course phy121
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14:39:17 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> `ds / `dt = vAve we then plug vAve into the equasion ( vI + vF ) / 2 = vAve where vI = 0 ( vI + vF ) = vAve * 2 vI + vF = (vAve*2) vF = (vAve*2)-vI This gives us the final velocity (vF) Divide vF by end clock time and this will give you the rate of change of acceleration which you can multiply by desired clock time to calculate acceleration at any given point.
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14:40:15 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 --> ok
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14:45:11 ** 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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14:47:41 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 --> Very little.
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14:49:26 ** 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 --> The order of the questions to the introduction of data confuses me profoundly...
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14:53:07 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 --> 4cm/sec = vAve at 1/2 `dt Therefor you can enter the values we have into the averaging equasion. (vI + vF) /2 =vAve vI = 0cm/sec vF = 8cm/sec vAve = 4cm/sec (0+8)/2=4
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14:53:32 ** 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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14:54:54 Why does a linear velocity vs. time graph give a curved position vs. time graph?
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RESPONSE --> Because of the rate at which the object is accelerating is linear so the distance traveled increases with each time interval. SO, the rise of the line increases relative to a steady run (time interval)
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14:54:58 ** 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 -->
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14:56:38 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 --> The average velocity for an interval is equal to the velocity of the object at the midpoint time of the interval.
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14:56:41 ** 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 -->
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