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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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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).
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The graph is a uniformly increasing straight line.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I drew a straight line between the two points.
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise is 30cm/s. (40cm/s -10cm/s = 30cm/s).
The run is 5sec. (9s - 4s= 5s).
The slope is 6. (30/5).
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@& 30 has units, as does 5, and as does the quotient 6. Units are essential.*@
• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The area of the graph beneath this segment can be calculated using the areas for a triangle and a rectangle. First I have to find the area of the triangle the slope of the graph makes if I line is drawn to the right of the lower point to the x-coordinate of the higher point. Then I have to find the area of the rectangle below the rest of the graph until it hits the x-axis.
A= (1/2*b*h) + (b*h)
A= (1/2*5s*30cm/s) + (5s*10cm/s)
A=(75cm) + (50cm) = 125 cm^2
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@& 5 dollars + 7 dollars = 12 dollars, not 12 dollars^2.
As used in the above, 'dollars' is an algebraic quantity. 5 dollars and 7 dollars are like terms, in the same way 5 x and 7 x would be like terms. They are added using the distributive law:
(5 + 7) * x = 5 x + 7 x.
or
(5 + 7) dollars = 5 dollars + 7 dollars.
Similarly, 75 cm + 50 cm = 125 cm.*@
*#&!
&#This looks good. See my notes. Let me know if you have any questions. &#