cq_1_022

PHY 201

Your 'cq_1_02.2' 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 graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds. The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).

• What is the clock time at the midpoint of this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

(5 + 13) / 2 = 9 seconds

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• What is the velocity at the midpoint of this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

(16 + 40) / 2 = 28 cm/s

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• How far do you think the object travels during this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

At an average velocity of 28 cm/s with a time interval of 8 seconds the object travels

distance = 28 cm/s * 8 s = 224 cm

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• By how much does the clock time change during this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

Changes by 13 - 5 = 8 seconds

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• By how much does velocity change during this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

Changes by 40 - 16 = 24 cm/s

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• What is the average rate of change of velocity with respect to clock time on this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

a = 24 cm/s / 8 s = 3 cm/s^2

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• What is the rise of the graph between these points?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

The run is 40 - 16 = 24

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• What is the run of the graph between these points?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

The rise is 13 - 5 = 8

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• What is the slope of the graph between these points?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

Slope is equal to the rise over the run which is

24 / 8 = 3

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• What does the slope of the graph tell you about the motion of the object during this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

The slope is the same as the acceleration of the object.

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• What is the average rate of change of the object's velocity with respect to clock time during this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

a = 24 cm/s / 8 s = 3 cm/s^2

#$&*

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30 min

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Very good.

Do note that rise, run and slope all have units, which should be included in calculations. This is important for interpretation.