cq_1_022

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Phy 121-

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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SEED Question 2.2

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):

(t1 + t2)/ 2 = midpoint

(5 +13) / 2 = 9sec

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

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

(v1 + v2) / 2 = midpoint velocity

right, but there are no 0 and 3 cm/s velocities involved, so your next line appears to be incorrect.

(0cm/s + 3cm/s) / 2 = 1.5cm/s

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

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

I’m guessing that this question refers to the previous question about the midpoint.

(d2 – d1) / 2 = midpoint distance

(40cm -16cm) / 2 = 12cm

the velocity is never less than 16 cm/s so it's going to move a lot further than that

40 cm and 16 cm are not quantities associated with this motion. 40 cm/s and 16 cm/s are.

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

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

(t2 –t1) / 2 = midpoint time interval

(13sec -5sec) / 2 = 4sec

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• By how much does velocity change during this interval?

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

The ball is traveling at a constant rate so the rate of change stays the same (0)

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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):

(dist 2 –dist 1) / total time = Ave velocity

(40cm -16cm) / 8sec = 3cm/sec

40 cm and 16 cm are not given quantities

40 cm/s and 16 cm/s are

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

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

(d2 –d1) = rise

40cm -16cm = 24cm

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

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

(t2 – t2) = run

13s – 5s = 8 sec

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

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

(d2 – d1) / (t2 – t1) = slope

(40 – 16) / (13 – 5) = 3

Right calculation, but all quantities have units.

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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):

that the average velocity of the ball is 3cm/sec

if 40 and 16 were positions the since 13 and 5 were clock times then this would be an average velocity

but the aren't so it isnt

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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):

There is no change the ball is moving at a constant rate.

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1 hr

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