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 velocity in meters/sec^2 and clock time in seconds, what is the area between the horizontal axis and the points ( 3 , 16 ) and ( 13 , 7 )?  What is the meaning of this area?

Solution

The area under the segment will consist of a trapezoid with altitudes 7 m/s^2 and 16 m/s^2, and uniform width ( 13 sec - 3 sec) = 10 sec.

• The average of the altitudes represents the average of the accelerations 7 m/s^2 and 16 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 ( 16 m/s^2 + 7 m/s^2) / 2 = 11.5 m/s^2.

• The area is thus 11.5 m/s^2 * 10 sec = 115 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)