Set 6 Problem number 5

The graph below depicts acceleration vs clock time, with acceleration in meters/s^2 and clock time in seconds. The gridlines depict units of .2 m/s^2 in the vertical direction and 4 seconds in the horizontal direction.

What is the area under the graph between t = .2 sec and t = 1.4 sec, and what is the specific meaning of this area?

Solution

The area under the curve could be broken into tiny trapezoids with altitudes representing acceleration in m/s^2 and widths representing time intervals. The average of the altitudes of each trapezoid represents the approximate average acceleration on that interval, and the width of the represents the time interval over which this approximate average acceleration is sustained.

The area of a trapezoid therefore represents a product of average acceleration and elapsed time, which represents change in velocity.
When summed the total area of the trapezoids therefore represents the approximate change in velocity from clock time t = .2 sec to t = 1.4 sec.
In the limit as the widths of the trapezoids approach zero the total area represents the exact change in velocity.

We therefore estimate the area under the curve.

We find the area by a trapezoidal approximation using several well-chosen trapezoids, or alternatively by counting the rectangles under the curve (the area of each rectangle representing the change in velocity .2 m/s^2 * 4 sec = .8 m/s).

University Physics note that the precise velocity change would be obtained by integrating the function represented by the graph between t = .2 and t = 1.4.