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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:
9 sec because that is the midpoint between 5 and 13
What is the velocity at the midpoint of this interval?
answer/question/discussion:
24cm/s because that is the intersection of the 9 s midpoint and the velocity
How far do you think the object travels during this interval?
answer/question/discussion:
24cm-16cm=8cm
By how much does the clock time change during this interval?
answer/question/discussion:
8 sec
By how much does velocity change during this interval?
answer/question/discussion:
24cm/s
What is the rise of the graph between these points?
answer/question/discussion:
24cm/s
What is the run of the graph between these points?
answer/question/discussion:
8s
What is the slope of the graph between these points?
answer/question/discussion:
3cm/s/s
What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion:
The motion is increasing
What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion:
3 cm/s/s
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20 minutes
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Most, but not all of your solution is correct.
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What is the clock time at the midpoint of this interval?
The graph is a straight line, so the midpoint lies halfway between the two clock times, and halfway between the two velocities.
The quantity halfway between two values is the mean of the two values, obtained by adding the values and dividing by 2.
The midpoint of the time interval between (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s) is the clock time midway between 5 sec and 13 sec; this occurs at clock time (5 sec + 13 sec) / 2 = 9 sec.
What is the velocity at the midpoint of this interval?
The midpoint of the velocity between 16 cm/s and 40 cm/s occurs midway between the two velocities, at (16 cm/s + 40 cm/s) / 2 = 28 cm/s.
How far do you think the object travels during this interval?
Note that the duration of the interval from t = 5 sec to t = 13 sec is 13 sec - 5 sec = 8 sec, which is not the same as the midpoint of the time interval.
In commonsense terms, we see that the object moves at average velocity 28 cm/s for 8 seconds, and so travels 28 cm/s * 8 s = 224 cm.
You should also understand this more formally, in terms of the relevant symbols and definitions:
If the graph is a straight line, the midpoint velocity will be the average velocity on the interval. The average velocity is the average rate of change of position with respect to clock time, i.e.,
vAve = `ds / `dt
We want to find the displacement `ds and we now know vAve. All we need is the value of `dt, the duration of the time interval. If we know `dt, we can rearrange the equation vAve = `ds / `dt to obtain
`ds = vAve * `dt
and simply substitute our values of vAve and `dt.
The time interval lasts from t = 5 sec to t = 13 sec; the duration of the interval is therefore 13 s - 5 s = 8 s. Thus we can find the displacement `ds:
`ds = vAve * `dt = 28 cm/s * 8 s = 224 cm.
By how much does the clock time change during this interval?
The change in clock time from t = 5 sec to t = 13 sec is
`dt = 13 sec - 5 sec = 8 sec.
By how much does velocity change during this interval?
The change in velocity during the interval is from 16 cm/s to 40 cm/s, a change of 24 cm/s.
This change occurs in the 8-sec duration of the 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 with respect to clock time is
ave rate = change in velocity / change in clock time = ( 24 cm/s ) / (8 s) = 3 (cm/s) / s = 3 (cm/s) * (1/s) = 3 cm/s^2.
What is the rise of the graph between these points?
The rise of the graph during the interval is from 16 cm/s to 40 cm/s, a rise of 24 cm/s.
What is the run of the graph between these points?
This run of the graph is the 8-sec duration of the time interval, which runs from 5 sec to 13 sec.
What is the slope of the graph between these points?
The slope of the graph is
graph slope = rise / run = ( 24 cm/s ) / (8 s) = 3 (cm/s) / s = 3 (cm/s) * (1/s) = 3 cm/s^2.
What does the slope of the graph tell you about the motion of the object during this interval?
The slope calculation is identical to the calculation of the average rate of change of velocity with respect to clock time. The slope therefore represents this average rate of change.