Kinematics: motion graphs and constant acceleration
Kinematics describes motion without asking what causes it. With constant acceleration, five quantities are linked by five equations, and graphs give a second way into the same problems.
Part 1 of 3: Learn it
In short
- Gradient of a velocity-time graph = acceleration; area under it = displacement.
- The suvat equations only apply when acceleration is constant.
- For vertical motion, take a = −9.8 m/s² if up is positive.
Where this is in your specification
Spec points: DfE P1 and Q1 to Q3 (AQA 7357 P1 and Q1 to Q3, Edexcel 9MA0 Mechanics topics 6 and 7, OCR A H240 3.01 and 3.02)
| Board | Topic: Kinematics with constant acceleration |
|---|---|
| DfE content | P1, Q1-Q3, Q5 (vertical motion) |
| AQA 7357 | P1, Q1-Q3 |
| Edexcel 9MA0 | Mechanics topics 6 and 7 |
| OCR H240 | 3.01, 3.02 |
Quantities and units
| Symbol | Quantity | Unit |
|---|---|---|
| s | displacement | m |
| u | initial velocity | m/s |
| v | final velocity | m/s |
| a | acceleration | m/s² |
| t | time | s |
Displacement, velocity and acceleration are vectors, so choose a positive direction at the start and stick to it. Convert km/h to m/s by dividing by 3.6.
The suvat equations
Write down the three quantities you know and the one you want, then pick the equation that contains exactly those four.
Motion graphs
On a displacement-time graph the gradient is velocity. On a velocity-time graph the gradient is acceleration and the area between the line and the time axis is displacement; area below the axis counts as negative displacement. Straight-line sections split the area into triangles, rectangles and trapeziums.
Vertical motion
An object moving freely up or down has constant acceleration g = 9.8 m/s² downwards (ignoring air resistance). At the highest point its velocity is 0 for an instant. The model treats the object as a particle, so its size and spin are ignored.
Convert 72 km/h to m/s.
Show the answer
72 ÷ 3.6 = 20 m/s.
Part 2 of 3: See it worked
Worked examples
Example 1
A car speeds up from 4 m/s to 22 m/s in 6 s with constant acceleration. Find the acceleration and the distance travelled.
- a = (v − u) ÷ t = (22 − 4) ÷ 6 = 3 m/s²
- s = ½(u + v)t = ½ × 26 × 6
Answer: 3 m/s² and 78 m.
Example 2
A ball is thrown straight up at 14.7 m/s. Find its greatest height and the time to reach it. (g = 9.8 m/s²)
- Up is positive: u = 14.7, v = 0, a = −9.8
- v² = u² + 2as: 0 = 216.09 − 19.6s, so s = 11.025
- v = u + at: 0 = 14.7 − 9.8t, so t = 1.5
Answer: 11.0 m (3 s.f.), reached after 1.5 s.
Common mistakes
- Using suvat when the acceleration is not constant.
- Giving g as positive in one part of a vertical problem and negative in another.
- Leaving speeds in km/h.
- Reading the area under a displacement-time graph as distance. Area only means displacement on a velocity-time graph.
What does the gradient of a velocity-time graph represent?
Show the answer
Acceleration.
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
Convert 72 km/h to m/s.
72 ÷ 3.6 = 20 m/s.
What does the gradient of a velocity-time graph represent?
Acceleration.
A particle starts from rest and accelerates at 2 m/s² for 5 s. How far does it travel?
s = ut + ½at² = 0 + ½ × 2 × 25 = 25 m.
Jobs that use this
- Aerospace engineer (opens a new tab)
- Mechanical engineer (opens a new tab)
- Civil engineer (opens a new tab)
Each link opens the job profile on the National Careers Service (England). In the rest of the UK: My World of Work (Scotland), Careers Wales, nidirect careers (Northern Ireland).
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