What Is the Area Under a Displacement–Time Graph?

If the vertical axis of a graph is a position coordinate (often called displacement in a displacement–time graph) and the horizontal axis is time, the area under the curve is found by integrating the position coordinate over time. This quantity is not the object’s net displacement or distance travelled. Its units are length units multiplied by time units, such as metre-seconds ( $\text{m}\cdot\text{s}$ ).

The distinction matters because displacement-time and velocity-time graphs encode motion differently. On a displacement-time graph, the slope represents velocity. On a velocity-time graph, the signed area represents displacement.

What the area means

Let $x(t)$ be an object’s position coordinate relative to a chosen origin at time $t$ . The signed area between its position-time graph and the time axis from $t=a$ to $t=b$ is

\int_a^b x(t)\,dt.

If the position coordinate is measured in metres and time in seconds, the integral has units of metres-seconds. This is a mathematically valid accumulated quantity, but it is not one of the usual quantities used to describe how far an object moves. In particular, it does not tell you the net displacement between the endpoints or the total distance travelled.

The word “signed” means that portions of the graph above the time axis contribute positively, while portions below it contribute negatively. If a problem asks for the total geometric area instead, add the absolute areas of the regions above and below the axis. That is different from the signed integral when the graph crosses the axis.

How to calculate it

Identify the time interval and read the position coordinate from the vertical axis.
Work out the signed area between the graph and the time axis. Use geometric shapes for straight-line sections, or integrate the position-coordinate function if it is given.
Keep track of regions below the axis as negative for the signed integral.
State the result with units of length multiplied by time.

For a straight-line section, the area can often be calculated as a rectangle or trapezoid. For curved graphs, use integration or an appropriate numerical estimate if the graph is supplied without a formula.

Worked example

Suppose an object’s position coordinate is

x(t)=3\,\text{m}+(2\,\text{m}/\text{s})t, \qquad 0\leq t\leq 4\,\text{s}.

At $t=0$ , its position coordinate is $3\,\text{m}$ ; at $t=4\,\text{s}$ , it is $11\,\text{m}$ . The graph is a straight line, so the area under it is a trapezoid with parallel sides of lengths $3\,\text{m}$ and $11\,\text{m}$ , and width $4\,\text{s}$ :

\begin{aligned} \text{signed area} &=\frac{3+11}{2}\times 4\\ &=28\,\text{m}\cdot\text{s}. \end{aligned}

Integration gives the same result:

\begin{aligned} \int_0^{4\,\text{s}} \left(3\,\text{m}+(2\,\text{m}/\text{s})t\right)\,dt &=\left[(3\,\text{m})t+(1\,\text{m}/\text{s})t^2\right]_0^{4\,\text{s}}\\ &=28\,\text{m}\cdot\text{s}. \end{aligned}

The object’s net displacement over the interval is instead $11\,\text{m}-3\,\text{m}=8\,\text{m}$ . Since it moves continuously in the positive direction, it travels $8\,\text{m}$ during this interval. Neither of those motion quantities is the area under the position-time graph.

Why the area depends on the origin

A position coordinate is measured relative to a chosen origin. If you shift the origin, every position-coordinate value changes by the same constant. That shifts the position-time graph vertically and changes its area, even though the object’s motion and its displacement between any two times have not changed.

For example, replacing $x(t)$ with $x(t)+c$ changes the integral over an interval of duration $T$ by $cT$ , where $c$ is a constant with units of length. This is another reason the area under a position-time graph is not generally treated as an intrinsic measure of distance travelled.

Common mistakes

Calling the area displacement. The net displacement between the endpoints is $x(b)-x(a)$ , not $\int_a^b x(t)\,dt$ .
Calling the area distance travelled. Distance depends on changes in position along the path; it is not the area under a position-time graph.
Mixing up slope and area. The slope of a position-time graph gives velocity. The signed area under a velocity-time graph gives displacement.
Ignoring negative regions. A signed integral subtracts area below the time axis. A request for total geometric area instead requires adding the positive magnitudes of all regions.
Omitting units. An area under a position-time graph has units such as $\text{m}\cdot\text{s}$ , not metres or square metres.