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Phy 201
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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Copy the problem below into a text editor or word processor.
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You might prefer for your own reasons to use a word processor (for example the formatting features might help you organize your answer
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As you will see within the first few assignments, there is an easily-learned keyboard-based shorthand that doesn't look quite as pretty as
word-processor symbols, but which gets the job done much more efficiently.
You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
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.
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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).
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Ok, graph is sketched. x2 = 9, y2 = 40 - x1 = 4, y1 = 10
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Line segment is sketched, linear line increasing at an increasing rate.
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
rise = 30 cm/s
run = 5 s
slope = m = 6 cm
you didn't show the calculation of the slope, which would have included units with every quantity, and if done correctly would have given you the correct units of your answer
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
y - 10 = 6(x-4)
f(x) = 6x - 14
Integral = (6x - 14 ) dx
= 3x^2 - 14x
a = 4
b = 9
f(b) = 117
f(a) = -8
117 - (-8) = 125
Area under the graph = 125 units (or) 125 cm
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30 minutes
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9/12 3pm
&#Check the link. No revision is necessary unless you have questions. If you do have questions or comments, you are of course welcome to submit a response. &#
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#