Set 6 Problem number 3

On a graph of velocity vs clock time, with velocity in meters/sec and clock time in seconds, what is the area between the horizontal axis and the points ( 18 , 19 ) and ( 20 , 4 )? What is the meaning of this area?

Solution

The area under the segment will consist of a trapezoid with altitudes 4 m/s and 19 m/s, and uniform width ( 20 sec - 18 sec) = 2 sec.

The average of the altitudes represents the average of the velocities 4 m/s and 19 m/s.

If velocity is changing at a uniform rate the segment in fact represents the velocity vs. clock time precisely; otherwise it is only an approximation to the behavior of a curving graph.

In general we therefore say that the average of the two velocities is the approximate, not the exact, average velocity on the interval.

The width of the trapezoid represents the time interval over which this approximate average velocity is sustained.
The area of the trapezoid is the product of the average altitude, representing the approximate average velocity on the interval, and the width, representing the time interval.

The area therefore represents approximate average velocity * time interval = displacement during the time interval.

The average altitude in the present example is ( 19 m/s + 4 m/s) / 2 = 11.5 m/s.

The area is thus 11.5 m/s * 2 sec = 23 meters.
We interpret this as the approximate displacement corresponding to the two points on the graph.

Generalized Solution

The graph below shows two points (t1, y1) and (t2, 2) on a graph of position vs. time.

The average altitude is ( y1 + y2 ) / 2, representing approximate average velocity.
The run is `dt = t2 - t1, the time interval between the points.

If we let v1 stand for y1 and v2 for y2, then the average altitude is (v1 + v2) / 2, standing again for approximate average velocity.

With this notation the area

average altitude * width = ( v1 + v2 ) / 2 * ( t2 - t1 )

thus represents approximate average velocity * time interval = approximate displacement.