course Phy 121 pËÎØÓàõ£ÛŠÔwµÛÿ²ðC¤PüWç÷ÐâÊassignment #004
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17:40:05 `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 change in speed is 20 m/s. The average rate of speed change is 20m/s / 4 sec = 5m/s/s confidence assessment: 1
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17:40:28 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 --> yeah self critique assessment: 2
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17:42:44 `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change?
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RESPONSE --> The rate of velocity change of an automobile is significant. You need to have a car that will ""get up and go"". For instance, going up an incline to get on the interstate requires the power of a good engine to get you there. If you have a slow engine, merging into traffic going 70 m/h with your car going 35 m/h can be very dangerous. confidence assessment: 2
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17:42:57 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: 2
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17:44:16 `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 --> The car's average velocity change is 20m/s. By dividing the time of 4 sec, we have the change in velocity m/s for every second. confidence assessment: 2
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17:44:37 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 --> correct. self critique assessment: 2
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17:45:47 `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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17:47:51 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 that the rate of change of velocity's units (using m and s) is the change in velocity each second that passes. Example 1m/s/s = the car accelerates 1meter/second every second that elapses. self critique assessment: 2
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17:49:19 `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 change in velocity is a total of 15m/s over 5 seconds. confidence assessment: 2
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17:50:22 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 got confused. Velocity can be negative (distance can't). I understand that the velocity change from 10 to -5 is 15m/s backwards, thus -15m/s. self critique assessment: 2
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17:51:15 `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes?
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RESPONSE --> 'dv/'dt confidence assessment: 2
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17:52:01 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'm finally getting it. self critique assessment: 2
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17:56:52 `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 --> The average acceleration from the start to 1.5 sec is the velocity change (6m/s) divided by the change in time (1.5s) = 4.5m/s/s. The average acceleration between the clock times 1.5 and 3.5 is the velocity change (9m/s - 6m/s = 3m/s) divided by the change in time (3.5s - 1.5s = 2s) = .64m/s/s confidence assessment: 1
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17:57:29 `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 had that in my answer confidence assessment: 2
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17:57:54 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> i think i had this in my answer confidence assessment: 2
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17:59:38 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 understand, I just wrote the quantities down wrong when I divided. self critique assessment: 2
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22:52:41 `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 = the elapsed time between 2 points = 3.5s - 1.5s = 2s. Rise = the displacement of the 2 points = 9m/s - 6m/s = 3m/s. The slope is the average acceleration or deceleration of these 2 points, rise/run = 3m/s / 2s = 1.5m/s/s. confidence assessment: 2
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22:53:36 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 --> rise = change in velocity not displacement. self critique assessment: 2
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22:57:10 `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 --> Greater slope does imply greater acceleration. If the velocity has a larger number and the clock time has a low number, when dividing the two they will have a larger number than if the velocity and clock time were in close range. For example, if the rise is 25cm/s and the run is 5 sec, the slope would be 5cm/s/s. That tells me that for every 25cm/s of velocity, the time moves only 5 sec. confidence assessment: 1
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22:57:34 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 --> ok. was my explanation correct? self critique assessment: 2
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23:01:27 `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 --> I think this graph (after the initial nearly constant velocity change indicating a rather straight graph line) would increase but at a decreasing rate. The car's velocity would increase but as it goes faster and has wind resistance it will continue to accelerate but at a slower pace. After the straight line, the graph would start to curve slightly downward. confidence assessment: 2
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23:03:28 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 assumed the graph would begin to curve down but realize that the car is still gaining velocity not losing resulting in a more level curve, not downward. To avoid this error, I should think about it a little more I guess. self critique assessment: 2
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23:06:03 `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 --> Again, in the beginning the constant velocity increase would result in a straight graph line. As wind resistance interferes, the graph should begin curving down and to the right. confidence assessment: 1
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23:06:40 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 --> got it. self critique assessment: 2
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