Velocity & acceleration graphs: slope and area
Connect signed graph areas to changes in physical quantities.
Three seconds at a steady velocity.
Need an earlier step? Start with position–time graphs →
A cart moves right at +2 m/s for 3 s. Each second adds 2 m to its position. The total displacement is +6 m. On a velocity–time graph, the same multiplication is the area of a rectangle.
Words and symbols you will use
- Graph height
- The value written on the vertical axis at a selected time.
- Signed area
- Area above the time axis counts positive; below counts negative.
- Slope
- How much the vertical quantity changes per second.
Read the picture, one step at a time.
- The graph height is 2 m/s and the interval width is 3 s.
- Rectangle area is (2 m/s)(3 s) = 6 m: displacement, not velocity.
- The line is horizontal, so its slope is zero. The acceleration is zero even though the cart moves.
| Compare | What it means | What follows |
|---|---|---|
| Position–time | Height: position; slope: velocity | Area is not displacement. |
| Velocity–time | Height: velocity; slope: acceleration | Signed area: displacement. |
| Acceleration–time | Height: acceleration | Signed area: velocity change. |
The idea to keep: Why area? A constant-velocity strip contributes velocity × time. Adding strips adds displacements. If velocity is negative, that strip’s displacement is negative too.
What does the area under a velocity–time graph mean?
Signed area gives displacement. Add absolute areas instead to find distance when velocity changes sign.
The slope of a velocity–time graph gives acceleration. A negative velocity with a positive slope describes an object moving in the negative direction while slowing. The vertical value of an acceleration–time graph is acceleration itself, not velocity.
Over an interval, signed area between a–t and the time axis gives the change in velocity. You still need an initial velocity to find a final velocity. Areas below the axis subtract. In algebra-based work, use rectangles, triangles and trapezoids when the graph permits.
For piecewise motion, analyze each time interval separately. Velocity can stay continuous while its slope changes at an idealized corner. Split signed areas at every zero crossing, and connect each interval back to the slope and curvature of the position graph.
A reliable approach
- Identify whether the graph is x–t, v–t or a–t before choosing slope or area.
- Split regions where the curve crosses the time axis.
- Use units to check: (m/s)·s = m and (m/s²)·s = m/s.
Work through one example.
Velocity decreases linearly from +4 m/s at 0 s to −4 m/s at 4 s. Find displacement and distance.
Follow the worked solution
- It crosses zero at 2 s. Each triangle has area ½(2 s)(4 m/s) = 4 m.
- Signed area = +4 − 4 = 0 m displacement.
- Absolute area = 4 + 4 = 8 m distance. Acceleration is −2 m/s².
Area under a–t gives Δv, not v. Area under x–t is not displacement.
Explain it without notes: Explain what extra information acceleration-time area needs to give final velocity.
Try explaining it now.
Velocity is −3 m/s for 2 s. Is the area a negative distance?
Compare with an explanation
The signed area is −6 m displacement. Distance is 6 m; it cannot be negative.
Another explanation, if you need oneOptional external lesson · Flipping Physics
Motion graphs: worked reasoning
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 0:00–6:54. 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 what extra information acceleration-time area needs to give final velocity.
Make a prediction. Test it.
Set v₀ = +4 m/s, a = −2 m/s², and t = 4 s. Compare zero net displacement with nonzero distance; identify the two cancelling areas.
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-checkAn ideal v–t graph joins (0 s, 0 m/s), (2 s, 6 m/s), (4 s, 6 m/s), and (6 s, −6 m/s) with straight segments.
- Find acceleration on each open interval.
- Find net displacement from 0 to 6 s.
- Find distance traveled.
- Sketch x–t starting from x = 0, showing the reversal and explaining the slopes.
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: Accelerations are +3 m/s², 0 m/s², and −6 m/s². At ideal sharp joins, a single instantaneous acceleration is not specified.
- 1 point: Signed areas: 6 + 12 + 0 = 18 m.
- 1 point: Absolute areas: 6 + 12 + 3 + 3 = 24 m; v crosses zero at 5 s.
- 1 point: x–t is concave up from 0–2 s, straight with positive slope from 2–4 s, then concave down. It peaks at x = 21 m at 5 s and ends at x = 18 m at 6 s.
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 1What does v–t crossing zero usually indicate?
A reversal if the velocity changes sign.
RECALL 2How do you get distance from v–t?
Add the magnitudes of all signed-area regions.
RECALL 3What is needed besides a–t area to get v_f?
The initial velocity: v_f = v_i + Δv.
Come back tomorrow: answer these with the cards closed. Try again a week later, especially the ones you missed.
Keep the key ideas handy.
Velocity & acceleration graphs: slope and area
Core idea: Signed area gives displacement. Add absolute areas instead to find distance when velocity changes sign.
- Slope of v–t → acceleration.
- Signed area under v–t → Δx.
- Signed area under a–t → Δv.
Avoid this: Area under a–t gives Δv, not v. Area under x–t is not displacement.
Remember why: Why area? A constant-velocity strip contributes velocity × time. Adding strips adds displacements. If velocity is negative, that strip’s displacement is negative too.
| 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 |
Split at zero crossings before adding signed areas or absolute areas.
Explain, don’t just substituteExplain what extra information acceleration-time area needs to give final velocity.
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 selection directly connects to this lesson. 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.
Need 1:1 help from a live tutor?
Bring your question about velocity & acceleration graphs: slope and area. Ask a Refresh Kid tutor to work through the idea and your next problem with you.