cq_1_022

PHY 201

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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?

The clock time is 9 seconds.

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

The velocity is 28cm/s.

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

Using the formula for change in velocity/change in seconds, and adding the change in cm/s each time the object would have traveled 212cm.

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

The clock time changes 8 seconds during this interval.

• By how much does velocity change during this interval?

The velocity changes by 24cm/s during this interval.

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

The average rate of change of velocity is 3cm/s.

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

The rise would be 24cm/s.

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

The run would be 8 seconds.

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

The slope would be 24/8 or 3/1.

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

The slope of the graph tells you that the object is accelerating.

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

The average rate of change of velocity is 3cm/s.

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I spent approximately 10 minutes on this question.

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Solution

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