Phy 231
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.
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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:
Sketched it. It's a linear line rising to the right.
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> scussion:
The line is rising to the right.
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
The run is (9s-4s) = 5 sec, since time is on the x axis, and the rise is (40 cm/s-10cm/s) = 30 cm/s since velocity is on the y axis, and the slope is (30 cm/s)/5sec = 6 cm/s^2
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
Well, you could actually take the area of the graph by adding the area of the rectangle and the triangle that make up the area under the segment, so
length = 9s - 4s = 5s
height (rectangle) = 10 cm/s - 0 cm/s
so area of rectangle is 10 cm/s * 5 sec = 50 cm
and since the length of the triangle is the same (5 s) and the height is 40 cm/s - 10 cm/s = 30 cm/s, and .5(30 cm/s)(5 sec) = 75 cm
then 50 cm + 75 cm = 125 cm under the line segment
or you could take the equation of the line:
y = mx + b
y = 6x + b
10 = 6(4) + b
b = -14
so y = 6x - 14
and then take the integral, 6x^2/2 - 14x, or 3x^2 - 14x
and then subtract:
(3(9)^2 - 14(9)) - (3(4)^2 - 14(4)) = 117 - (-8), or 125 cm
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10 minutes
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&#Very good work. Let me know if you have questions. &#