What are the key learning points about speed-time graphs?
A horizontal line means the object is moving at a constant speed.
The gradient of the line = the rate of change of speed.
The steeper the line, the greater the increase in speed.
The area under the graph = the distance travelled.
What do speed-time graphs show?
Speed-time graphs show how the speed of a moving object changes with time.
Key facts
Gradient of speed-time graph = rate of change of speed (m/s2).
Area under speed-time graph = distance travelled (m).
Summary
| Speed-time graph | |
|---|---|
| Gradient of graph | Rate of Change of Speed (m/s2) |
| Area under graph | Distance (m) |
What is an example of a speed-time graph?
This is a speed-time graph for a car moving between two sets of traffic lights.
Question
What is the rate of change of speed of the car between 0 s and 10 s?
Answer
The rate of change of speed is the slope or gradient of the speed-time graph
Rate of change of speed = gradient of the speed-time graph
= \(\frac{\text{(final speed – initial speed)}}{\text{time taken}}\)
Rate of change of speed = (16 m/s – 0 m/s) ÷ 10 s = 1.6 m/s2
Question
What is the rate of change of speed of the car between 20 s and 25 s?
Answer
The rate of change of speed = gradient of the speed-time graph = \(\frac{\text{(final speed – initial speed)}}{\text{time taken}}\)
The rate of change of speed = (0 m/s – 16 m/s) ÷ 5 s
= -3.2 m/s2.
The car is slowing down at a rate of 3.2 m/s2.
Question
What is the total distance between the two sets of traffic lights?
Answer
The total distance between the two sets of traffic lights = the total area under the graph.
There are 2 triangles and a rectangle. Find the area of each.
From 0 s – 10 s
distance travelled = area of the triangle = ½ x 10 s x 16 m/s = 80 m.
From 10 s – 20 s
distance travelled = area of the rectangle = 10 s x 16 m/s = 160 m.
From 20 s – 25 s
distance travelled = area of the triangle = ½ x 5 s x 16 m/s = 40 m.
Total distance travelled = 80 m + 160 m + 40 m = 280 m.
Question
What is the average speed of the car between the traffic lights?
Answer
Average speed = \(\frac{\text{total distance moved}}{\text{time taken}}\)
average speed = 280 m ÷ 25 s = 11.2 m/s.
What do velocity-time graphs show? (Higher tier only)
Velocity-time graphs show how the The rate of change of displacement. The distance travelled in one second in a specified direction. Measured in m/s. of a moving object changes with time.
- Constant The rate of change of velocity. It is measured in metres per second squared (m/s²). Acceleration = change of velocity ÷ time taken. is shown by a straight rising line, A.
- Constant retardation (or deceleration) is shown by a straight falling line, C.
- Constant velocity is shown by a horizontal line, B.
- A horizontal line along the X-axis shows the speed is zero, meaning that the vehicle has stopped, or is stationary, D.
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 | Zero | Stationary (at rest, v = 0) | Zero |
The slope or gradient of a velocity-time graph =\(\frac{\text{(final velocity – initial velocity)}}{\text{time taken}}\) = acceleration
Key fact
The gradient of a velocity-time graph is the acceleration of the object.
The steeper the line the greater the acceleration.
What is an example of a velocity-time graph?
The velocity-time graph above is for a racing car accelerating from rest.
Question
What is the acceleration in the first 10 s?
Answer
Acceleration = the gradient of the graph = \(\frac{\text{final velocity – initial velocity}}{\text{time taken}}\)
= (40 m/s – 0 m/s) ÷ 10 s
= 40 m/s ÷ 10 s
= 4 m/s2
The acceleration of the car in the first 10 s is 4 m/s2.
Question
What is the acceleration of the car between 30 s and 50 s?
Answer
Acceleration is the gradient of the graph = \(\frac{\text{final velocity – initial velocity}}{\text{time taken}}\)
= (0 m/s – 60 m/s) ÷ 20 s
= -60 m/s ÷ 20 s
= -3 m/s2
The car has an acceleration of -3 m/s2 (or a retardation of 3 m/s2).
How to calculate displacement using a velocity-time graph
(Higher tier only)
Key facts
- The A distance measured in a specified direction. of an object can be calculated from the area under a velocity-time graph.
The area under the graph can be calculated by:
Using geometry (if the lines of the graph are straight).
Counting the squares beneath the line (particularly if the lines of the graph are curved).
Examples
Calculate the total displacement of the object, whose motion is represented by the velocity-time graph below.
The displacement can be found by calculating the total area of the shaded sections between the line and the time axis.
There is a triangle and a rectangle – the area of both must be calculated and added together to give the total displacement.
To find the area of the triangle:
area = \(\frac{\text{1}}{\text{2}}\) x base x height
area = \(\frac{\text{1}}{\text{2}}\) x 4 s x 8 m/s = 16m
To find the area of the rectangle:
area = base × height
area = (10 - 4) s × 8 m/s = 48 m
Add the areas together to find the total displacement:
Total displacement = (16 m + 48 m) = 64 m
| Velocity-time graph | Speed-time graph | |
|---|---|---|
| Gradient | Acceleration (m/s2) | Rate of Change of Speed (m/s2) |
| Area under graph | Displacement (m) | Distance (m) |
The graph above is for an object which is thrown vertically upwards with an initial velocity of 20 m/s.
In the time period 0-2 seconds – the object decelerates from 20 m/s to 0 m/s.
At 2 seconds it reaches the highest point.
In the time period 2-4 seconds – the object accelerates from 0 m/s top 20 m/s in the opposite direction.
It hits the ground and comes to rest at 4 seconds.
Can you identify the parts of a velocity-time graph?
Questions
Example
Describe this journey in as much detail as possible by answering the questions that follow.
Give values for velocities, acceleration and displacements.
Question
Describe the motion of the car during the first 20 s.
Answer
The car has an initial velocity of 12 m/s.
From 0 to 20 seconds it moves at a constant velocity of 12 m/s.
Question
What is the displacement of the car after 20 s?
Answer
In 20 seconds, the displacement of the car is equal to the area of the rectangle.
Displacement = area of rectangle = 20 s × 12 m/s = 240 m.
Question
Calculate the deceleration of the car from 20 s to 30 s.
Answer
At 30 seconds the car stops, its velocity is 0 m/s.
Acceleration of the car is the gradient of the graph =
\(\frac{\text{final velocity – initial velocity}}{\text{time taken}}\)
acceleration = (0 m/s – 12 m/s) ÷ 10 s = -1.2 m/s2
The car has an acceleration of -1.2 m/s2 (or a retardation of 1.2 m/s2).
Question
How far does the car move while decelerating?
Answer
The displacement while decelerating equals the area of triangle
displacement = \(\frac{\text{1}}{\text{2}}\) 10 s x 12 m/s = 60 m
The car moves 60 m while decelerating.
Question
What is the average velocity of the car during its entire journey?
Answer
The total displacement of the car = total area under the graph = 240 m + 60 m = 300 m.
The average velocity of the car over the whole journey = \(\frac{\text{total displacement}}{\text{time}}\)
average velocity = 300 m ÷ 30 s = 10 m/s.
The average velocity of the car is 10 m/s.
How much do you know about speed-time graphs?
More on Unit 1: Motion
Find out more by working through a topic
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