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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.
** CQ_1_02.2_labelMessages **
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):
To find midpoint on line I will find the average of the beginning and end pints for the x and y axis.
X midpoint=(t_1+t_0)/2=(13s+5s)/2=18s/2=9s
Y midpoint=(v_f+v_0)/2=(40cm/s+16cm/s)/2=(56cm/s)/2=28cm/s
The midpoint of this line is (9s, 28cm/s)
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
28cm/s
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
With an average velocity of 28cm/s for the time duration of (13s-5s)=8s, the object traveled approx (28cm/s)*(8s)=224cm
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The clock time changes from (13s-5s)=8s.
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity changes 40cm/s-16cm/s=24cm/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):
aAve=’dv/’dt=(v_f-v_0)/(t_1-t_0)=( 40cm/s-16cm/s)/(13s-5s)=(24cm/s)/(8s)=3cm/s^2
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Rise is 40cm/s-16cm/s=24cm/s
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Run is 13s-5s=8s
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Slope is rise/run=(24cm/s)/(8s)=3cm/s^2
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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):
It appears the slope tells us the average rate of change in the velocity or the average acceleration, aAve.
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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):
aAve=’dv/’dt=(v_f-v_0)/(t_1-t_0)=( 40cm/s-16cm/s)/(13s-5s)=(24cm/s)/(8s)=3cm/s^2
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*#&!
&#Very good responses. Let me know if you have questions. &#