course PHY201-4 I hope I'm getting this transferring down a little bit better. ýñHï¬ó¤g㵃q©ó¿TôîØöƒ“çñÞÌÆðýáðõassignment #002
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19:35:41 `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 --> 12 meters/4 seconds = 3 meters per sec. I got this by dividing the time into the distance. The formula is average speed=distance traveled/time elapsed confidence assessment: 3
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19:36:34 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 --> That was my answer self critique assessment: 3
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19:37:50 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> The rate in instance is an average of distance per time confidence assessment: 2
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19:40:39 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: 2
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19:43:08 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> The object is dependent on time confidence assessment: 2
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19:45:10 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 --> ok self critique assessment: 2
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19:50:29 `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 --> I think I got it right by dividing the distance by the time to get the average rate. As mentioned before the object position is dependent on time. confidence assessment: 3
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19:53:42 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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20:09:55 `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 6 meters/3 seconds = 2 meters per second. The average velocity is -6 meters /3 seconds = -2 meters per second. The average speed was easyby dividing the time into the distance. I had to look in the book about average velocity because of the negative displacement. I'm not sure if the negative would be correct for the velocity. confidence assessment: 1
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20:10:57 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 --> I didn't know if a velocity could be negative but now I know. self critique assessment: 3
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20:17:56 `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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you make, 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. confidence assessment: 2
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20:20:37 `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 --> OK I see the 'ds standing for change in distance or position and 'dt standing for the change in time. The 'd states the same as delta s but we can't use the delta symbols on the keyboard. confidence assessment: 2
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20:22:28 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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20:29:21 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> I normally use the delta triangle which means change between two quanities but while using typewritter notation it's probably best to use the apostrophe then the proper letter designation. confidence assessment: 2
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20:30:23 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 self critique assessment: 3
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20:35:48 `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 --> The object moves 50 meters. It's related becaused an object has moved a given distance in a given time period. confidence assessment: 3
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20:46:32 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 --> I'm not too sure I understand this question. I know if an object is traveling at some rate of speed and then you ask the question how far will it travel in some timeframe. The resulting math reveals the total distance traveled. self critique assessment: 2
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20:52:41 `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 or the delta distance equals the average velocity times the delta time. confidence assessment: 3
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20:54:16 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 roughly my thoughts self critique assessment: 3
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21:04:04 `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 --> Rate is the change in an object's position during some time interval. If you know the time and the distance the rate can be computed. If you know the rate and the time interval, you can compute the distance, and if you know the rate and the distance, you can compute the time interval. confidence assessment: 2
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21:06:35 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 worded maybe a little different than mine but I think I said the same thing self critique assessment: 3
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21:13:18 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 --> The base formula is simple math of positioning the unknown on the left side of the equation and multiplying both side by a known variable or dividing both sides by a known variable self critique assessment: 3
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21:18:23 `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 displacement has to be the change in distance, the clock time has to be the change in time and the average velocity has to be some rate measured in some distance per time confidence assessment: 2
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21:20:43 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 --> I don't think I perceived the question very well in this respect. My think was to translate the words into a more common thought. self critique assessment: 1
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21:26:54 `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 --> vAve = 'ds / 'dt. first multiply both sides of the equation by 'dt to get 'dt * vAve = 'ds * 'dt / 'dt which results in 'dt * vAve = 'dt. Now divide both sides by vAve. 'dt * vAve / vAve = 'ds / vAve which results in 'dt = 'ds / vAve confidence assessment: 3
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21:27:17 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 that's what I got. self critique assessment: 3
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21:32:27 `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 --> It probably can be used by the automobile again. If you are traveling 60 mph and you are going to a town 120 miles away, this would help you figure about how long it take you to get to the town. With this example it would take about 2 hours of travel time confidence assessment: 3
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21:34:25 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 --> That is close to what to what I was thinking. self critique assessment: 3
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