Phy 121
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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: ->->->->->->->->->->->-> :
Both of these points are in Quadrant I. The point (9 s, 40 cm/sec) is higher and further to the right than the other point.
Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> :
The line between this two points has an increasing slope.
What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> :
The rise is:
40 cm/sec- 10 cm/sec= 30 cm/sec
The run is:
9 s- 4 s= 5 sec
The slope is the rise over run:
30 cm/sec/ 5 sec= 6
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> :
We use the area equation for a triangle. This is:
A= ½ bh
The base would be the run while the height would be the rise:
A= ½*(5 sec)*(30 cm/sec)= 75 cm^2
Then we must use the area equation for a rectangle and add that to the triangle:
The height from the x-axis to the bottom of the triangle is:
4 sec- 0= 4
A= bh
A= 5* 4= 20
The total area is the area of the rectangle added to the area of the triangle:
Total Area= 20 + 75= 95 cm^2
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30 minutes
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Solution
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