course PHY201-4 I'm having some trouble in the explanation part of the assignments. I understand formulae but transposing these into words for what various questions are asking for is somewhat confusing. This video assignment wasn't too bad to understand but the one on the class notes was somewhat confusing. I'll be sending that files later
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14:25:18 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> We reason out the process by making a flow diagram., but I'm not exactly sure how with what's given
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14:34:59 ** 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 but I think if I could see this in the flow chart I might understand a little better
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14:41:42 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 --> Where did the approximate velocity come from when all we have talked about is change, initial, and final velocities. The positions can be plotted on the graph knowing the x axis is the time and the y axis is the displacement. These points reveal the average velocity The average acceleration will how long it takes the velocity to change positions.
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14:43:49 ** 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 --> I think I've missed the point again in understanding what's being asked for. This is going to take some work.
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14:47:51 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 --> By the observation of the stream of water exiting from the hole while compairing to a specific point you can visually realize the acceleration being constant
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14:51:17 ** 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 --> I need to calculate the rate of change in the bottle as the water leaves while timing the points?
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14:53:03 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 --> On the graph, the plotted points will make it real clear especially when drawing the trapezoid (triangle) because the average velocity is the sum of the initial and final velocities.
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14:53:35 ** 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 I'm good with that
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14:54:36 Why does a linear velocity vs. time graph give a curved position vs. time graph?
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RESPONSE --> Linear velocity shows an increasing velocity per time interval
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14:56:24 ** 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 but I don't think I'm associating the material in the notes very well
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14:58:54 ** 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 --> I associate the avg velocity to the midpoint because the midpoint of the time is the same point as an average would be taken from. The avg velocity is the line drawn upward from the midpoint on the time intervalline
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_vHͼ~ assignment #002 Ӎ|܆Ꝫwcz Physics I Vid Clips 02-03-2008
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15:55:02 Given a graph of velocity vs. time how do we determine the rate which velocity increases?
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RESPONSE --> You determine the velocity increase or changes by finding the difference between the two velocity points in queston for the time frame. avg rate of velocity equals change in velocity divided by the change in time
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15:57:04 ** 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 close to what I said
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15:59:16 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 --> vAve='dv/'dt This formula will comopute the avg change of velocity for the two given points.
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16:00:23 ** We divide the difference `dv in velocities by the duration `dt of the time interval. **
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RESPONSE --> OK I think that's about what I said
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16:02:14 Does the rate at which a speedometer needle moves tell us how fast the vehicle is moving?
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RESPONSE --> I really don't think so because if you have been driving 60 mph for 2 hours, the speedometer isn't going to be moving any but the vehicle will still be moving
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16:03:08 ** 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 I'm good
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16:04:38 Does a stationary speedometer needle implies stationary vehicle?
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RESPONSE --> Yes if the vehicle speedometer is zero but if the needle is stationary at 15 mph then the vehicle is moving.
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16:06:02 ** 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 I'm doing good on this set so far
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16:08:05 Does a quickly moving speedometer needle imply a quickly moving vehicle?
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RESPONSE --> I would say that it's an indication of increasing acceleration of the vehicle.
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16:12:21 ** 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 I guess I was thinking that if the needle moved quickly from 0 to 60 mph that would indicate an increasnig acceleration
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16:14:03 What does it feel like inside a car when the speedometer needle is moving fast?
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RESPONSE --> I woiuld say that a force would be pushing you backwards into the seat.
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16:15:22 ** 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 I'm good with that
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16:17:51 What does the speed of the speedometer needle tell us?
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RESPONSE --> The speedometer indicates the average velocity the vehicle is traveling. 60 mph would mean the vehicle would travel 60 miles in an hour timespan
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16:21:19 ** 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 --> OK but with everything going digital how does this compare?
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}̝E assignment #002 Ӎ|܆Ꝫwcz Physics I Vid Clips 02-03-2008 Ky }Od assignment #002 Ӎ|܆Ꝫwcz Physics I Vid Clips 02-03-2008