course Phy 121
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18:39:57 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> We must first obtain the average velocity by dividing the displacement by the time duration. Since we began from rest, we can assume that the final velocity is 2 times the average velocity. In this case the final velocity would be the change in velocity since we started from rest. Now we determine average acceleration by taking the change in velocity and dividing by the time duration.
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18:40:48 ** 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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RESPONSE --> ok
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18:50:49 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 --> By analyzing the position vs clock time, we note the displacement and time duration associated with it. We can then calculate the average velocities for each of these pairs of data. A graph of velocity vs. clock time can be constructed whichj can than be divided into equal time intervals. The midpoint of each of these blocks (trapezoids) will correspond to the average velocity associated with that time interval. next we use the change in velocity and divide by the change in time to arrive at the acceleration.
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18:52:01 ** 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 --> ok
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18:53:40 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 --> Much certainty
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18:54:45 ** 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 --> This is uniform acceleration.
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18:55:56 06-06-2006 18:55:56 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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NOTES -------> The average velocity is at the midpoint of the trapezoid, implying that it is one half of the final velocity.
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18:55:57 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 -->
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18:56:00 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 -->
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18:56:17 ** 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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18:57:30 Why does a linear velocity vs. time graph give a curved position vs. time graph?
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RESPONSE --> constant acceleration implies that position changes at an increasing rate.
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18:58:22 ** 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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RESPONSE --> ok
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19:01:28 Why do we associate the average velocity over an interval with the midpoint clock time for that interval? I don't know the answer to this question.
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RESPONSE --> because the slope of the line corresponds to the velocity for that time increment. It would follow that the midpoint of the interval would also correspond to the average between the two point on the graph.
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19:01:46 ** 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 --> ok "