Area Under a Velocity–Time Graph = Displacement

Guiding Question

A velocity–time graph never shows where an object is. So how can you tell how far it moved, and in which direction, just by looking at one?

Pick a motion (or drag the points to build your own v–t graph), then slide the time t or press Play. Watch the shaded area under the velocity curve grow, and compare it to the change in position on the motion diagram and the position–time graph.

Why area = displacement

For constant velocity, displacement is velocity × time: Δx = v · Δt On a v–t graph, that product is the area of a rectangle whose height is v and whose width is Δt.

Check the units: (m/s) × (s) = m Area on a v–t graph is measured in meters, the unit of displacement. When the velocity changes, split the time into thin slices; each slice is nearly a rectangle, and adding them up gives the total area. Turn on Show rectangles below to see this.

  • Area above the t-axis (v > 0) counts as positive displacement (+x direction).
  • Area below the t-axis (v < 0) counts as negative displacement (−x direction).
  • The net (signed) area is the displacement Δx. Adding the areas without signs gives the distance traveled.
  • Area gives the change in position, not the position itself: x(t) = x₀ + Δx.
Velocity vs. time → displacement each second
t = 0.00 s
Each 1-second slice of area has its own shade. The motion map underneath shows the displacement during each second as its own arrow, laid tip to tail. The highlighted slice funnels down onto its displacement segment. Hover over or tap a slice, a row, or a chip to highlight that second; tap again to release. Drag the dots to reshape the graph (each grid box = 1 m).
positive area → +x displacement negative area → −x displacement velocity v(t)
Area above axis
v > 0 from 0 to t
moved in +x direction
Area below axis
v < 0 from 0 to t
moved in −x direction
Net area = displacement
Δx = (area above) + (area below)
Total area = distance
d = |area above| + |area below|
no direction
Position at t
x(t) = x₀ + Δx
Position vs. time
t = 0.00 s
The small arrows show how much the position rises (or falls) during each second, the same amounts as the area slices. The long arrow at time t shows x(t) − x₀, the net area from 0 to t.

Check your understanding

Find each answer by counting grid boxes or using rectangle and triangle area formulas, then check with the simulation.

1. “Speed up, cruise, slow down”: what is the displacement from 0 to 10 s?

Triangle (0–3 s): ½ × 3 s × 4 m/s = 6 m. Rectangle (3–7 s): 4 s × 4 m/s = 16 m. Triangle (7–10 s): 6 m. Total = +28 m.

2. “Ball rolled up a ramp”: what are the displacement and distance from 0 to 10 s?

Area above = ½ × 5 s × 6 m/s = +15 m; area below = −15 m. Displacement = 0 m (it returns to its start), but distance = 30 m.

3. “Walk out and back”: find the area above, the area below, the displacement, and the distance for 0–10 s.

Above: 0.75 + 2.25 + 9 + 1.5 = +13.5 m. Below: −1 − 4 − 1 = −6 m. Displacement = +7.5 m; distance = 19.5 m.

4. In question 3, if the walker starts at x₀ = +2 m, where do they end up?

x = x₀ + Δx = 2 m + 7.5 m = 9.5 m. Set x₀ to 2 m to check. The area didn’t change, because area only tells you the change in position.

5. Why does area below the t-axis count as negative?

Below the axis the velocity is negative, so each slice v·Δt is negative: the object is moving in the −x direction, reducing its position.

6. Turn on “Show rectangles” for “Car speeding up from rest.” Why is the estimate too small, and what happens as Δt shrinks?

Each rectangle uses the velocity at its left edge, which is the slowest velocity in that slice, so it misses a small triangle every time. As Δt gets smaller, the missing triangles shrink and the estimate approaches the exact area.

7. “Ball rolled up a ramp”: find the area of the 0–1 s slice and of the 9–10 s slice. What do they tell you about the motion?

0–1 s: the slice is a trapezoid, ½ × (6 + 4.8) m/s × 1 s = +5.4 m. 9–10 s: ½ × (−4.8 + −6) m/s × 1 s = −5.4 m. The ball covers the most ground in the seconds when it is fastest, in opposite directions at the start and end. Compare the lengths of those two arrows on the motion map.

8. Build your own: drag the points to make a graph whose displacement from 0 to 10 s is zero, even though the object moves the whole time.

One answer: set the dots at t = 0–4 s to +2 m/s, the dot at 5 s to 0, and the dots at 6–10 s to −2 m/s. Area above = 8 + 1 = +9 m, area below = −9 m, so Δx = 0 m while the distance is 18 m. Any shape works as long as the area above the axis equals the area below.