Set 6 Problem number 4

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

On a graph of acceleration vs clock time, with acceleration in meters/s^2 and clock time in seconds, we find the points ( 10 , 18 ) and ( 20 , 4 ).

• What does the area under the line segment connecting these points (i.e., the area between the line segment and the horizontal axis) represent?

Solution

The area under the segment will consist of a trapezoid with altitudes 4 m/s^2 and 18 m/s^2, and uniform width ( 20 sec - 10 sec) = 10 sec.

• The average of the altitudes represents the average of the accelerations 4 m/s^2 and 18 m/s^2.

• If acceleration is changing at a uniform rate the segment in fact represents the acceleration 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 accelerations is the approximate, not the exact, average acceleration on the interval.

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

• The area therefore represents approximate average acceleration * time interval = change in velocity during the time interval.

The average altitude in the present example is ( 18 m/s^2 + 4 m/s^2) / 2 = 11 m/s^2.

• The area is thus 11 m/s^2 * 10 sec = 110 meters/sec.
• We interpret this as the approximate velocity change corresponding to the two points on the graph.

Generalized Solution

On a graph of acceleration vs. clock time t, two points will have coordinates (t1, a1) and (t2, a2).

• The average altitude of the trapezoid defined by these points is (a1 + a2) / 2, representing the approximate average acceleration aAve of an object between clock times t1 and t2.
• The run is from t1 to t2, a run of `dt = t2 - t1. This run represents the difference in clock time between t1 and t2, or the time interval between t1 and t2.
• The area is the product of the average altitude aAve = ( a1 + a2 ) / 2 and the time interval `dt = t2 - t1.  The area is thus

• area = approximate velocity change = ( a1 + a2 ) / 2 * ( t2 - t1 ).

Explanation in terms of Figure(s), Extension

The graph below shows two points (t1, y1) and (t2, y2) on a graph of position vs. time.

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

If we let a1 stand for y1 and a2 for y2, average altitude is (a1 + a2) / 2 and represents average acceleration. 

With this notation the area  

• average altitude * width = ( a1 + a2 ) / 2 * ( t2 - t1 )

thus represents approximate average acceleration * time interval = approximate change in velocity.

Figure(s)