The area is the displacement
A velocity–time graph gives you more than the acceleration. It also gives the displacement — as the area of the region under the line.
1. Uniform velocity — a rectangle
An object moving at a uniform 6 m s−1 for 8 s. The velocity does not change, so the graph is a straight line parallel to the x axis.
| t (s) | 0 | 8 |
|---|---|---|
| v (m s⁻¹) | 6 | 6 |
Two methods, one answer
By formula: displacement = velocity × time = 6 × 8 = 48 m
By graph: area of the rectangle = 6 × 8 = 48
The area is found by multiplying the length along the x axis (the time) by the height along the y axis (the velocity) — which is the formula doing the same work.
2. Uniform acceleration — a triangle
An object starting from rest and reaching 12 m s−1 in 4 s. Now the line slopes, so the region beneath it is a triangle.
| t (s) | 0 | 4 |
|---|---|---|
| v (m s⁻¹) | 0 | 12 |
Two methods, one answer
By formula: average velocity × time = (12 ÷ 2) × 4 = 24 m
By graph: area of the triangle = ½ × 4 × 12 = 24
The 12 ÷ 2 inside the triangle-area formula is the average velocity itself. So the two methods really are one calculation.
The general rule
An object's displacement is equal to the numerical value of the area enclosed by its velocity–time graph. It makes no difference whether that area is a rectangle or a triangle.