Displacement is how far an object is from its start in a stated direction. Velocity is how fast the displacement changes, and acceleration is how fast the velocity changes. Motion graphs show all three, and each one is read by either gradient or area.
This lesson is part of SPM Physics force and motion I. The next lesson, using equations of uniformly accelerated motion, turns these readings into calculations.
Which graph gives which quantity?
Match the quantity to the graph before you calculate anything.
| Graph | Gradient gives | Area gives |
|---|---|---|
| Displacement-time | velocity | no standard meaning |
| Velocity-time | acceleration | displacement |
| Acceleration-time | no standard meaning | change in velocity |
If you read acceleration from a displacement-time graph, the answer will be a velocity. Name the axes first, then decide.
Worked example: one journey, three readings
A car moves in a straight line. Its velocity rises from 0 to 12 m s⁻¹ in 4 s, stays at 12 m s⁻¹ for 6 s, then falls to 0 in 3 s.
- First 4 s: gradient = (12 − 0) ÷ 4 = 3 m s⁻².
- Steady stage: the gradient is 0, so the acceleration is 0.
- Last 3 s: gradient = (0 − 12) ÷ 3 = −4 m s⁻², a deceleration of 4 m s⁻².
- Displacement: add the three areas under the graph, ½ × 4 × 12 = 24 m, then 12 × 6 = 72 m, then ½ × 3 × 12 = 18 m. The total is 24 + 72 + 18 = 114 m.
The velocity stayed positive throughout, so distance and displacement are equal here. The car never reversed.
When are distance and displacement different?
They differ whenever the direction changes. A runner goes 30 m east along a track and then 10 m back west.
The distance is 30 + 10 = 40 m. The displacement is 30 − 10 = 20 m east. Speed and velocity follow the same split: average speed uses distance and average velocity uses displacement.
The mistake to avoid
The common mistake is to treat the steepness of any graph as acceleration. The table shows what each slope means.
| Graph drawn | Student says | Correct reading |
|---|---|---|
| Displacement-time, straight slope | constant acceleration | constant velocity |
| Velocity-time, horizontal line | not moving | moving at constant velocity |
| Velocity-time, line below axis | slowing down | moving in the negative direction |
Check what the vertical axis measures before you take a gradient. The motion graph explorer lets you change one graph and watch the other two respond.
Check yourself
A cyclist’s velocity rises from 2 m s⁻¹ to 10 m s⁻¹ in 4 s, then stays at 10 m s⁻¹ for 5 s. Find the acceleration in the first 4 s and the total displacement.
Answer
Acceleration = (10 − 2) ÷ 4 = 2 m s⁻².
Displacement: the first stage is a trapezium, ½ × (2 + 10) × 4 = 24 m. The second stage is a rectangle, 10 × 5 = 50 m. Total = 24 + 50 = 74 m.
What to study next
Continue with using equations of uniformly accelerated motion to get the same results from algebra. Check your units with the units and significant figure checker, and record graph errors in the mistake log and paper-error review.
If you want a teacher to go through graph questions with you, see online one-to-one Physics tuition.