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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?
Clock time is 8 seconds at the midpoint. This was obtained by subtraction:
13 sec 5 sec = 8 seconds
What is the velocity at the midpoint of this interval?
The velocity at the midpoint is 3 cm/s. This was obtained by first subtracting 40cm-16cm = 24. Next we take 24cm/8sec (from the previous question) and the result is 3 cm/s.
How far do you think the object travels during this interval?
The object travels 24 cm. This was obtained by either subtracting the 40cm-16cm or by taking 3cm/s and multiply by 8 seconds to get a result of 24 cm.
By how much does the clock time change during this interval?
The clock time changes from 13 seconds to 5 seconds = 8 seconds
By how much does velocity change during this interval?
The velocity changes from 40 cm/s to 16 cm/s = 24 cm/s
What is the rise of the graph between these points?
The rise is 24. This is obtained by taking the vertical coordinates and subtracting.
What is the run of the graph between these points?
The run is 8. This is obtained by taking the horizontal coordinates and subtracting.
What is the slope of the graph between these points?
The slope is 3. This is obtained by taking the rise/run. 24/8 = 3
What does the slope of the graph tell you about the motion of the object during this interval?
Im not completely sure about this question but I think that with a slope of 3, the motion of the object isnt moving very fast. Obviously the steeper the slope the faster the object will move. From observing the graph, the line looks steep but if the slope is 3 I dont think it is moving very fast so Im a little uncertain. Also, for smaller slopes it is apparent that friction has a constant influence on the acceleration.
What is the average rate of change of the object's velocity with respect to clock time during this interval?
Average rate of change = 24 cm/8 seconds = 3cm/s
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30 min
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I was a little confused about some of the answers so I'm interested to see if my answers are correct.
Note first that 24 cm/s / (8 s) = 3 cm/s^2, not 3 cm/s. This result is the average rate of change of velocity with respect to clock time.
Compare your remaining responses to the following. I'll be glad, if you ask, to answer specific questions or clarify specific points.
&#The clock time midway between 5 sec and 13 sec is (5 sec + 13 sec) / 2 = 9 sec.The velocity midway between 16 cm/s and 40 cm/s is 28 cm/s.The average velocity on the interval would be expected to lie between the initial and final velocities. Without additional information it is not certain that this is the case, but without additional information the most reasonable assumption would be that the average velocity is halfway between the initial and final velocities.Basing a distance estimate on the assumption that the midpoint velocity of 28 cm/s is the average velocity, we would conclude that the object moves 28 cm/s * 9 s = 224 cm.If it is known that acceleration is uniform, the average velocity is equal to the midpoint velocity and this estimate would be accurate. If acceleration is not uniform, then it probably isn't completely accurate.The change in velocity from 16 cm/s to 40 cm/s is 24 cm/s, and this corresponds to the rise of the graph.The change in clock time is 13 sec - 5 sec = 8 secThe average rate of change of velocity with respect to clock time isave rate = (change in velocity) / (change in clock time) = 24 cm/s / (8 s) = 3 cm/s^2.
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