cq_1_041

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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&#Please compare your solutions with the expanded discussion at the link

Solution

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