course Phy201 Æi˜~Î¥ô}ÞÀ€ŽèèTêÂÊ¥assignment #004
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21:38:48 `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 --> 5 meters/second is the average rate that the car is changing with respect to clock time. confidence assessment: 1
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21:40:03 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 answered the question correctly and understood this concept. self critique assessment: 3
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21:44:24 `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 --> Yes, i believe a car with a more powerful engine would be most capable of a greater rate of velocity change. In terms that we are talking in speed, in which we are looking at how fast the object is moving and in what direction. confidence assessment: 1
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21:46:31 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 --> I understand this....to me we are talking more about acceleration anyways. I agree you would want a car that could change his velocity very drastically in a short amount of time. self critique assessment: 3
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21:49:39 `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 --> You look at the intial given meters/second over the ending distance and time interval to get the calculation ( the change). confidence assessment: 1
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21:50:57 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 --> Thats what I was trying to explain....the change in velocity is the initial divided by the time elapsed. self critique assessment: 3
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21:52:22 `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 --> ???Meters/second^2 confidence assessment: 0
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21:52:53 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 answered the question correctly. self critique assessment: 3
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21:54:37 `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 --> The average rate of the changing velocity is -3m/s. confidence assessment: 1
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21:55:09 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 --> self critique assessment: 3
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21:59:11 `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 --> Ave'dv ='dv/ 'dt confidence assessment: 0
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22:00:49 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 believe I was pretty close to that.....i didn't know that was going to be the formula for Average acceleration. self critique assessment: 3
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22:03:30 `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= 3/2 m/s confidence assessment: 1
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22:04:26 `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= 2 s confidence assessment: 1
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22:05:10 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> I thought the aAve= 3/2 m/s confidence assessment: 1
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22:05:49 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 believe I understood the question pretty well and answered it correctly. self critique assessment: 3
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22:08:04 `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 in this experiment will be 3 meters, represents the x axis on the graph (horizontal). The rise represents the vertical y axis of 2 seconds on this graph. the slope between these two points for the graph is 2/3. confidence assessment: 1
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22:09:35 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 --> Yes, I got the rise and run mixed up on the graph that I was constructing in my head. I had the concept of the two down just used the wrong values, got the 'dv and 'dt mixed up. self critique assessment: 3
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22:11:16 `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 --> If I am correct I believe the slope and the aAve were the same values. A steeper slope is showing that the speed increase in a shorter amount of time. confidence assessment: 1
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22:12:02 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 --> The definiton of average acceleration is 'dv/ 'dt self critique assessment: 3
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22:15:25 `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 --> Well at first you're going to see a line moving right and up with a slight incline to it on the graph then the line is going to flatten out a bit but still moving towards the right but only slight up now and that pattern continues throughout. So at first...increasing at an increasing rate and then increasing at a constant rate to the end confidence assessment: 1
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22:17:33 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 --> My wording was a little off in this one, but I understood the concept of the scenario. I too never saw the graph as decreasing even thought the velocity changes due to air resistance. I did in my views saw the graph increasing at an increasing rate at first, being that it was going down an incline. self critique assessment: 2
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22:20:45 `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 --> With there being a constant down hill slope the acceleration remains at a constant velocity mostly thoughout the distance traveled until the car reaches a certain velocity, which then becomes affected by air resistance, but never decreases in velocity due to the constant slope of the downhill. Increasing at a constant rate through-out confidence assessment: 1
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22:23:42 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 guess in some ways i can see the graph somewhat decreasing at somepoint. The car should remain a constant velocity through a good part of the down hill trip due to the slope of the incline. As the air resistance built up due to the speed of the car, I guess i can see that the aAve decreasing. self critique assessment: 2
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