course phy 231 004. Acceleration Physics I 02-02-2008
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17:28:20 `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 --> 25m/sec - 5 m/sec divided by 4sec = 5m/sec for the average rate confidence assessment: 2
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17:36:31 `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 --> because the engine is powerful,it wouldbe able to cause displacement in a shorter amount of time; compared to a less powerful engine. so the velocity would be greater confidence assessment: 3
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17:38:50 `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 --> meters could be calculated as the displacement caused by the car. second would be amount of time it takes to cause this displacement. and the average velocity would in turn be in m/ sec as the amount of time the car takes to move from one point to another and the distance traveled in that time. confidence assessment: 3
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17:45:00 `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 --> 1 m/sec squared confidence assessment: 3
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17:47:53 `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 --> 10m/s-(-5m/s)/ 5sec= 3m/s as the average rate of velocity confidence assessment: 3
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17:49:39 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 --> forgot to put the negative sign. and mention it can also be written as -3m/s^2 self critique assessment: 2
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17:51:21 `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 confidence assessment: 3
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17:52:22 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 --> average acceleration would have a in the front. while average velocity has v in the front self critique assessment: 2
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17:55:32 `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 --> 9m/sec-6m/sec divided by 3.5-1.5= 1.5 m/sec aAve so distance over time. confidence assessment: 3
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17:57:50 `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/t =aAve confidence assessment: 2
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17:59:24 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> dv/dt confidence assessment: 2
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18:13:40 `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 the line is v/t which equals acceleration. the greater the velocity, the more time is taken to reach it confidence assessment: 2
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18:16:44 `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 --> from the top left of the graph tothe bottom, velocity is increasing. and as it rolls down and continues on a straight line, it decreases confidence assessment: 3
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