course Phy 201 zPރsȟĿыassignment #004 004. Acceleration Physics I 02-01-2008
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15:13:26 `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 --> 20 m/s confidence assessment: 1
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15:14:05 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 --> speed changes by 5 meters/second self critique assessment: 3
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15:16:19 `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 --> If you are driving and you speed up to 45 mi/hr and then must slow down for a car coming off teh off-ramp then the car is going to have a rate of which its velocity changes since it isnt going to start at 45 mi/hr or slow to a STOP when you slow down...it is gradual. I feel a more powerful engine would be capable of a greatter rate of velocity change since you have more horse power to get to a specific speed faster. confidence assessment: 2
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15:16:35 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 --> self critique assessment: 3
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15:28:04 `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 obtain meters/second by dividing the distance traveled by the elapsed time. confidence assessment: 1
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15:28:14 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 --> self critique assessment: 3
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15:29:06 `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/s confidence assessment: 3
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15:29:36 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 --> meters * 1 / second^2, or just meters / second^2 self critique assessment: 3
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15:30:30 `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 --> 1 m/s confidence assessment: 0
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15:30:54 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 --> -3 m/s every second self critique assessment: 3
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15:31:59 `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 --> triange V confidence assessment: 3
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15:33:26 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 thought it would have been triangle v since on the cahrt it says change in vel. which I made an assumption since it did not indicate it being a rate of change. self critique assessment: 3
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15:36:50 `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-6m=3m 3.5m-1.5m=2m 3/2 m/s confidence assessment: 3
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15:37:22 `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 --> 3/2 confidence assessment: 3
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15:39:23 `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 --> The run between the two points represents the change in position. The rise between the two points represents change in velocity. Slope between two points represents the average acceleration for the time intervals. confidence assessment: 3
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15:39:38 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 --> understood. Idea 3 self critique assessment: 3
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15:40:53 `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 --> Yes correspoonding to that time interval. Greater slope indicates greater acceleration because of the rise (greater change in velocity). confidence assessment: 3
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15:41:10 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 --> Understood self critique assessment: 3
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