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Phy 121
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.
** CQ_1_04.1_labelMessages **
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).
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Sketch a straight line segment between these points.
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What are the rise, run and slope of this segment?
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The rise is 40cm/s - 10cm/s = 30cm/s
The run is 9s - 4s = 5 s
The slope is 30cm/s / 5s = 6cm/s^2
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
To find the area under the segment, we need to know the change in x value(base) and the and the average of the change in position(height).
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The change in heights would have numerical value 30. However the change in height isn't relevant to the area.
The average height, which does have numerical value 25, is relevant to the area.
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A = b * h, or b = 9 - 4 = 5, h = 40 - 10 = 30 / 2 = 15 + 10 = 25, so A = 5 * 25 = 125cm/s^2
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The area does have numerical value 5 * 25.
What are the units of the quantity you've represented by 5?
What are the units of the quantity you've represented by 25?
What is the meaning of each of these quantities?
What therefore are the units and the meaning of their product?
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You need to include units in your calculation, and also the meaning of each quantity and each result.
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You've got the right idea, but your units aren't right and you haven't quite interpreted your result.
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