ʐ|NPȂ assignment #004 Wq}PsӲqʧp Physics I Class Notes 01-29-2006
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21:08:36 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> Acceleration is calculated by dividing displacement by time duration.
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21:11:47 01-29-2006 21:11:47 ** COMMON ERROR: By first finding the average velocity. Also we must understand this gives us the change in velocity which gives us the acceleration. INSTRUCTOR COMMENT: Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. ANSWER TO QUESTION: Average acceleration is defined to be the average rate at which velocity changes, which is change in velocity/change in clock time, or equivalently change in velocity/time duration. However we do not know the change in velocity, nor from the given information can we directly determine the change in velocity. So we have to look at what we can reason out from what we know. Given displacement from rest and time duration we calculate average velocity: vAve = `ds / `dt = displacement/time duration. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Now that we know the initial velocity 0 and the final velocity we can find change in velocity by subtracting initial vel from final vel. Dividing change in velocity by a time duration we finally obtain the average acceleration. **
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NOTES -------> Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Then divide the change in velocity by the time duration.
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21:15:59 01-29-2006 21:15:59 How do we obtain an approximate velocity vs. clock time table and graph from a series of position vs. clock time observations? How do we then obtain an approximate acceleration vs. clock timetable and graph?
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21:16:07 How do we obtain an approximate velocity vs. clock time table and graph from a series of position vs. clock time observations? How do we then obtain an approximate acceleration vs. clock timetable and graph?
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RESPONSE -->
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21:26:43 ** If we start with a position vs. clock time graph we do divide the graph into intervals. On each interval velocity, which is rate of change of position, is represented by the slope of the graph. So we have to calculate the slope of the graph on each interval. If we then graph the slopes vs. the midpoint times of the intervals we get a good approximation of the velocity vs. clock time graph. The velocity at a given instant is the slope of the position vs. clock time graph. STUDENT ANSWER: We first calculate the time interval and displacement between each pair of data points. We use these calculations to calculate the average velocity. To obtain an approximate accelerations. Clock timetable and graph by associating the average velocity on a time interval with the midpoint clock time on that interval. INSTRUCTOR CRITIQUE: Using your words and amplifying a bit: We first calculate the time interval and displacement between each pair of data points. We use these results to calculate the average velocity, dividing displacement by time interval for each interval. Then we make a table, showing the average velocity vs. the midpoint time for each time interval. To obtain approximate accelerations we use the table and graph obtained by associating the average velocity on a time interval with the midpoint clock time on that interval. We find the time interval between each pair of midpoint times, and the change in average velocities between those two midpoint times. Dividing velocity change by time interval we get the rate of velocity change, or acceleration. **
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RESPONSE --> Hit the wrong button too soon. First calcualate the time interval and displacement between each pair of data points, and then calculate average velocity. Clock timetable and graph by associating the average velocity on a time interval with the midpoint clock time on that interval. To obtain approximate accelerations we use the table and graph obtained by associating the average velocity on a time interval with the midpoint clock time on that interval. We find the time interval between each pair of midpoint times, and the change in average velocities between those two midpoint times. Dividing velocity change by time interval we get the rate of acceleration
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21:30:00 For water flowing from a uniform cylinder through a hole in the bottom, with how much certainty can we infer that the acceleration of the water surface is uniform?
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RESPONSE --> We can be certain that the acceleration of the water surface is uniform because the hole in the bottom is uniform that the water is flowing through.
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21:31:15 ** If we first calculate velocities from the position vs. clock time data we get decreasing velocities. If we graph these velocities vs. midpoint clock times we get a graph which appears to be well-fit by a straight line. This is evidence that the acceleration of the water surface is uniform. If we calculate average accelerations based on average velocities and midpoint clock times we get a lot of variation in our results. However since acceleration results depend on velocities and changes in clock times, and since the velocities themselves were calculated based on changes in clock times, our results are doubly dependent on the accuracy of our clock times. So these fluctuating results don't contradict the linearity of the v vs. t graph. We also find that the position vs. clock time data are very well-fit by a quadratic function of clock time. If the position vs. clock time graph is quadratic then the velocity is a linear function of clock time (University Physics students note that the derivative of a quadratic function is a linear function) and acceleration is constant (University Physics students note that the second derivative of a quadratic function is constant). **
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RESPONSE --> Calculate the velocities first and then draw a graph.
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21:33:28 How does the graph make it clear that an average velocity of 4 cm / s, and initial velocity 0, imply final velocity 8 cm / s?
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RESPONSE --> The graph shows that the average between 0 cm and 8 cm is 4 cm because if you add 0 and 8 and then divide by 2 you get 4.
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21:33:43 ** If the graph is linear then the average velocity occurs at the midpoint clock time, and is halfway between the initial and final velocities. In this case 4 cm/s would be halfway between 0 and the final velocity, so the final velocity would have to be 8 cm/s (4 is halfway between 0 and 8). **
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RESPONSE --> ok
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21:39:58 Why does a linear velocity vs. time graph give a curved position vs. time graph?
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RESPONSE --> A linear velocity vs. time graph would give a curved position vs. time graph because the postion of an object will change even if the velocity is constant regardless of the time.
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21:40:48 01-29-2006 21:40:48 ** Assuming uniform time intervals, a linear v vs. t graph implies that over every time interval the average velocity will be different that over the previous time intervals, and that it will be changing by the same amount from one time interval to the next. The result is that the distance moved changes by the same amount from one time interval to the next. The distance moved is the rise of the position vs. clock time graph. If over uniform intervals the rise keeps changing, by the same amount with each new interval, the graph has to curve. **
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NOTES -------> If over uniform intervals the rise keeps changing, by the same amount with each new interval, the graph has to curve.
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21:48:02 ** For an object with positive acceleration, at the beginning of a time interval the velocity is less than at the end of the interval. We expect that the average velocity is between the beginning and ending velocities; if acceleration is uniform, in fact the average velocity is equal to the average of initial and final velocities on that interval, which is halfway between initial and final velocities. If the interval is short, even if acceleration is not uniform, then we expect the average velocity occur near the midpoint clock time. **
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RESPONSE --> The midpoint is halfway between the initial and final velocity and the average velocity is found at the midpoint.
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yɵx֭h assignment #005 Wq}PsӲqʧp Physics I Class Notes 01-29-2006
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21:54:14 For small slopes does the corresponding acceleration seem to change by an amount that is proportional to the change in slope?
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RESPONSE --> For small slopes, the corresponding acceleration does seem to change by an amount that is proportional to the change in slope.
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21:54:25 ** For small slopes, the data indicate that the change in acceleration would be nearly proportional to change in slope. **
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RESPONSE --> ok
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21:55:58 Do we expect the small-slope behavior of acceleration vs. slope to continue when the slope becomes large?
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RESPONSE --> Yes, we expect the small-slope behavior of acceleration vs. slope to continue when the slope becomes large.
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21:56:15 ** Large slope means slope of large magnitude. For the accel vs. ramp slope experiment the slopes were all small, less than .1. The small-slope behavior appears to be linear, in the sense that acceleration appears to be proportional to ramp slope. As the ramp approaches vertical the slope approaches infinity, while acceleration approaches simply the acceleration of gravity. If the graph was linear then as slope approaches infinity we would have acceleration approaching infinity. We conclude that the linear acceleration vs. slope relationship will not continue. ** As we approach vertical slope approaches infinity, but acceleration simply approaches the acceleration of gravity. So the graph has to start leveling off at some point. **
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RESPONSE --> ok"