course Phy 121 ¢ÍËõÐx«ÆþÃzêј£Ö¼Õèw¬assignment #004
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13:24:30 `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 --> it changed 20m/s in 4 seconds so it is changing 5m/s per second confidence assessment: 3
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13:25:38 `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 --> a more powerful car would be capable of a greater rate of velocity change of course we don't know its weight or the wind or incline or anything but we're assuming... confidence assessment: 3
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13:26:29 `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 take the change in velocity and divide by the change in time interval confidence assessment: 3
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13:28:01 `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 --> we get the meters divided by the seconds squared confidence assessment: 1
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13:29:10 `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 --> it changed -15 m/s in 5 seconds so it would be -3m/s per second confidence assessment: 3
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13:30:51 `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: 3
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13:32:19 `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 --> in 2 seconds the runner accelerated by 3m/s per second so his avg acceleration is 1.5m/s per second confidence assessment: 3
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13:33:02 `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 per sec `dt= 2 seconds confidence assessment: 3
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13:33:22 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> 1.5m/s per second confidence assessment: 3
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13:34:18 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 forgot the multiplication and therefore forgot to squared the seconds. self critique assessment: 2
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13:37:44 `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 --> rise is 3m/s run is 2 sec slope represents the rate at which the runner is averaging on acceleration slope is 1.5......seems like there should be more to this... confidence assessment: 2
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13:38:24 `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 would indicate a greater rate of avg acceleration confidence assessment: 3
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13:40:19 `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 --> it would be increasing steadily at first, and then increasing still but at a decreasing rate confidence assessment: 3
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13:42:12 `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 --> its acceleration would be increasing steady first, then its avg acceleration would be decreasing confidence assessment: 2
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