Position–time graphs: slope & curvature
Read velocity from slope and acceleration from changing slope.
Turn a short walk into a graph.
Need an earlier step? Start with position, distance & displacement →
At 0 s you are 2 m right of a door. After 1 s you are at 4 m; after 2 s you are at 6 m. Assume you walk steadily. Each extra second adds 2 m to your position.
Words and symbols you will use
- Horizontal axis t
- Time, in seconds. Moving right on the graph means looking later.
- Vertical axis x
- Position, in meters. Graph height tells you where the object is.
- Slope Δx/Δt
- Position change divided by time change: velocity.
Read the picture, one step at a time.
- Read one point: at t = 1 s, x = 4 m.
- Read a change: from 0 to 2 s, position rises from 2 to 6 m.
- Divide the rise by the run: (6 − 2) m ÷ 2 s = +2 m/s.
| Compare | What it means | What follows |
|---|---|---|
| Height | Where is the object? | At 1 s: x = 4 m. |
| Slope | How is position changing? | v = +2 m/s throughout. |
| Horizontal segment | Does position change? | No: velocity is zero. |
The idea to keep: This is a graph of position against time, not a picture of a hill. The cart can move along a flat floor while its graph rises.
How do you find velocity from a position–time graph?
Use its slope, not its height. A rising graph means positive velocity; a falling graph means negative velocity.
The vertical coordinate tells you position. The slope tells you how position changes with time. A line above zero can still slope downward: the object is on the positive side of the origin while moving in the negative direction.
A straight segment has constant velocity. A curved graph has changing slope and therefore changing velocity. Concave-up curvature means velocity is increasing algebraically, so acceleration is positive; this can mean slowing down if the velocity is negative.
A reliable approach
- Read axis labels and units.
- Compute Δx/Δt between two points for a straight segment.
- For a curve, distinguish the tangent at one time from the secant over an interval.
Work through one example.
A straight x–t line connects (1 s, 8 m) to (5 s, 0 m). Find velocity.
Follow the worked solution
- Slope = (0 − 8)/(5 − 1) = −2 m/s.
- The velocity is constant and negative.
- The object starts on the positive side of the origin and reaches x = 0 at 5 s.
A position–time graph is not a picture of the path. A rising curve does not mean the object climbs a hill.
Explain it without notes: Explain why a position-time curve is not a drawing of a hill.
Try explaining it now.
A position graph is above zero and slopes down. Which way is the object moving?
Compare with an explanation
In the negative direction. Graph height tells position; the negative slope tells velocity.
Another explanation, if you need oneOptional external lesson · Flipping Physics
Connect graph slope to real movement
This lesson is complete without a video. For another teacher’s explanation, open the original resource below. Pause after a diagram and explain the idea in your own words.
Suggested section 3:04–6:25. Video by Flipping Physics / Jonathan Thomas-Palmer. Refresh Kid is not affiliated with or endorsed by Flipping Physics. These links open another website. Some videos use g = 9.81 m/s²; our examples state g = 10 m/s². Follow the value given in each problem.
After watching: Explain why a position-time curve is not a drawing of a hill.
Make a prediction. Test it.
Change x₀ while keeping v₀ and a fixed. Watch the x–t graph shift vertically while v–t and a–t remain unchanged.
Motion graphs · same clock, different quantities
Try two questions.
Two original Refresh Kid questions. Choose an answer, explain it to yourself, then check the reasoning. These are not released AP exam questions.
Show your reasoning.
Original mini-FRQ · 4 points · Self-checkObject A follows x_A = 2t and B follows x_B = 12 − t, with x in meters and t in seconds.
- Determine each velocity.
- Find their meeting time.
- Find their meeting position.
- Explain whether their velocities match there.
Your response stays on this page and is not submitted or automatically graded. Copy it before leaving.
Compare with the worked solution & scoring guide
- 1 point: v_A = +2 m/s; v_B = −1 m/s.
- 1 point: 2t = 12 − t gives t = 4 s.
- 1 point: x = 8 m.
- 1 point: The lines intersect but their slopes remain different, so their velocities do not match.
Accept an equivalent correct method. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Does x = 0 imply rest?
No. Rest concerns slope, not position.
RECALL 2What does an x–t intersection mean?
The objects occupy the same position at the same time.
RECALL 3Does an intersection imply equal velocities?
No. Compare the slopes separately.
Come back tomorrow: answer these with the cards closed. Try again a week later, especially the ones you missed.
Keep the key ideas handy.
Position–time graphs: slope & curvature
Core idea: Use its slope, not its height. A rising graph means positive velocity; a falling graph means negative velocity.
- Slope of x–t → velocity.
- Horizontal x–t segment → rest.
Avoid this: A position–time graph is not a picture of the path. A rising curve does not mean the object climbs a hill.
Remember why: This is a graph of position against time, not a picture of a hill. The cart can move along a flat floor while its graph rises.
| Graph | Height tells you | Slope tells you | Signed area gives |
|---|---|---|---|
| x–t | Position | Velocity | No standard Unit 1 motion quantity |
| v–t | Velocity | Acceleration | Displacement |
| a–t | Acceleration | Not needed here | Velocity change |
Read the graph axes before interpreting height, slope or area.
Explain, don’t just substituteExplain why a position-time curve is not a drawing of a hill.
Refresh Kid · AP Physics 1 · Unit 1 · 1.3.A · Check units, direction and model assumptions.
Connect to released AP practice.
2026 · Question 1 · Version J
Use Part A(i) for component velocity graphs and Part A(ii) for a kinematics derivation. The remaining parts use fluid concepts from Unit 8; they are not Unit 1-only practice.
This is a Unit 1 synthesis task to revisit after learning projectile motion. Historical papers may use different timing or course coverage. Official questions remain on College Board’s site; our questions below are original practice.
Browse released years and scoring information ↗Framework alignment: College Board CED, Topic 1.3. Current exam corrections. Checked September 16, 2026. These explanations and practice items are independently authored by Refresh Kid. The simulation is a mathematical model, not experimental data.
About the videos and learning approach
Optional videos are linked to the publisher’s website and YouTube channel with credit. No external video player is loaded on this lesson page. The written lessons, simulations and practice here are independently authored by Refresh Kid.
We combine worked examples, visual models, explanation and recall practice. See the Institute of Education Sciences study guide ↗ for the underlying learning recommendations.
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