If you graph an object's velocity versus time, the area between the graph and the horizontal axis equals the object's displacement. The horizontal axis represents time, t.

The velocity graph shown in Concept 1 is a horizontal line that indicates an object moving at a constant velocity of +5.0 m/s. To determine the object’s displacement between 2.0 and 6.0 seconds from the graph, we calculate the area of the rectangle shown in Concept 1, which equals 20 m.

Why does the area equal the displacement? Because it equals the product of velocity and elapsed time, which is displacement. (Some algebra, using the definition of velocity, yields this result: v =  = Δx/Δt, so Δx = vΔt.)

Remember that displacement can be positive or negative. When the velocity graph is above the horizontal axis, the velocity is positive and the displacement is positive. When it is below, the velocity is negative and the displacement during that interval is negative. You see this point emphasized in Concept 2.

The graph in Equation 1 represents the motion of an object whose velocity changes with time. The area under a graph can be determined using calculus. It equals the integral of the velocity function over a given time interval, in this case from 0 to 5 seconds. The fundamental theorem of calculus shows that this area also equals the change in position, i.e., the displacement, of the object over that time interval. You learned in a previous section that instantaneous velocity is the derivative of the position function with respect to time. By rearranging and integrating both sides of that equation, we get the equation shown in Equation 1.

Example 1 shows the calculation of the displacement between 0 and 5 seconds for the velocity function in Equation 1.