course Phy 231 Again, this is the 2nd Send file that was created for Feb 15. Thanks. ÜüÝÒ¯«ä÷˜©¬ØÃòŸþ“È~a‹S’¸assignment #002
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11:53:36 `q001. Note that there are 14 questions in this assignment. If an object moves 12 meters in 4 seconds, then at what average rate is the object moving? Explain how you obtained your result in terms of commonsense images.
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RESPONSE --> 3 m/s I obtained this answer by dividing 12 m by 4 s. If an item travels 12 m over the course of a 4 second time interval, then the result is an average velocity of 3 m/s. confidence assessment: 3
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11:53:46 Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will have 3 meters, corresponding to the distance moved in 1 second, on the average.
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RESPONSE --> OK self critique assessment: 3
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11:54:39 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> A rate is dividing a change in some unit by a change in some other unit. Determing average velocity fits this definition. confidence assessment: 3
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11:54:50 A rate is obtained by dividing the change in a quantity by the change in another quantity on which is dependent. In this case we divided the change in position by the time during which that change occurred.
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RESPONSE --> OK self critique assessment: 3
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11:56:54 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> time is dependent on object position when determining average velocity. When looking at the definition of a rate, the dependent object is located in the denominator. Therefore, time is the dependent factor in determining average velocity. confidence assessment: 2
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11:59:27 Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else.
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RESPONSE --> Interesting. I was wrong on that one. The explanation makes sense, though it didn't occur to me. Time moves regardless of whether the object moves, therefore time is independent of the position. self critique assessment: 2
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12:01:18 `q004. So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations.
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RESPONSE --> OK - this makes sense. The only concept that I missed, related to whether time was dependent on position or position dependent on time. Position is dependent on time because time moves regardless of whether the object is moving or not. confidence assessment: 2
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12:01:39 You should always self-critique your work in this manner. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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RESPONSE --> OK self critique assessment: 3
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12:06:30 `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object what is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas.
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RESPONSE --> The average speed is the change in distance travelled divided by the change in time interval, regardless of the direction. The average speed is therefore 6 m / 3 s = 2 m/s. The average velocity is the change in distance divided by the change in time, with direction taken into account. The average velocity is therefore: -6 m / 3 s = -2 m/s. If you imagine a body of water, with the surface being at 0 m and diving down into the water, it takes you 3 seconds to travel 6 meters downward. confidence assessment: 2
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12:06:44 Speed is the average rate at which distance changes, and distance cannot be negative. Therefore speed cannot be negative. Velocity is the average rate at which position changes, and position changes can be positive or negative.
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RESPONSE --> OK! self critique assessment: 3
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12:08:36 `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which the position changes, then what expression stands for the average velocity vAve of the object during this time interval?
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RESPONSE --> vAve = 'ds / 'dt confidence assessment: 3
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12:08:42 Average velocity is rate of change of position. Change in position is `ds and change in clock tim is `dt, so vAve = `ds / `dt.
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RESPONSE --> OK self critique assessment: 3
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12:09:43 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> the expression 'ds on paper would be written like: triangle s; and the expression 'dt would be written as: triangle t confidence assessment: 3
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12:10:31 You use the Greek capital Delta symbol Delta. `d is often used here because the symbol for Delta is not interpreted correctly by some Internet browsers. You should get in the habit of thinking and writing Delta when you see `d. You may use either `d or Delta when submitting work and answering questions.
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RESPONSE --> OK - I didn't use the Greek term delta, though that is what I was trying to explain - delta looks like a triangle. I will stick to the Greek terminology in the future. self critique assessment: 2
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12:12:28 `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move? How is this problem related to the concept of a rate?
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RESPONSE --> 50 m This answer was obtained by multiplying 5 m/s by 10 s. To be honest, I'm not completely sure how to answer this, as I believed rate had to do with division-type problem. However, I imagine this is related to the concept of rate because there is a change in the time that is being multiplied by a rate change (5 m/s). confidence assessment: 2
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12:16:34 In this problem you are given the rate at which position changes with respect to time, and you are given the time interval during which to calculate the change in position. Given the rate at which one quantity changes with respect to another, and the change in the second quantity, how do we obtain the resulting change in the first?
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RESPONSE --> To obtain the resulting change, we multiply the rate at which position changes with respect to time by the time interval. self critique assessment: 3
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12:18:30 `q009. If vAve stands for the rate at which the position of the object changes (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position?
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RESPONSE --> 'ds = vAve * 'dt This answer was obtained by using the formula aAve = 'ds / 'dt and the multiplying both sides by 'dt to isolate 'ds confidence assessment: 3
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12:18:41 To find the change in a quantity we multiply the rate by the time interval during which the change occurs. We therefore obtain the change in position by multiplying the velocity by the time interval: `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: Just as when we multiply pay rate by time (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour), when we multiply vAve, in cm / sec or meters / sec or whatever, by `dt in seconds, we get displacement in cm or meters, or whatever, depending on the units of distance used.
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RESPONSE --> OK! self critique assessment: 3
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12:20:07 `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem problem.
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RESPONSE --> These are related to the definition of a rate because to find vAve were are calculating the change in position of an object over the change in time. confidence assessment: 3
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12:20:17 vAve is the average rate at which position changes. The change in position is the displacement `ds, the change in clock time is `dt, so vAve = `ds / `dt.
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RESPONSE --> OK self critique assessment: 3
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12:21:45 `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result?
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RESPONSE --> To solve the equation for 'ds, we simply multiply both sides by 'dt to give us the equation: vAve * 'dt = 'ds confidence assessment: 3
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12:21:58 To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt.
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RESPONSE --> OK self critique assessment: 3
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12:24:44 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> The result makes sense because we expect the change in object position to be equal to the avg velocity multplied by the time it took the object to travel that distance. Just as we saw in the previous problem, where we multiplied 5 m/s by 10 s to get 50 m. confidence assessment: 3
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12:26:11 Our most direct intuition about velocity probably comes from watching an automobile speedometer. We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea. Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval.
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RESPONSE --> OK - I didn't use a commonsense explanation. I like the explanation of the speedometer as this is easy to relate to. self critique assessment: 2
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12:28:32 `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?
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RESPONSE --> to solve for 'dt, we can multiply both sides by 'dt to get: 'dt * vAve = 'ds and then divide both sides by vAve, to get: 'dt = 'ds / aAve confidence assessment: 3
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12:28:43 To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve.
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RESPONSE --> OK self critique assessment: 3
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12:29:51 `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> The result of 'dt = 'ds / vAve makes sense because we expect the time interval to be equal to how far the object travelled (the distance) with respect to how fast on average the object travelled over the course of that distance. confidence assessment: 3
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12:30:29 If we want to know how long it will take to make a trip at a certain speed, we know to divide the distance in miles by the speed in mph. If we divide the number of miles we need to travel by the number of miles we travel in hour, we get the number of hours required. We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval.
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RESPONSE --> OK - again, I like your presentation of using a car and distance travelled with respect to speed of the car. self critique assessment: 2
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