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Distance, speed and acceleration – WJECVelocity-time graphs of motion

The motion of an object can be predicted by analysing the forces that act on the object. Balanced forces have no effect, while unbalanced forces can lead to change in acceleration.

Part of Physics (Single Science)Forces, space and radioactivity

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.
A velocity-time graph. The line is split into 4 parts. Part 1 shows uniform acceleration. Part 2 shows steady speed. Part 3 shows uniform acceleration. Part 4 shows deceleration.

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.

A velocity-time graph. The line is split into 4 parts. Part 1 is labelled accelerating. Part 2 is labelled constant speed/velocity. Part 3 is labelled accelerating. Part 4 is labelled decelerating.

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.

A velocity-time graph. The area under the graph line is split into 4 parts and shaded.

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.

A velocity-time graph. The line is split into 2 parts. It starts at time 0 and velocity 12 and continues at the same velocity for 20 seconds. From 20 to 30 seconds the velocity decreases from 12 to 0.