course Phy 121 狫x{ۥ\assignment #004
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11:43:01 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 --> The symbol for final velocity is vf, and in this particular situation the units of vf are meters per second (m/s). The symbol for displacement is 'ds (or delta s), and in this particular case the units of 'ds are meters. confidence assessment: 1
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11:45:02 **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 was confused as to what the question was asking. I thought it was asking me for the symbols of variables and the units the resulting answer of that variable are in. Instead you wanted me to say how the problem was supposed to be solved in terms of symbols. self critique assessment: 2
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13:11:53 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> Average acceleration is change in velocity over change in time. Displacement is associated with determining the average velocity. These are related because the average velocity and the change in velocity are closely linked. confidence assessment: 0
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13:13:23 ** 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 didn' t associate the first equation listed in the solution as uniform acceleration. I didn't realize I needed to include equations in my answer. self critique assessment: 2
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13:17:24 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> The top of the flow chart begins with v0, vf, and 'dt. There are black lines connecting the v0 and vf to vAve in a triangular fashion (vAve = (vo+vf)/2). There are blue lines connecting 'dt and vAve to 'ds in a triangular fashion ('ds = vAve*'dt = (vf+v0)/2 * 'dt). This chart describes uniform acceleration. confidence assessment: 1
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13:17:41 ** The first level in the diagram would contain `dt, v0 and vf. Then v0 and vf would connect to `dv in the second level. The second level would also contain vAve, connected from vf in the first level to v0 in the first level. The third level would contain an a, connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, connected to vAve in the fourth level and `dt in the first level. **
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RESPONSE --> I got the correct answer. self critique assessment: 3
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13:18:58 Query Add any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The relationship between acceleration and velocity is quite complicated. I would be most appreciative of comments concerning better ways of distinguishing between the two. confidence assessment: 2
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13:19:32 ** Student Response: I think I really confused what information stood for what in the Force and Pendulum Experiment. However, I enjoy doing the flow diagrams. They make you think in a different way than you are used to. INSTRUCTOR NOTE: These diagrams are valuable for most people. Not all--it depends on learning style--but most. **
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RESPONSE --> Diagrams are definitely helpful. self critique assessment: 2
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QͲ[Oӕ}ؙw assignment #004 004. Acceleration Physics I 06-14-2007
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11:50:55 `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 --> Here, I think we are talking in terms of average acceleration. Therefore the formula for average acceleration is: Average accel. = (v2-v1)/(t2-t1) or change in velocity over time Therefore, I plug in the numbers of the information I have been given into this formula. Average accel. = (25m/s-5m/s)/(4s) = 5 m/s confidence assessment: 1
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11:51:29 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 got the correct answer. self critique assessment: 3
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12:09:07 `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 --> Automobiles serve as a primary form of transportation for most people in this area. Therefore we need them to reach our desired destination in an efficient and quick manner. So generally, when we put our foot down on the accelerator, we want to go at a quicker speed than the one at which we are currently traveling. For example if your car can go from 0mph to 60mph in about 3 seconds then that car is considered to have an efficient engine. We highly value the rates and which we can speed up and slow down quickly because on occasion this can prevent an accident. In general I would assume a more powerful engine would be capable of producing a greater rate of velocity change. But, this also depends on the make of the car. Having a larger engine could actually slow you down unless the body of the car is built to compensate for this sort of thing. confidence assessment: 1
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12:10:49 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 --> Although I did not state everything that is shown in this answer, I think I was on the right track of the correct point that was to me made. I got the part right about the more powerful engine would produce a greater change in velocity. self critique assessment: 2
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12:18:26 `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 --> Take, for example, the following problem: An object is observed for 8 seconds, during which time its velocity is seen to increase steadily from 5 meters per second to 11 meters per second. On the average, by how many meters per second does velocity increase, per second, while being observed? Of course to solve this problem we would calculate the change in velocity, and say that the velocity changed 6 m/s in 8 seconds. Per second, this is an increase of 6/8 meters per second, or .75 meters per second per second. confidence assessment: 1
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12:19:21 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 --> I knew there was a simpler way of explaining this. The best way for me to explain it here was in terms of a problem. self critique assessment: 2
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12:21:11 `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 --> I am pretty sure that when you multiply (meter/second) * (1/second) the resulting answer is meter/seconds squared (m/(s^2)). confidence assessment: 1
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12:21:36 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 --> I got the correct answer. self critique assessment: 3
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12:27:01 `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 --> vAve = (vf+vo)/2 = (-5+10)/2 = 2.5 m/sec This change happens over a period of 5 seconds. Therefore 2.5/5 = .5 The average rate the velocity is changing is .5m/s/s. confidence assessment: 1
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12:29:43 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 --> I'm not quite sure where I went wrong here. Any assistance you can provide would be most appreciated. Where did -15 come from, because I think my calculations went downhill from that part of the problem. self critique assessment: 2
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12:35:58 `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 --> To determine the average velocity we use: vAve = 'ds/'dt OR vAve = (Vf+Vo)/2 So I think the correct expression is: Average rate of change of velocity = vAve/time confidence assessment: 0
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12:37:21 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 --> My first thought was that this was supposed to be the acceleration equation, but the question was asked in terms of velocity so I thought my answer had to be in terms of velocity and that was where I made my mistake. self critique assessment: 2
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12:40:33 `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 --> aAve = 'dv/'dt = (9m/s-6m/s)/(3.5s-1.5s) = 1.5 m/s/s confidence assessment: 1
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12:43:48 `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 = 3m/s 'dt = 2s confidence assessment: 1
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12:45:02 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> aAve = 'dv/'dt = (9-6)/(3.5-1.5) = 1.5 m/s/s confidence assessment: 1
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12:45:38 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 messed up the units, but I think everything else was right. self critique assessment: 3
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12:50:20 `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 --> Run is 2 and represents change in time. Rise is 3 and represents change in velocity. Slope is 3/2 and represents the change in velocity over time between these two points. confidence assessment: 1
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12:51:07 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 did not include units in my answer, but I had all of the numbers right. self critique assessment: 2
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12:53:32 `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 definition of acceleration is change in velocity over change in time. Therefore, a greater slope implies greater acceleration because the change in velocity is occuring more quickly over a shorter amount of time. confidence assessment: 1
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12:53:50 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 --> My answer is at least partially correct. self critique assessment: 3
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12:56:32 `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 --> Overall, the velocity is increasing at a decreasing rate. The graph will look like a hill that plateaus. As the car gets closer to the bottom of the hill, it is going to slow down and eventually stop because of the air resistance. confidence assessment: 1
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12:59:12 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 --> I should have made this a more detailed description and described each section of the line as opposed to the relationship as a whole. To avoid error, I need to consider what the question is asking, and then consider what is happening with the line. self critique assessment: 2
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13:05:35 `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 --> Okay, very generally, the graph will start with a slanted line ( / ); and at the top of this line will start another slanted line (of course it is a continuous graph) that isn't angled nearly as much as the first segment. Think of the letter L rotated 180 degrees to the right with the top portion of the letter pulled slightly upward. The first part of the L increases at a constant rate, and the second part of the L increases at a decreasing rate. confidence assessment: 1
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13:08:08 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 --> I'm not sure where I messed up here. Any assistance you can provide would be most appreciated. I think my description is accurate, but now after reading the solution I'm not sure. self critique assessment: 2
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course Phy 121 狫x{ۥ\assignment #004
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11:43:01 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 --> The symbol for final velocity is vf, and in this particular situation the units of vf are meters per second (m/s). The symbol for displacement is 'ds (or delta s), and in this particular case the units of 'ds are meters. confidence assessment: 1
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11:45:02 **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 was confused as to what the question was asking. I thought it was asking me for the symbols of variables and the units the resulting answer of that variable are in. Instead you wanted me to say how the problem was supposed to be solved in terms of symbols. self critique assessment: 2
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13:11:53 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> Average acceleration is change in velocity over change in time. Displacement is associated with determining the average velocity. These are related because the average velocity and the change in velocity are closely linked. confidence assessment: 0
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13:13:23 ** 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 didn' t associate the first equation listed in the solution as uniform acceleration. I didn't realize I needed to include equations in my answer. self critique assessment: 2
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13:17:24 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> The top of the flow chart begins with v0, vf, and 'dt. There are black lines connecting the v0 and vf to vAve in a triangular fashion (vAve = (vo+vf)/2). There are blue lines connecting 'dt and vAve to 'ds in a triangular fashion ('ds = vAve*'dt = (vf+v0)/2 * 'dt). This chart describes uniform acceleration. confidence assessment: 1
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13:17:41 ** The first level in the diagram would contain `dt, v0 and vf. Then v0 and vf would connect to `dv in the second level. The second level would also contain vAve, connected from vf in the first level to v0 in the first level. The third level would contain an a, connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, connected to vAve in the fourth level and `dt in the first level. **
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RESPONSE --> I got the correct answer. self critique assessment: 3
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13:18:58 Query Add any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The relationship between acceleration and velocity is quite complicated. I would be most appreciative of comments concerning better ways of distinguishing between the two. confidence assessment: 2
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13:19:32 ** Student Response: I think I really confused what information stood for what in the Force and Pendulum Experiment. However, I enjoy doing the flow diagrams. They make you think in a different way than you are used to. INSTRUCTOR NOTE: These diagrams are valuable for most people. Not all--it depends on learning style--but most. **
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RESPONSE --> Diagrams are definitely helpful. self critique assessment: 2
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QͲ[Oӕ}ؙw assignment #004 004. Acceleration Physics I 06-14-2007
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11:50:55 `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 --> Here, I think we are talking in terms of average acceleration. Therefore the formula for average acceleration is: Average accel. = (v2-v1)/(t2-t1) or change in velocity over time Therefore, I plug in the numbers of the information I have been given into this formula. Average accel. = (25m/s-5m/s)/(4s) = 5 m/s confidence assessment: 1
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11:51:29 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 got the correct answer. self critique assessment: 3
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12:09:07 `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 --> Automobiles serve as a primary form of transportation for most people in this area. Therefore we need them to reach our desired destination in an efficient and quick manner. So generally, when we put our foot down on the accelerator, we want to go at a quicker speed than the one at which we are currently traveling. For example if your car can go from 0mph to 60mph in about 3 seconds then that car is considered to have an efficient engine. We highly value the rates and which we can speed up and slow down quickly because on occasion this can prevent an accident. In general I would assume a more powerful engine would be capable of producing a greater rate of velocity change. But, this also depends on the make of the car. Having a larger engine could actually slow you down unless the body of the car is built to compensate for this sort of thing. confidence assessment: 1
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12:10:49 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 --> Although I did not state everything that is shown in this answer, I think I was on the right track of the correct point that was to me made. I got the part right about the more powerful engine would produce a greater change in velocity. self critique assessment: 2
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12:18:26 `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 --> Take, for example, the following problem: An object is observed for 8 seconds, during which time its velocity is seen to increase steadily from 5 meters per second to 11 meters per second. On the average, by how many meters per second does velocity increase, per second, while being observed? Of course to solve this problem we would calculate the change in velocity, and say that the velocity changed 6 m/s in 8 seconds. Per second, this is an increase of 6/8 meters per second, or .75 meters per second per second. confidence assessment: 1
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12:19:21 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 --> I knew there was a simpler way of explaining this. The best way for me to explain it here was in terms of a problem. self critique assessment: 2
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12:21:11 `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 --> I am pretty sure that when you multiply (meter/second) * (1/second) the resulting answer is meter/seconds squared (m/(s^2)). confidence assessment: 1
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12:21:36 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 --> I got the correct answer. self critique assessment: 3
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12:27:01 `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 --> vAve = (vf+vo)/2 = (-5+10)/2 = 2.5 m/sec This change happens over a period of 5 seconds. Therefore 2.5/5 = .5 The average rate the velocity is changing is .5m/s/s. confidence assessment: 1
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12:29:43 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 --> I'm not quite sure where I went wrong here. Any assistance you can provide would be most appreciated. Where did -15 come from, because I think my calculations went downhill from that part of the problem. self critique assessment: 2
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12:35:58 `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 --> To determine the average velocity we use: vAve = 'ds/'dt OR vAve = (Vf+Vo)/2 So I think the correct expression is: Average rate of change of velocity = vAve/time confidence assessment: 0
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12:37:21 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 --> My first thought was that this was supposed to be the acceleration equation, but the question was asked in terms of velocity so I thought my answer had to be in terms of velocity and that was where I made my mistake. self critique assessment: 2
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12:40:33 `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 --> aAve = 'dv/'dt = (9m/s-6m/s)/(3.5s-1.5s) = 1.5 m/s/s confidence assessment: 1
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12:43:48 `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 = 3m/s 'dt = 2s confidence assessment: 1
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12:45:02 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> aAve = 'dv/'dt = (9-6)/(3.5-1.5) = 1.5 m/s/s confidence assessment: 1
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12:45:38 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 messed up the units, but I think everything else was right. self critique assessment: 3
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12:50:20 `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 --> Run is 2 and represents change in time. Rise is 3 and represents change in velocity. Slope is 3/2 and represents the change in velocity over time between these two points. confidence assessment: 1
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12:51:07 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 did not include units in my answer, but I had all of the numbers right. self critique assessment: 2
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12:53:32 `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 definition of acceleration is change in velocity over change in time. Therefore, a greater slope implies greater acceleration because the change in velocity is occuring more quickly over a shorter amount of time. confidence assessment: 1
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12:53:50 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 --> My answer is at least partially correct. self critique assessment: 3
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12:56:32 `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 --> Overall, the velocity is increasing at a decreasing rate. The graph will look like a hill that plateaus. As the car gets closer to the bottom of the hill, it is going to slow down and eventually stop because of the air resistance. confidence assessment: 1
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12:59:12 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 --> I should have made this a more detailed description and described each section of the line as opposed to the relationship as a whole. To avoid error, I need to consider what the question is asking, and then consider what is happening with the line. self critique assessment: 2
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13:05:35 `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 --> Okay, very generally, the graph will start with a slanted line ( / ); and at the top of this line will start another slanted line (of course it is a continuous graph) that isn't angled nearly as much as the first segment. Think of the letter L rotated 180 degrees to the right with the top portion of the letter pulled slightly upward. The first part of the L increases at a constant rate, and the second part of the L increases at a decreasing rate. confidence assessment: 1
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13:08:08 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 --> I'm not sure where I messed up here. Any assistance you can provide would be most appreciated. I think my description is accurate, but now after reading the solution I'm not sure. self critique assessment: 2
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