Areas in Velocity-Time Graphs

Imagine an object moving at a constant velocity of 4 m/s for 3 s. It travels 4 m each second, so it covers $4 \times 3 = 12\text{ m}$. On its velocity–time graph, the line is horizontal at 4 m/s. The area beneath it is a rectangle with a height of 4 m/s and a width of 3 s: the same multiplication gives 12 m.

The area therefore has a physical meaning, not just a geometrical one. Its units are $(\text{m/s})\times\text{s}=\text{m}$.

Strictly, the signed area gives displacement: the change in position, including direction. When velocity stays positive, the object does not reverse direction, so this area also gives the total distance travelled.

Splitting the area into simple shapes

Changing velocity does not prevent us from finding distance using area. A straight sloping line represents a uniform change in velocity, and the region beneath it can be divided into triangles and rectangles.

For a rectangle, use:

For a triangle, use:

The base is a time interval, measured along the horizontal axis. The height is read from the velocity axis. A triangular region occurs when velocity rises uniformly from zero or falls uniformly to zero.

If a sloping section starts and ends above zero velocity, the region beneath it is a trapezium. You can split it into a rectangle beneath the lower velocity and a triangle above that rectangle. The triangle’s height is the difference between the two velocities.

Finding the distance for a whole journey

Suppose an object starts from rest, reaches 6 m/s uniformly after 4 s, continues at 6 m/s until 9 s, then slows uniformly to rest at 12 s.

Data for Distance from a velocity–time graph

Divide the area at 4 s and 9 s, where the slope changes. This gives a triangle, a rectangle and another triangle.

• First triangle, 0–4 s: $\frac{1}{2}\times4\times6=12\text{ m}$.
• Rectangle, 4–9 s: the time interval is $9-4=5\text{ s}$, so its area is $5\times6=30\text{ m}$.
• Final triangle, 9–12 s: the time interval is $12-9=3\text{ s}$, so its area is $\frac{1}{2}\times3\times6=9\text{ m}$.

Adding these areas gives the total distance:

All three regions lie above the time axis, so the object travels 51 m in the positive direction without reversing.

What if the graph goes below the time axis?

Negative velocity means motion in the opposite direction. A region below the time axis counts as negative displacement, but it still represents distance travelled.

For total distance, split the graph wherever it crosses the time axis and add the sizes of all the areas. For displacement, add the areas with their signs.

For example, suppose the area above the axis represents 8 m forwards and the area below it represents 3 m backwards. The total distance is $8+3=11\text{ m}$, but the displacement is $8-3=5\text{ m}$ forwards. Moving back towards the starting point reduces displacement; it does not undo the distance already travelled.