A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
Sketch a straight line segment between these points.
What are the rise, run and slope of this segment?
The rise is from 10 cm/s to 40 cm/s, which is a rise of (40 cm/s - 10 cm/s) = 30 cm/s.
Note that 10 cm/s and 40 cm/s have the same units and are therefore like terms. As like terms they can be subtracted, and their difference is therefore a like term with the same units.
Similarly the run is from 4 sec to 9 sec, a run of (9 s - 4 s) = 5 s.
Slope is therefore rise / run = (30 cm/s) / (5 s) = 6 cm/s^2.
What is the area of the graph beneath this segment?
The region beneath this segment forms a trapezoid, with 'altititudes' representing the initial and final velocities 10 cm/s and 40 cm/s. The average 'altitude' of the trapezoid is therefore (10 cm/s + 40 cm/s) / 2 = 25 cm/s, and represent the average of the initial and final velocities. Since the graph is a straight line, the average altitude therefore represents the average velocity on the interval.
The width of the trapezoid is equal to the 'run' between the two points, which as seen earlier is 5 s. This represents duration of the time interval.
The area of the trapezoid is equal to average altitude * width = 25 cm/s * 5 s = 125 cm.
This area represents the product of the average velocity and the duration of the time interval. The product therefore representsthe displacement of the object during the time interval.