course Phy201 Radius questionapprox. uncertainty in area of circle given radius 2.8 * 10^4 cm
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09:26:00 `q001. Note that there are 11 questions in this assignment. vAve = `ds / `dt, which is the definition of average velocity and which fits well with our intuition about this concept. If displacement `ds is measured in meters and the time interval `dt is measured in seconds, in what units will vAve be obtained?
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RESPONSE --> vAve will be obtained in the units m/sec confidence assessment: 3
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09:27:39 vAve = `ds / `dt. The units of `ds are cm and the units of `dt are sec, so the units of `ds / `dt must be cm / sec. Thus vAve is in cm/s.
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RESPONSE --> I think the question stated that the units of 'ds were m not cm. However if the units of 'ds were cm and the units of 'dt were sec then vAve would be obtained in cm/sec. self critique assessment: 3
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09:28:22 `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured?
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RESPONSE --> 'ds must be measured in cm confidence assessment: 3
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09:28:42 Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm.
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RESPONSE --> seconds are eliminated leaving only cm self critique assessment: 3
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09:29:27 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> when we multiply cm/sec by sec the sec cancel each other out leaving only cm confidence assessment: 2
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09:30:43 When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. When we multiply fractions we will multiply numerators and denominators. We obtain cm * sec / ( sec * 1). This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm.
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RESPONSE --> so they end up canceling each other self critique assessment: 2
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09:31:20 `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt = `ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured?
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RESPONSE --> 'dt must be measured in sec confidence assessment: 3
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09:31:49 Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds.
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RESPONSE --> km is canceled out self critique assessment: 2
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09:33:27 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> km/sec * km/1 we then multiply our numerators km*km and our denominators sec*1 our km cancel out leaving us with only sec confidence assessment: 3
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09:34:33 The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec.
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RESPONSE --> I understand this process self critique assessment: 3
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09:37:02 `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity?
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RESPONSE --> vAve = 'ds/'dt vAve = 6m/3sec vAve = 2cm/sec change in 'ds is 6m change in 'dt is 3sec confidence assessment: 3
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09:37:36 We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and your reasoning.
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RESPONSE --> my answer was correct self critique assessment: 3
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09:38:59 `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time?
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RESPONSE --> change in position 'ds s2-s1 = 'ds change in clock time 'dt t2-t1 = 'dt confidence assessment: 2
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09:40:10 We see that the change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. What expression therefore symbolizes the average velocity between the two clock times.
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RESPONSE --> average velocity between the two clock times vAve = 'ds/'dt self critique assessment: 2
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09:43:45 `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent?
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RESPONSE --> the rise of the triangle is 6 and it represents 6 meters of displacement the run of the triangle is 3 and it represents 3 seconds of time confidence assessment: 2
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09:44:03 The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s.
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RESPONSE --> pretty close to what I said self critique assessment: 2
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09:45:18 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> the slope of the triangle is 2 and it represents the vAve of the object which is 2cm/sec confidence assessment: 3
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09:46:27 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> This is what I said. Question: I thought slope was to be listed without units? self critique assessment: 2
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09:47:54 `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity?
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RESPONSE --> well in logical terms the greater the slope, the more speed the object will be able to gather. mathmatically, the slope represents the vAve confidence assessment: 2
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09:48:10 Since the rise between two points on a graph of velocity vs. clock time represents the change in `ds position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position /change in clock time, or `ds / `dt. This is the definition of average velocity.
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RESPONSE --> this is what I stated self critique assessment: 3
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09:50:28 `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time. If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing?
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RESPONSE --> a graph of the position of a car from rest down a hill would be increasing at an increasing rate. the slope of the graph is increasing confidence assessment: 2
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09:50:45 The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point). The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing slope is said to be increasing at an increasing rate.
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RESPONSE --> I was correct self critique assessment: 3
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\[ٚކq˕֬P assignment #004 004. Acceleration Physics I 11-07-2008
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11:49:18 `q001 Note that there are 10 questions in this assignment. At a certain instant the speedometer of a car reads 5 meters / second (of course cars in this country generally read speeds in miles per hour and km per hour, not meters / sec; but they could easily have their faces re-painted to read in meters/second, and we assume that this speedometer has been similarly altered). Exactly 4 seconds later the speedometer reads 25 meters/second (that, incidentally, indicates very good acceleration, as you will understand later). At what average rate is the speed of the car changing with respect to clock time?
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RESPONSE --> 25m/s - 5m/s = 20m/s 4s - 0s = 4s 20m/s / 4s = 5m/s confidence assessment: 3
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11:49:54 The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time? The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second.
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RESPONSE --> I understand this self critique assessment: 3
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11:53:49 `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change?
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RESPONSE --> The significance for an automobile's rate of velocity change to be greater rather than less is huge in our society. It seems everyone wants to be the most powerful. Yes, I do think that a car with a more powerful engine would be capable of a greater rate of velocity change providing you kept all variables the same such as weight of the automobile. confidence assessment: 3
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11:54:17 A car whose velocity changes more rapidly will attain a given speed in a shorter time, and will be able to 'pull away from' a car which is capable of only a lesser rate of change in velocity. A more powerful engine, all other factors (e.g., weight and gearing) being equal, would be capable of a greater change in velocity in a given time interval.
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RESPONSE --> exactly what I said self critique assessment: 3
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11:55:24 `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed.
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RESPONSE --> because we divide our vAve (m/s) by 'dt (s) giving us (m/s) / (s) confidence assessment: 3
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11:56:10 When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second.
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RESPONSE --> My mistake. I wrote vAve instead of the change in velocity which is what I meant. self critique assessment: 3
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11:57:15 `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions?
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RESPONSE --> (m/s) * (1/s) = m/s^2 confidence assessment: 2
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11:57:35 Multiplying the numerators we get meters * 1; multiplying the denominators we get second * second, which is second^2. Our result is therefore meters * 1 / second^2, or just meters / second^2. If appropriate you may at this point comment on your understanding of the units of the rate of change of velocity.
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RESPONSE --> This is what I stated self critique assessment: 3
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11:58:34 `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing?
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RESPONSE --> 15m/s / 5s = 3m/s^2 confidence assessment: 3
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11:59:29 We see that the velocity changes from 10 meters/second to -5 meters/second, a change of -15 meters / second, during a five-second time interval. A change of -15 m/s during a 5 s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2. This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3 m/s every second.
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RESPONSE --> -3 not 3 self critique assessment: 2
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12:00:47 `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes?
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RESPONSE --> 'dv / 'dt confidence assessment: 1
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12:01:17 The average rate would be expressed by [ave rate of velocity change with respect to clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to clock time] is pretty cumbersome so we give it a name. The name we give it is 'average acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write down the definition of average acceleration. The definition of average acceleration is aAve = `dv / `dt. Please make any comments you feel appropriate about your understanding of the process so far.
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RESPONSE --> I understand this self critique assessment: 3
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12:03:05 `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times?
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RESPONSE --> 9m/s - 6m/s = 3m/s 3.5s - 1.5s = 2s 3m/s / 2s = 1.5ms^2 confidence assessment: 3
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12:03:49 `q006a. What is the change `dv in the velocity of the runner during the time interval, and what is the change `dt in clock time during this interval?
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RESPONSE --> 'dv is 3m/s 'dt is 2s confidence assessment: 3
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12:04:17 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> 3m/s / 2s = 1.5m/s^2 confidence assessment: 3
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12:06:55 We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval their runs from t = 1.5 sec to t = 3.5 sec so that the duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s. The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2. Please comment if you wish on your understanding of the problem at this point.
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RESPONSE --> I understand this self critique assessment: 3
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12:09:06 `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent? What is the rise between these points what does it represent? What is the slope between these points what does it represent?
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RESPONSE --> The run is 2 and it represents 2 seconds of time passing The rise is 3 and it represents 3m/s of increased velocity The slope is 3m/s / 2s = 1.5m/s^2 and it represents the vAve confidence assessment: 3
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12:10:20 The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point.
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RESPONSE --> I understand this self critique assessment: 3
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12:11:31 `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration?
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RESPONSE --> The slope of the graph is the vAve Greater slope implies greater acceleration because it allows the object to build velocity faster confidence assessment: 3
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12:11:54 Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration.
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RESPONSE --> I understand this self critique assessment: 3
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12:14:13 `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> The first part of the graph would be increasing at a constant rate, the second part of the graph would be increasing at a constant rate as well, but the slope of this part of the graph would not be as great as the first part. confidence assessment: 2
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12:15:45 Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors.
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RESPONSE --> This makes sense, the fact that it would eventually level off. I understand where I went wrong in my thinking. self critique assessment: 3
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12:16:35 `q010. An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of acceleration vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> At first increasing at a constant rate, then decreasing at a constant rate. confidence assessment: 3
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12:23:06 Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis. In answer to the following question posed at this point by a student: Can you clarify some more the differences in acceleration and velocity? ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec. Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2. Velocity is the slope of a position vs. clock time graph. Acceleration is the slope of a velocity vs. clock time graph. **
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RESPONSE --> This all makes sense self critique assessment: 3
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qxnݷ亮Jɪs assignment #003 003. `Query 3 Physics I 11-07-2008
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11:28:19 Query Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> 1.80m + 142.5cm + 5.34 * 10^5 micro m 1.80m + (142.5cm/(100cm/1m)) + 5.34 * 10^5 micro m 1.80m + 1.43m + 534000 micro m 3.23m + (534000micro m/(1000000 micro m/1m)) 3.23m + .53m 3.76m confidence assessment: 3
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11:32:42 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore nothing smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, accurate to within .00000001 m. When these are added you get 3.22500534 m; however the 1.80 m is not resolved beyond .01 m so the result is 3.23 m. Remaining figures are meaningless, since the 1.80 m itself could be off by as much as .01 m. **
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RESPONSE --> This all makes sense, however, why would you not convert the micro m to m just as you did with the cm giving 3.76 instead of 3.23? self critique assessment: 3
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yJc۳\N assignment #004 004. `Query 4 Physics I 11-07-2008
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Intro Prob 6 given init vel, accel, `dt find final vel, dist If initial velocity is v0, acceleration is a and time interval is `dt, then in symbols what are the final velocity vf and the displacement `ds?
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RESPONSE --> a * 'dt = 'dv 'dv + v0 = vf (vf + v0) / 2 = vAve vAve * 'dt = 'ds confidence assessment: 3
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12:53:43 **You would use accel. and `dt to find `dv: a * `dt = `dv. Adding `dv to initial vel. vo you get final vel. Then average initial vel. and final vel. to get ave. vel.: (v0 + vf) / 2 = ave. vel. You would then multiply ave. vel. and `dt together to get the distance. For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s: 3 m/s^2 * 5 s = 15 m/s = `dv 15 m/s + 3 m/s = 18 m/s = fin. vel. (18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve 10.5 m/s * 5 s = 52.5 m = dist. In more abbreviated form: a * `dt = `dv v0 + `dv = vf (vf + v0) /2 = vAve vAve * `dt = `ds so `ds = (vf + v0) / 2 * `dt. **
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RESPONSE --> I understand this self critique assessment: 3
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12:55:14 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> (v0 +vf) / 2 = vAve vAve * 'dt = 'ds confidence assessment: 2
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12:55:46 ** Since accel is uniform vAve = (v0 + vf) / 2. Thus displacement is `ds = vAve * `dt = (v0 + vf) / 2 * `dt, which is the first equation of uniformly accelerated motion. **
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RESPONSE --> I think this is pretty much what I had written self critique assessment: 2
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12:57:54 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> top level 'dt, v0, and vf second level dv from v0 and vf, vAve from v0 and vf third level a from 'dt and dv, 'ds from 'dt and aAve confidence assessment: 2
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