Coaching Centre
Motion graphs
Drawing and interpreting position–time and velocity–time graphs, gradient as velocity and acceleration, area under a velocity–time graph as displacement, and describing a journey from its graphs.
Engineers testing a new train, coaches studying a sprinter's start and police reconstructing a crash all do the same thing first: they draw a graph. A single line on a motion graph tells you where something was, how fast it was going and how quickly it sped up or slowed down, at every instant.
Position–time graphs
A position–time graph (also called a displacement–time or distance–time graph) plots position on the vertical axis against time on the horizontal axis.
- The gradient (rise ÷ run) of a position–time graph is the velocity: metres ÷ seconds.
- A flat line means the object is stationary (velocity zero).
- A straight sloping line means constant velocity. The steeper the line, the faster.
- A line sloping down has a negative gradient: the object is moving back towards the reference point.
- A curve means the velocity is changing: getting steeper means speeding up; flattening out means slowing down.
A section of a position–time graph rises from 2 m to 23 m between t = 1 s and t = 4 s. Gradient = (23 − 2) ÷ (4 − 1) = 21 ÷ 3 = +7 m/s.
Another section falls from 23 m to 3 m between t = 6 s and t = 11 s. Gradient = (3 − 23) ÷ 5 = −4 m/s.
If positive means south, the object first moves south at 7 m/s, then north at 4 m/s.
Velocity–time graphs
A velocity–time graph (or speed–time graph if direction is ignored) plots velocity against time. Be careful: it looks similar to a position–time graph but means something different.
- A flat line means constant velocity (not stationary, unless it is on the time axis).
- The gradient is the acceleration.
- The area between the line and the time axis is the displacement (or distance travelled).
Area under a velocity–time graph
For constant velocity, distance = velocity × time, which is the area of a rectangle under the graph. This works for any shape of graph: split the area into rectangles and triangles (or trapeziums) and add them up.
From 0 to 10 s: a = 20 ÷ 10 = 2 m/s². Distance = area of triangle = ½ × 10 × 20 = 100 m.
From 10 to 70 s: a = 0. Distance = rectangle = 60 × 20 = 1200 m.
From 70 to 100 s: a = −20 ÷ 30 = −0.67 m/s². Distance = ½ × 30 × 20 = 300 m.
Total distance = 100 + 1200 + 300 = 1600 m (1.6 km). Average speed = 1600 ÷ 100 = 16 m/s.
If a bike slows steadily from 8 m/s to 4 m/s in 3 s, the area is a trapezium: ½ × (8 + 4) × 3 = 18 m.
- In Challenge 1 the target rises from 0 m to 3 m in the first 3 s. Set the first velocity to 3 ÷ 3 = +1 m/s. Work out the other three sections, then press Walk.
- Look at the velocity–time graph you made. What does a flat section of the position graph look like on it? What about a section sloping down?
- Try Challenge 2, which starts at the far end of the track. Why is the first velocity negative?
- In Challenge 3, use the velocity–time graph to check the total displacement: add up velocity × 3 s for each section. Does it match where the walker finishes?
Test yourself
On a position–time graph, a horizontal line means the object is
The position isn't changing, so the velocity (gradient) is zero.
On a velocity–time graph, a horizontal line above the time axis means the object is
The velocity stays the same, so the acceleration (gradient) is zero.
A position–time graph is a straight line from (0 s, 0 m) to (5 s, 30 m). The velocity is
Gradient = 30 ÷ 5 = 6 m/s.
A velocity–time graph rises in a straight line from 0 to 12 m/s in 4 s. The distance travelled in that time is
Area of triangle = ½ × 4 × 12 = 24 m.
The gradient of the same graph (0 to 12 m/s in 4 s) is
Gradient = acceleration = 12 ÷ 4 = 3 m/s².
Label each column with the motion it shows.
Which graph does each statement describe?
The gradient of a position–time graph is the . The gradient of a velocity–time graph is the . The under a velocity–time graph is the displacement. A horizontal line on a position–time graph shows the object is .
A car starts from rest and accelerates steadily for 10 s to reach 20 m/s. It cruises at 20 m/s for 60 s, then slows steadily to a stop over 30 s. (a) Sketch the velocity–time graph, labelling key values. (b) Calculate the acceleration in each stage. (c) Calculate the total distance in metres and kilometres. (d) Describe the shape of the matching position–time graph.
(a) The velocity–time graph rises in a straight line from (0 s, 0 m/s) to (10 s, 20 m/s), is horizontal at 20 m/s until 70 s, then falls in a straight line to (100 s, 0 m/s). (b) Stage 1: a = (20 − 0) ÷ 10 = 2 m/s². Stage 2: velocity is constant, so a = 0. Stage 3: a = (0 − 20) ÷ 30 = −0.67 m/s² (a deceleration of 0.67 m/s²). (c) Distance = area under the graph: triangle ½ × 10 × 20 = 100 m, rectangle 60 × 20 = 1200 m, triangle ½ × 30 × 20 = 300 m. Total = 1600 m = 1.6 km. (d) On a position–time graph, the first 10 s is an upward curve that gets steeper as the car speeds up; from 10 s to 70 s it is a straight sloping line because the velocity is constant; from 70 s to 100 s it curves over, getting less steep, and becomes horizontal at 100 s when the car stops at 1600 m.