course phy201 9:00 pm 2/11/2010 Query 4.1
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Given Solution: 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. STUDENT QUESTION Would we not have s^2 since we are multiplying s by s? INSTRUCTOR RESPONSE That is correct. However in this question I've chosen not to confuse the issue by simplifying the complex fraction m/s/s, which we address separately. To clarify, m / s / s means, by the order of operations, (m/s) / s, which is (m/s) * (1/s) = m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I think that a car with a more powerful engine would be capable of a greater rate of velocity change. It will make the car accelerate at a faster rate. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
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Given Solution: 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. STUDENT COMMENT: The significance for an automobile of the rate at which its velocity changes is the amount of speed it takes to travel to a place in a certain amount of time. If one car is traveling along side another, and they are going to the same location the velocity will be how long it take this car to get to this location going at a speed other than the other car. If a car with a more powerful engine were to travel the same distance its velocity would be capable of a greater rate if increased speed occurred. INSTRUCTOR RESPONSE: It's necessary here to distinguish between velocity, which is a pretty intuitive concept, and rate of change of velocity, which is much less intuitive and less familiar. An object can change velocity at a constant rate, from rest to a very high velocity. All the while the rate of change of velocity with respect to clock time can be unchanging. So the rate of change of velocity with respect to clock time has nothing to do with how fast the object is moving, but rather with how quickly the velocity is changing. Moving from one location to another, the displacement is the change in position. If the displacement is divided by the time required we get the average rate of change of position with respect to clock time, or average velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the units of meters / second / second are obtained because the change in velocity, normally in untis of m/s is divided by clocktime normally in seconds so m/s / s is like m/s * 1/s or m/s^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ @@@@@@@@@@@@@@@@@@@@@@@@@@@
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m/s^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10 m/s - -5 m/s = 15m/s 15 m/s / 5 s = 3 meters confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ @@@@@@@@@@@@@@@@@@@@@@@@@@@
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ?????????ok, I undersstand we are to add this, but should I have added -5 cm/s + -10 cm/s to = - 15 cm/s so that my sign would be correct???????????? How do I know in what order to add if it is going from 10 and ending up at -5 it seems it would be 10- -5 or 10 +5 ?????????????????????? And I forgot to add the m/s^2 onto the solution
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Given Solution: 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. STUDENT COMMENT: It’s average velocity so it would be aAve. INSTRUCTOR RESPONSE: Good, but note: It’s average acceleration (not average velocity) so it would be aAve. In most of your course acceleration is constant, so initial accel = final accel = aAve. In this case we can just use 'a' for the acceleration. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times? Before answering, click for the next question &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `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? Before answering, click for the next question &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q006b. What therefore is the average rate at which the velocity is changing during this time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: avave = change in velocity/ change in clocktime avave = 6 m/s - 9 m/s / 1.5s- 3.5s avave= 3m/s / 2 s avave = 1.5 m/ s^2 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
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Given Solution: 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. I understand this at this point &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent? What is the rise between these points what does it represent? What is the slope between these points what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: run = change in clocktimes. Clocktimes is on the horizontal axis 1.5s -3.5s = 2.0s Rise = change in velocity. Velocity is on the vertical axis rise = 6m/s - 9 m/s = 3 m/s Slope is the rise/run 3m/s / 2.0s = 1.5m/s^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope represent the accleration of the any graph of velocity vs clocktime because avave= 'dv/ 'ds the slope is the rise/run or 'dv/ 'ds. As the slope increases the velocity also increases as does the acceleration in respect to the increas in time as it passes. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
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Given Solution: 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. Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This graph would be velocity would be increasing at an decreasing rate. on a graph of velocity vs clocktime the velocity would be increasing but the slope of the graph wold be decreasing confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok so the graph would be increasing until it get to the point at which resistance of the wind start to effect the rate, and then the graph would start to level off, but not necessarily decrease
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Given Solution: 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. STUDENT QUESTION: Can you clarify some more the differences in acceleration and velocity? INSTRUCTOR RESPONSE: ** 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. ** STUDENT QUESTION: In the problem it states that velocity continues to increase even though the rate at which velocity changes decreases. I don’t understand your the slope will decrease if this is true. I can understand a diminish in velocity and time, but not a down turn of the slope, which is what your solution leans to. INSTRUCTOR RESPONSE Your thinking is good, but you need carefully identify what it is you're describing. The question here concerns the acceleration vs. clock time graph, whereas most of your comments apply to the velocity vs. clock time graph. Under these conditions the slope of the velocity vs. clock time graph will decrease, this will occur as long as the acceleration vs. clock time graph decreases, regardless of whether that decrease is at a constant, an increasing or a decreasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok so we are dealing with acceration here as opposed to velocity. since acceleration is the change in velocity / clocktime????????It is hard to imagine that the graph would not be increasing since the velocity is increasing???????????????