Velocity-time graphs of motion
Velocity-time graphs
Velocity-time graphs show how the velocity (or speed) of a moving object changes with time.
These graphs also show if the object is moving at a constant speed or accelerating, decelerating, or still.
What is shown on the velocity-time graph?
- Acceleration is shown by a straight rising line.
- Deceleration is shown by a straight falling line.
- Constant speed is shown by a horizontal line.
- A horizontal line along the \(\text{x}\)-axis shows the speed is zero, meaning that the vehicle has stopped, or is stationary.
The area under the graph is the distance moved.
The gradient or the steepness of the graph can be used to work out the acceleration. A steep line means high acceleration.
A line upwards curving to become steeper means that the vehicle is accelerating at an increasing rate – non-uniform acceleration.
A line upwards curving to become flat means that the vehicle is accelerating at a decreasing rate to constant speed.
Remember, acceleration is the same as gradient, and acceleration = velocity change ÷ time.
Look at the graph. The acceleration of the vehicle in the first 10 seconds is:
= (40 m/s – 0 m/s) ÷ 10 s
= 4 m/s2
The acceleration of the vehicle between 20 and 30 seconds is:
= (60 m/s – 40 m/s) ÷ 10 s
= 2 m/s2
The acceleration of the vehicle between 30 and 50 seconds is:
= (0 m/s – 60 m/s) ÷ 20 s
= -3 m/s2 (decelerating)
To find the distance travelled, look at the area under the graph.
In the first 10 seconds this is the area of the triangle, ½ × 10 s × 40 m/s = 200 m.
From 10 to 20 seconds this is the area of the rectangle, 10 s × 40 m/s = 400 m.
From 20 to 30 seconds this is the area of the rectangle and triangle, (10 s × 40 m/s) + (½ × 10 s × 20 m/s) = 500 m.
From 30 to 50 seconds, this is the area of the triangle, ½ × 20 s × 60 m/s = 600 m.
Total distance travelled = 200 m + 400 m + 500 m + 600 m = 1700 m.
Average speed = total distance ÷ total time = 1,700 m ÷ 50 s = 34 m/s.
Look at this velocity-time graph and answer the question. This question requires an extended response, ie your answer must be fairly long.
Question
Describe this journey in as much detail as possible. Give values for speeds, acceleration and distances travelled.
From 0 to 20 seconds the vehicle is at a constant speed of 12 m/s.
In this time it covers a distance of 20 s × 12 m/s = 240 m, the area of the rectangle.
From 20 to 30 seconds the car decelerates or slows to a stop at 30 seconds.
The deceleration = velocity change/time = (0 m/s – 12 m/s) ÷ 10 s = -1.2 m/s2
The distance travelled while decelerating = area of triangle = ½ × 10 s × 12 m/s = 60 m
The total distance travelled = 240 m + 60 m = 300 m
The mean speed of the vehicle over the whole journey is the total distance divided by the total time, ie
300 m ÷ 30 s = 10 m/s