Reconstruct displacement and distance from velocity
A velocity–time graph answers two different questions. Its slope gives acceleration, with units (m/s)/s. Its signed area gives displacement, with units (m/s)×s. A point on the vertical axis is velocity itself, not displacement or acceleration. Before calculating, write the quantity and the units you need.
Learning outcome: Compare (v₀,a)=(3,−1), (4,−1), and (−3,1). Calculate each area before checking the table. Locate any interior reversal and explain the boundary case where velocity reaches zero exactly at 4 s. Record both displacement and distance.
Starting recap
Use signed quantities, algebraic substitution, graph slopes and areas. State the system, coordinate axes and units before calculating.
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Try
For v₀=3 m/s and a=−1 m/s², the graph crosses zero at 3 s. From 0 to 3 s the area is +4.5 m. From 3 to 4 s it is −0.5 m. Displacement is 4 m, distance is 5 m, average velocity is 1 m/s and average speed is 1.25 m/s.
Open the experimentPredict
For a graph with equal positive and negative triangular areas, can displacement vanish while distance remains positive?
Experiment
Compare (v₀,a)=(3,−1), (4,−1), and (−3,1). Calculate each area before checking the table. Locate any interior reversal and explain the boundary case where velocity reaches zero exactly at 4 s. Record both displacement and distance.
Reconstruct signed area and total distance
Worked teaching model · synthetic values · Synthetic linear velocity v=v₀+at over 0–4 s. Position starts at zero. Acceleration continues through v=0; no stop controller. Exact area uses any interior zero crossing.
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Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.
| Quantity | Value | Unit |
|---|---|---|
| Signed displacement | 4 | m |
| Total distance | 5 | m |
| Final velocity | -1 | m/s |
| Average velocity | 1 | m/s |
| Interior reversal time | 3 | s |
Accessible plot coordinates
| time (s) | velocity (m/s) |
|---|---|
| 0 | 3 |
| 0.25 | 2.75 |
| 0.5 | 2.5 |
| 0.75 | 2.25 |
| 1 | 2 |
| 1.25 | 1.75 |
| 1.5 | 1.5 |
| 1.75 | 1.25 |
| 2 | 1 |
| 2.25 | 0.75 |
| 2.5 | 0.5 |
| 2.75 | 0.25 |
| 3 | 0 |
| 3.25 | -0.25 |
| 3.5 | -0.5 |
| 3.75 | -0.75 |
| 4 | -1 |
The negative-velocity region contributes negative displacement but positive distance.
Area under velocity is signed. Average speed uses total distance/time, not the absolute value of average velocity when the motion reverses. A zero velocity at an endpoint is not an interior reversal.
The negative-velocity region contributes negative displacement but positive distance.
Area under velocity is signed. Average speed uses total distance/time, not the absolute value of average velocity when the motion reverses. A zero velocity at an endpoint is not an interior reversal.
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Why is |average velocity| insufficient to reconstruct distance? Include a counterexample with an actual direction change.
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