Your 'cq_1_04.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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The problem:
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 of the segment is 30 cm/s, the run is 5s, and the slope is 6cm/s^2.
What is the area of the graph beneath this segment? Acceleration. 75cm/s^2.
This is not the area, and the area does not have units of cm/s^2.
The region beneath the segment is a trapezoid, with 'altitudes' measured parallel to the vertical axis; the 'altitudes' are 10 cm/s and 40 cm/s.
The width of the trapezoid is measured parallel to the horizontal axis.
The area of a trapezoid oriented in this way is equal to the product of the average altitude and the width, because the trapezoid can easily be 'cut-and-pasted' into a rectangle of this altitude and width.
What therefore is the area, what does it mean and why, and what are its units?
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20 min.
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Please respond relative to goal of C.
You got the slope, which you could have identified as the average acceleration.
The area is not the acceleration, and your area is in any case incorrect.
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