course Phy 201 ~ԢUUassignment #004
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21:55:42 `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 --> The speed of the car is changing at an average rate of 5 m/s. Average rate = 25 m -5 m = 20 m/ 4s = 5 m/s confidence assessment: 2
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21:56:46 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 --> Ok! I forgot to mention that is is changing by 5m/s every second. self critique assessment: 3
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21:58:27 `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 a car with a more powerful engine is capable of a greater rate of velocity change because it can cover more meters in less time as opposed to an old car that is barely creeping along because the engine has little power. confidence assessment: 2
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21:58:57 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 --> ok self critique assessment: 3
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22:03:28 `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 --> We obtained meters/second/second because the car changed by a speed of 20 m/s in 4 seconds. When we calculate this out we see that the car's speed is changing at a rate of 5 m/s/s. This means that every second the car is changing by a speed of 5 meters per second. confidence assessment: 3
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22:03:41 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 --> ok self critique assessment: 3
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22:04:57 `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^2 confidence assessment: 1
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22:05:42 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 understand the units of the rate of change of velocity. self critique assessment: 3
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22:07:14 `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 --> 3 m/s/s. The average rate of velocity change is 3 meters/second every second. 10 m/s - -5m/s = 15 m/s / 5 s = 3 m/s/s confidence assessment: 2
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22:08:41 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 don't really understand why the answer is negative because when you subtract a negative number, two negatives make a positive. self critique assessment: 1
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22:10:57 `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 --> `ds = `dv/ `dt confidence assessment: 2
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22:12:07 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 --> oh ok. . I understand! self critique assessment: 3
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22:14:20 `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 = 1.28 m/s/s 9m/s - 6m/s = 3 m/s 3.5s -1.15s = 2.35s 3m/s / 2.35 s = 1.28 m/s/s confidence assessment: 3
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22:15:25 `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 = 9 m/s -6m/s = 3m/s `dt = 3.5 s - 1.15s = 2.35 s confidence assessment: 3
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22:16:11 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> `dv/`dt = 3m/s / 2.35s = 1.28 m/s/s confidence assessment: 3
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22:17:03 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 misread the data. Instead of 1.5s I thought it said 1.15 seconds. However, I did the concept right. self critique assessment: 2
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22:22:01 `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 = 3.5 sec - 1.5 sec = 2.0 sec The run represents the difference in the two hoorizontal coordinates. This tells us the change in velocity. Rise = 9 meters - 6 meters = 3 meters The rise represents the difference in the two vertical coordinates. This tells us the change in time The slope is 3/2 = 1.5 m/s The slope represents the average rate of change of velocity in respect to clock time. confidence assessment: 3
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22:22:24 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 self critique assessment: 3
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22:24:44 `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 higher the slope the greater the acceleration because the slope is telling us the average rate of change in velocity or the average acceleration of the object. A higher slope tells us that the object is changing at a faster rate in respect to clock time. This concept was also seen several times previously with the marble experiments. confidence assessment: 3
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22:24:58 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 --> ok self critique assessment: 3
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22:28:09 `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 graph would show the velocity increasing at a decreasing rate. This would look like a curved line going upward from left to right. confidence assessment: 2
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22:30:24 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 failed to mention that I labeled the x-axis at time and y-axis as velocity. I simply took for granted that this was known because we have set up every graph in this manner. I also need to be more precise about the constant rate in the beginning instead of giving an overall explanation of the graph. self critique assessment: 2
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22:34:02 `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 --> When drawing the graph the x-axis is labeled as time and the y-axis is labeled as velocity. The graph begins with the point 0. At first the graph shows a constant increase in velocity over time. Then the graph shows velocity increasing at a decreasing rate. This looks like a curved upward line from left to right. confidence assessment: 3
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22:35:56 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 understand. I again read the question wrong. I think I'm having a bad day! self critique assessment: 2
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