA Synergy
- Equilibria - AQA Synergy
- Electrons and chemical reactions - AQA Synergy
- Sample exam questions - movement and interactions - AQA Synergy
