course Phy121 I had a question about the experiment for notes 05 and so have not done those questions. I emailed the question to you.
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07:56:54 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> We can determine acceleration from rest given displacement and time duration, assuming it is accelerating at a uniform rate. From the change in displacement and time duration, we can figure out its average velocity. From the average velocity, we can determine its final velocity (this will be 2 * vAve) since it is increasing uniformly. Since it is starting from rest, initial velocity will be 0, and we then subtract 0 from the final velocity to get the change in velocity. We then divide this change by the change in time to find acceleration.
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07:57:36 ** 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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08:04:27 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 --> Position and clock time observations provide us with the base of the trapezoid for that time interval, as well as the area of the same trapezoid for that time interval. From this information, we can determine its height, which is the velocity of the the object at that particular point in time. Once we have velocities at these various points, we can determine the change in velocity, over the time duration, and obtain an approximate acceleration at these same time points along the graph.
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08:26:43 ** 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 --> Ok. Was I correct in stating that since we are provided the displacement, we can calculate the altitude at that particular time interval?
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08:35:01 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 --> I'm not too sure I understand how to answer this. From looking at the table in the class notes, it appears the displacement is changing at -10 cm per time interval, but the time intervals are increasing in length by 1 second. The last 3 however, show the change in average velocity remaining at -.06cm/s. Would all of this indicate it is accelerating at a uniform pace?
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08:36:54 ** 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 --> Could you provide another example of this? I think it would help me grasp it a bit better. Thanks.
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08:37:45 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 --> If we look at the graph, the 4cm/s is at the midpoint of the initial and final velocity. The final velocity is double the size of the midpoint, hence 8cm/s.
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08:37:55 ** 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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08:39:50 Why does a linear velocity vs. time graph give a curved position vs. time graph?
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RESPONSE --> I was pondering this from the notes. A linear velocity vs. time graph states the acceleration is uniform. Even though the acceleration is uniform, the velocity is still increasing at an increasing rate. This increase will therefore effect the displacement. As the velocity increaes, so does the effect on the displacement. Rather than a straight line, it curves upward, since the changes in displacement are greatly increasing.
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08:41:59 ** 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 --> I feel my explanation touched this, but that it needs some fine tuning. I will have to think about this a bi tmore, but I do think I understand what is going on here.
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08:45:20 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 --> This is done when the acceleration is uniform, b/c the average velocity is the sum of the final velocity and initial velocity(in that time interval) divided by 2, thus, the midpoint clock time for that interval.
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08:45:39 ** 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.
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͖߆wñ assignment #003 ̐輻Iyʴ Physics I Vid Clips 02-04-2006 稤ǖ assignment #002 ̐輻Iyʴ Physics I Vid Clips 02-04-2006
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10:15:58 Given a graph of velocity vs. time how do we determine the rate which velocity increases?
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RESPONSE --> In order to find the rate at which velocity changes, we would need to divide the change in velocity over the change in time. We can do this at a particular time interval or over the entire time period for the event.
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10:16:24 ** To find the average rate of change of velocity we use the slope of the graph between two points, which indicates rise / run = change in velocity/change in clock time. This is equivalent to using the slope of the graph, which indicates rise / run = change in velocity/change in clock time. COMMON ERROR: rate = Velocity/time INSTRUCTOR COMMENT: If you divide v by t you do not, except in certain special cases, get the average rate at which velocity increases. You have to divide change in velocity by change in clock time. For example if you're traveling at a constant 60 mph for 1 hour and divide 60 mph by 1 hour you get a nonzero acceleration of 60 mph / hr while in fact you are traveling at a constant rate and have acceleration 0. **
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RESPONSE --> ok
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10:17:50 If we know the velocities at two given clock times how do we calculate the average rate at which the velocity is changing during that time interval?
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RESPONSE --> We can take the difference between the velocities at these two given times and divide them by the difference in the times to find the average rate at which velocity is changing during that time interval. This corresponds to the slope between the two velocities, the rise/run.
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10:17:57 ** We divide the difference `dv in velocities by the duration `dt of the time interval. **
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RESPONSE --> Ok.
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10:18:37 Does the rate at which a speedometer needle moves tell us how fast the vehicle is moving?
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RESPONSE --> The rate at which it moves does not tell us the speed, but how the speed is changing.
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10:19:35 ** The rate at which the speedometer needle moves does not tell us how fast the vehicle is moving. The fact that the speedometer needle moves tells us that the car is either speeding up or slowing down, not at what rate the car is moving. The reading on the speedometer tells us how fast we are moving. A speedometer that simply sits there on 60 mph doesn't move at all, but presumably the vehicle is moving pretty fast. If the speedometer moves, that indicates that velocity is changing. If the speedometer moves quickly, the velocity is changing quickly. If the speedometer moves slowly, velocity is changing slowly. A quickly moving speedometer needle doesn't imply a quickly moving vehicle. The speedometer might go past the 5 mph mark very fast. But at the instant it passes the mark the car isn't moving very quickly. The needle can move quickly from 1 mph to 2 mph, or from 100 mph to 101 mph. The speed of the needle has little to do with the speed of the vehicle. Of course if the needle is moving quickly for an extended period of time, this implies a large change in the velocity of the vehicle. However this information still does not tell you what that velocity actually is. **
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RESPONSE --> Ok.
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10:20:14 Does a stationary speedometer needle implies stationary vehicle?
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RESPONSE --> No, a stationary speedometer indicates the speed of the vehicle is not changing. A speedometer can stay at 100mph over a period of time which would indicate a very fast moving vehicle.
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10:20:21 ** A stationary speedometer needle does not imply a stationary vehicle. The speedometer could be constant at the 60 mph mark while the car is moving 60 mph. **
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RESPONSE --> ok
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10:21:13 Does a quickly moving speedometer needle imply a quickly moving vehicle?
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RESPONSE --> It does not imply a quickly moving vehicle at all. A quickly moving speedmeter means the rate of the speed is chaning very quickly, but it could drop quickly from 60mph to 20mph, or from 1mph to 5mph in a short period of time, but the vehicle would not be moving quickly.
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10:21:25 ** A quickly moving speedometer needle does not imply a quickly moving vehicle. It merely implies that the velocity of car is changing quickly. **
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RESPONSE --> ok.
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10:22:47 What does it feel like inside a car when the speedometer needle is moving fast?
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RESPONSE --> When it's moving fast you will feel a sudden push or pull depending on whether it is quickly slowing down, or speeding up.
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10:23:00 ** When the speedometer is moving fast, as for example when you are first starting out in a hurry, in a powerful vehicle, you feel yourself pushed back in your seat. The speedometer can also move quickly when you press hard on the brakes. In this case you tend to feel pressed forward toward the steering wheel. **
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RESPONSE --> ok
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10:23:19 What does the speed of the speedometer needle tell us?
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RESPONSE --> It tells us how fast the car is currently moving.
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10:24:17 ** The position on the speedometer tells us how fast we are moving--e.g., 35 mph, 50 mph. However the speed of the needle tells us at what rate we are speeding up or slowing down--specifically the speed of the speedometer needle gives us the rate at which velocity changes. **
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RESPONSE --> I think I answered too quickly. I obviously responded what the speedometer tells us, but yes, the needle would indicate the rate at which the velocity of the vehicle is changing.
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