📈 Upward slope
The velocity is going up as time passes, so the object is accelerating. The go-kart accelerates for the first 4 s.
Movement and Position » Velocity-Time Graphs
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 1.7, 1.8
- How to plot a velocity-time graph from a table of data
- What a horizontal line, an upward slope, a downward slope and a steeper line each tell you
- How to determine acceleration from the gradient, including a negative gradient
- How the shapes differ from those on a distance-time graph
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Unlock This CourseWhat is a velocity-time graph?
A velocity-time graph shows how the velocity of an object changes during a journey. Time goes along the bottom (the horizontal axis) and velocity goes up the side (the vertical axis). Each point on the line answers one question: how fast was the object moving at this moment?
That is the big difference from the graphs in the lesson Distance-Time Graphs. The height of the line no longer tells you how far the object has gone. It tells you how fast it is going.
Key terms:
- Velocity-time graph: a graph with time on the horizontal axis and velocity on the vertical axis, showing how the velocity changes during a journey.
- Constant velocity: a steady velocity. The object is moving, but it is not speeding up or slowing down.
Plotting a graph from a table
A go-kart is driven along a straight track. A sensor records its velocity every 2 seconds.
| Time (s) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|---|---|---|---|
| Velocity (m/s) | 0 | 4 | 8 | 8 | 8 | 6 | 4 | 2 | 0 |
You plot it in the same way as any other graph.
- Draw the axes with a ruler and label them "Time (s)" along the bottom and "Velocity (m/s)" up the side.
- Choose even scales that let the points fill more than half of the grid. Here, 2 s for each large square along the bottom and 1 m/s for each large square up the side work well.
- Plot each pair of values as a small, neat cross. For the pair (2 s, 4 m/s), go along to 2 s, then up to 4 m/s.
- Join the points. These points lie on straight lines, so use a ruler.
The finished graph has three straight sections. The line slopes up from (0, 0) to (4, 8). It is then flat from (4, 8) to (8, 8). Finally it slopes down from (8, 8) to (16, 0), where it meets the time axis.
What the shape of the line tells you
Each section of the line shows a different kind of motion.
🚗 Horizontal line
The velocity is not changing, so the object is moving at a constant velocity. The go-kart moves at a steady 8 m/s from 4 s to 8 s.
📉 Downward slope
The velocity is going down as time passes, so the object is decelerating. The go-kart decelerates from 8 s to 16 s.
🛑 Line along the time axis
The velocity is 0 m/s, so the object is stationary. If the go-kart stays still after 16 s, the line carries on along the time axis.
The steepness matters too. A steeper line means the velocity changes by more in each second, so the acceleration is greater. Picture two go-karts that both start from rest. Kart A reaches 10 m/s after 5 s. Kart B takes 10 s to reach 10 m/s. Both lines end at the same height, but the line for kart A is twice as steep because kart A has twice the acceleration.
The same rule works for a downward slope. A steeper downward line shows a greater deceleration, such as braking hard.
Now you can explain a whole graph in words. For the first go-kart: it accelerates steadily from rest for 4 s, moves at a constant velocity of 8 m/s for 4 s, then decelerates steadily for 8 s until it stops.
Finding acceleration from the gradient
The gradient of a velocity-time graph is the acceleration. The rise of the line is a change in velocity and the run along the bottom is the time taken. Divide one by the other and you have the equation from the lesson Acceleration.
Acceleration from a straight section
acceleration = gradient = change in velocity ÷ time taken
Pick two points on the straight section, as far apart as you can. Subtract the velocities (later value minus earlier value), subtract the times, then divide. The unit is m/s2.
Worked example 1: the go-kart speeding up
Use the points (0 s, 0 m/s) and (4 s, 8 m/s) from the table.
change in velocity = 8 − 0 = 8 m/s
time taken = 4 − 0 = 4 s
acceleration = gradient = 8 ÷ 4 = 2 m/s2
Worked example 2: a negative gradient
Now use the points (8 s, 8 m/s) and (16 s, 0 m/s) for the last section.
change in velocity = 0 − 8 = −8 m/s
time taken = 16 − 8 = 8 s
acceleration = gradient = −8 ÷ 8 = −1 m/s2
The line slopes down, so the gradient is negative. A negative acceleration is a deceleration: the go-kart loses 1 m/s of velocity every second.
Worked example 3: a table of plotted values
A coach brakes on a straight road. These values are plotted and give one straight line.
| Time (s) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Velocity (m/s) | 20 | 16 | 12 | 8 | 4 |
Use the first and last points: (0 s, 20 m/s) and (4 s, 4 m/s).
change in velocity = 4 − 20 = −16 m/s
time taken = 4 − 0 = 4 s
acceleration = gradient = −16 ÷ 4 = −4 m/s2
Notice that this line does not start at the origin. The coach was already moving at 20 m/s when the timing began.
What about the flat section of the go-kart graph? The velocity does not change, so the change in velocity is 0 m/s and the gradient is zero. The acceleration is 0 m/s2, which is exactly what constant velocity means.
You will not always be given a table. You can read any two points off a straight line. If the line passes through (2 s, 5 m/s) and (10 s, 25 m/s), the change in velocity is 25 − 5 = 20 m/s and the time taken is 10 − 2 = 8 s. The acceleration is 20 ÷ 8 = 2.5 m/s2.
Do not mix up the two graphs
Velocity-time graphs and distance-time graphs can look the same, but the same shape means something different on each one. Always check the label on the vertical axis first.
| Shape of the line | On a distance-time graph | On a velocity-time graph |
|---|---|---|
| Horizontal line above the time axis | Stationary | Constant velocity |
| Straight line sloping upwards | Constant speed | Constant acceleration |
| Steeper upward line | Faster | Greater acceleration |
| The gradient gives the | Speed | Acceleration |
Common mistakes
- Saying a horizontal line means stationary. On a velocity-time graph a flat line above the time axis means constant velocity. The object is only stationary when the line is along the time axis, at 0 m/s.
- Thinking a steep line means fast. Steepness shows the acceleration. To find how fast the object is moving, read the height of the line on the velocity axis.
- Dividing one point's values. For the go-kart's last section, 8 ÷ 16 is wrong. Use the change in velocity and the time taken between two points.
- Losing the minus sign. Always do later velocity minus earlier velocity. A downward slope must give a negative gradient.
- Calling a downward slope "going backwards". The object is still moving forwards. It is just slowing down.
- Giving the wrong unit. The gradient here is an acceleration, so the unit is m/s2, not m/s.
Exam-style question
A tractor moves along a straight farm road. The table shows its velocity at different times.
| Time (s) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| Velocity (m/s) | 0 | 4 | 8 | 8 | 8 | 4 | 0 |
A student plots a velocity-time graph using this data and joins the points with straight lines.
(a) Describe the motion of the tractor between 10 s and 20 s. [1 mark]
(b) Use the gradient of the graph to determine the acceleration of the tractor during the first 10 s. [3 marks]
(c) Without doing a calculation, explain how the graph shows that the deceleration in the last 10 s is the same size as the acceleration in the first 10 s. [2 marks]
(d) The tractor then stays still for 5 s. Describe the line the student should add to the graph. [1 mark]
Model answer
(a) The tractor moves at a constant velocity (of 8 m/s). (1)
(b) change in velocity = 8 − 0 = 8 m/s and time taken = 10 − 0 = 10 s (1)
acceleration = gradient = 8 ÷ 10 (1)
acceleration = 0.8 m/s2 (1)
(c) The two sloping lines are equally steep. (1) The first slopes upwards and the last slopes downwards, so the gradients are the same size but one is positive and one is negative. (1)
(d) A horizontal line along the time axis (at 0 m/s) from 30 s to 35 s. (1)
Exam tip
In part (a), "stationary" is the trap answer. The line is flat, but it is flat at 8 m/s, so the tractor is still moving. In part (b), the word "gradient" in the question means you must show the change in velocity and the time taken from the graph, not just write down an answer.