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.
See graph above
• What are the rise, run and slope of this segment?
Rise = 30 s, run = 5cm/s, slope = 6 cm/s / s
• What is the area of the graph beneath this segment?
Area = .5h(b1 + b2) = .5*5cm/s*(4s+9s) = 32.5 cm^2
Your units are incorrect. cm/s * s = cm, not cm^2.
5 cm/s/s is the rate of change of velocity with respect to clock time, represented by the slope of the graph. However the slope of the v vs. t graph has nothing directly to do with the area beneath the graph.
The bases of a trapezoid are the parallel side and the altitude is the perpendicular distance between them. This terminology is not particularly helpful when analyzing and interpreting graph trapezoids. However you are welcome to use it if you choose, provided you translate the quantities correctly. For this trapezoid the bases would represent the quantities 10 cm/s and 40 cm/s, both bases being parallel to the y axis. The altitude of the trapezoid is the distance between the bases; one altitude runs along the horizontal axis from 4 sec to 9 sec so the altitude of the trapezoid represents 5 sec. The standard formula for the area of a trapezoid, using these quantities, gives you 125 cm.
The standard formula, incidentally, represents the fact that you multiply the altitude by the average of the two bases. The form .5 h (b1 + b2) obscures this fact; the form h * (b1 + b2) / 2 better represents the meaning of the formula, which is important for interpretation.
The standard terminology is cumbersome when applied to the trapezoidal approximation to a graph. I recommend that you think of a graph trapezoid as having two altitudes measured perpendicular to the x axis, and a width measured along the x axis. The average of the altitudes represents the average value of the quantity represented on the vertical axis, and the width represents the change in the quantity represented on the horizontal axis. With this orientation we say that the area of the trapezoid is the product of its average altitude and width, and represents the product of the average value of the 'vertical' quantity and the change in the value of the 'horizontal' quantity.
There is no need to take several minutes to use a computer to create this graph, which you can sketch and label in a few seconds, and describe in a couple of short typewritten lines.
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20 minutes
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I am also submitting this assignment to you in an email since it does contain a graph.
You did not calculate the area of the graph correctly. Be sure you understand how the area is related to the calculation of displacement.
I believe your use of the standard formula for the area of a trapezoid was one of the causes of your error. Be sure you understand my notes.
Let me know if you have questions. You should submit a modified solution if you are not sure.