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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 **
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: ->->->->->->->->->->->-> scussion (start in the next line):
The clock time is 9 seconds at the midpoint. The midpoint is the average of the two points. (5 sec + 13 sec )/2 = 18 sec/2 = 9 sec
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
To determine velocity at the midpoint we find the mid point value for y which is velocity.
(16cm/s +40cm/s ) =(56c m/s)/2 = 28 cm/s
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
`ds = vAve * `dt
`ds = 28cm/s * 8s = 224 cm
( I almost used 9 s there, but caught it. We don't use the mid point, we use change in time which is 13 sec - 5 seconds = 8 seconds)
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Very good. You do have to be careful, since it's very easy to use the wrong quantity. This is especially so with velocities, since we have initial, final, average, instantaneous velocities and change in velocity, all with the same units.
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The clock time changes 8 seconds because 13 sec - 5 sec = 8 sec
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The velocity changes 40cm/s - 16cm/s = 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: ->->->->->->->->->->->-> scussion (start in the next line):
Tha average rate of change of velocity or aAve is change in velocity/ change in clock time
(24cm/s)/8 s = 3cm/s^2
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The rise of the graph is 24 m/s. 40cm/s - 16cm/s = 24 cm/s
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The run of the graph is 8 sec. 13 sec - 5 sec = 8 sec
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The slope of the graph is the rise/run
(24cm/s)/8s = 3 cm/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: ->->->->->->->->->->->-> scussion (start in the next line):
The slope of the graph tells you that the velocity of the object is increasing at a rate of 3cm per second for every second of elapsed clock time.
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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: ->->->->->->->->->->->-> scussion (start in the next line):
The average rate of change in velocity with respect to clock time is 3 cm/s per second or 3cm/s^2
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Excellent work. Check my one note.
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