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16:47:59 `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 --> this vehicle is traveling at 5meters/sec confidence assessment: 0
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16:48:49 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 --> got that but didnt show you how i subtracted 25-5 =20 /4 = 5meters/sec self critique assessment: 0
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17:10:56 `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 --> well the more power you have in a car the more torque and stamina you will have in pulling. as for does it give you more velocity. I don't know anything about motors and cars except how to turn one on and relatively how to drive one, but i do know the more power you have the faster you go. the greater your acceleration the greater your velocity. confidence assessment: 1
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17:11:30 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 --> im not sure if i got that right or not self critique assessment: 0
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17:13:13 `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 --> if you mulitple by meters/sec and sec then you get meters/sec/sec confidence assessment: 0
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17:13:43 02-08-2008 17:13:43 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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NOTES -------> 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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17:14:18 `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/sec^2 confidence assessment: 1
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17:14:41 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 --> yes self critique assessment: 2
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17:18:13 `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 --> 1m/sec 10-5=5 5/5=1m/s or you could say 10-(-5)=15 15/5=3m/s not really sure which of these would be right but my gut says the first answer confidence assessment: 0
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17:19:36 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 sure why the 15 is neg because 10 -(-5) is 10+5 so thats a positive 15 not a neg how did you get a neg? self critique assessment: 1
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17:24:55 `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 --> vAve='dv/'dt confidence assessment: 1
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17:26:11 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 got that but i used vAve instead of aAve self critique assessment: 1
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17:29:22 `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=the change in velocity /the elapsed time so 9-6=3m/s 3.5-1.5=2s so 3m/s / 2s = 1.5m/s^2 confidence assessment: 1
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17:30:58 `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 --> i think i answered that as 9-6=3m/s 3.5-1.5=2s confidence assessment: 1
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17:32:02 `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: 1
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17:33:57 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 --> o so i did get it self critique assessment: 2
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17:46:31 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 --> well since the rise/run is the slope and the slope is the change in y over the change in x we find the rise is 3 m/s and the run is 2 s (3,2) the slope then is 1.5 m/s^2 Im going to read to the end next time before answering any questions but at least i know i got it right self critique assessment: 3
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17:57:48 `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 a v vs t graph represents the accelreation of an object because the area under the graph is can i simply say because it does it just does confidence assessment: 0
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17:58:34 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 --> well i knew that i have to stop over thinking these things its giving me a headache self critique assessment: 1
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18:34:42 `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 --> a graph of a constant is a straight line then as the velocity decreases at an increasing rate we we dip into the right side of th egraph if it were a decrease at decreasing rate it would be to the left side. confidence assessment: 2
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18:35:49 02-08-2008 18:35:49 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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NOTES -------> 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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18:36:05 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 believe i misunderstood what you were asking for self critique assessment: 1
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19:04:14 `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 --> is this where the graph is increasing at an increasing rate then it decreases at an increasing rate. confidence assessment: 1
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19:04:53 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 --> ok self critique assessment: 0
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