Phy121
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q 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:
If the first time is 5sec and the second is 13 the midpoint would be
(dt1 + dt2) / 2 =
(5 + 13) / 2 = 9sec
What is the velocity at the midpoint of this interval?
answer/question/discussion:
(dv1 + dv2) / 2 =
(16cm/sec + 40cm/sec) / 2 = 28cm/sec
• How far do you think the object travels during this interval?
answer/question/discussion:
vAve = (dt*dv)
vAve = (9sec * 28cm/sec)
252 cm = (9sec * 28cm/sec)
9 seconds is the midpoint of the time interval, but the midpoint of the interval is not the same thing as the duration of the interval.
How long does the time interval last and how does this affect the present calculation?
• By how much does the clock time change during this interval?
answer/question/discussion:
With 9 sec being the midpoint and the first point being 5sec there is a difference 0f 4 sec
(dt2 – dt1) = dist interval = 4sec
•
•
• By how much does velocity change during this interval?
answer/question/discussion:
By averaging the two coordinates (y1-y2) / (x1- x2) for time interval between the midpoint and first coordinate ( 9-5 ) / ( 28-16) = 4/12
= 1/3cm/sec
The interval lasts from the beginning to the end, not from the beginning to the midpoint.
Neither your numerator nor denominator are expressed in units.
The question here asks for the change in the velocity during the interval, not for a rate of change.
•
• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion:
If we take the amount of velocity increase = 12cm/sec and divide by the amount of time increase of (4sec) we then get the velocity increase between the two intervals =
(.3333cm/sec)
• What is the rise of the graph between these points?
answer/question/discussion:
The rise from (5sec) to point (9sec) = 4sec
On a graph of v vs. t, the velocity is the 'vertical' coordinate. What you have given here is the run from the initial instant to the midpoint. What is needed is the rise from the initial instant the final instant.
• What is the run of the graph between these points?
answer/question/discussion:
The run from point (16cm/sec) to point (28cm/sec) = 12cm/sec
This is the rise from the midpoint instant to the final instant. We need to run from the initial instant to the final instant.
• What is the slope of the graph between these points?
answer/question/discussion:
the slope = m or the (y1 – y2) / (x1 – x2) = (9-5) / (28-16) = 4/12 =
1/3 = .33333
• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion:
The slope is increasing at an increasing rate
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion:
If we add the two intervals and divide by 2 we obtain the average velocity between the two int
The question asked about the average rate of change of the object's velocity with respect to clock time, not its average velocity. Start with the definition of average rate of change of velocity with respect to clock time.
Form Confirmation
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45 min
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You are far off on many these procedures, but you do have at least one error on most.
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