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Distance, velocity and acceleration
Change a vehicle’s motion and compare two graphs of the same journey. Follow the quantities on the axes carefully.
Change the model
Change a value, then use Update model or leave the field to redraw. The table gives the same quantities as the graph. Your entries are not saved to progress; changing the model resets the check question.
Try these comparisons
- Set acceleration to zero and choose a non-zero initial velocity. Compare the shapes of the two graphs.
- Start from rest with acceleration 2 m/s² for 5 s. Calculate final velocity and distance before reading the table.
- Keep initial velocity at zero and double the duration. Explain why distance becomes four times as large.
Read distance against time
The distance–time graph shows how far the vehicle has travelled. Its gradient gives speed. With a constant forward velocity, distance increases by the same amount each second and the line is straight.
When positive acceleration increases the velocity, distance grows more in each successive second. The distance–time curve becomes steeper. The curve’s height gives distance, not speed.
Read velocity against time
The velocity–time graph shows speed in a stated direction. Its gradient gives acceleration. A straight sloping line represents constant acceleration; a horizontal line represents constant velocity.
Area under a velocity–time graph gives change in position. In this model the vehicle always moves forwards, so this also equals distance travelled. A model with reversals would need more care: negative areas would subtract from change in position.
Connect a calculation to the graph
For constant acceleration, mean velocity is (initial velocity + final velocity) ÷ 2. Multiply by elapsed time to find distance. On the graph this is the area of a trapezium.
For a vehicle starting from rest, the area is a triangle. For constant velocity it is a rectangle. If both initial velocity and acceleration are zero, the vehicle stays still and both graphs lie on zero. Graph scales update when you change the controls.
Worked example
Starting at 2 m/s with acceleration 1 m/s² for 4 s gives final velocity 6 m/s. The velocity–time area is (2 + 6) ÷ 2 × 4 = 16 m. The velocity–time gradient is (6 − 2) ÷ 4 = 1 m/s².
What the model assumes
A one-dimensional journey with constant acceleration and no reversal. Initial position is zero. These controls specify the motion; they do not model a particular vehicle, force or source of energy.