course Phy 201 ZyÖð’eáÛÓž‘Ž…±ƒ¨úS§Yê‡î÷²ê°|assignment #002
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19:40:58 `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 --> I would say that the object is moving 3 m/s. Well the object went a total of 12 meters, it took 4 seconds to go that distance. Divide the distance by the seconds to get the answer. confidence assessment: 2
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19:45:10 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> For every second, the object traveled 3 meters. This rate was constant for 4 seconds up to 12 meters. Rate in this problem is the ratio of distance traveled to elapsed time. confidence assessment: 1
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19:48:24 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> I would say in this experiment that it is time dependent on object position. Because the rate would have changed if the position of the object traveled would have been different. confidence assessment: 1
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19:51:05 `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 --> Yeah, i missed that concept on that one. I see what you mean, even though the position does matter it doesn't control the experiment. The clock starts and ends no matter how far the object moves or doesn't move. confidence assessment: 1
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19:52:19 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 --> I have no problems at this instance; I stated the few errors I had on the previous problems. self critique assessment: 2
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19:58:41 `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 --> Average speed=distance traveled/time Avergae speed=2m/s Average velocity=final position-initial position/time Average velocity=-2m/s The object displaced -6 meters, this means to me, 6 units away from origin of zero. The time traveled in both was 3 seconds. confidence assessment: 1
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19:59:51
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RESPONSE --> I think I understood that question and answered it correctly. self critique assessment: 3
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20:03:46 `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 --> I believe the expression that shows the average velocity of the object during a time interval is: vAve=displacement/time elapsed=(final position-initial position)/(time elapsed). confidence assessment: 1
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20:07:33 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 --> I believe i misunderstood the question there--- I look at average velocity as the distance an object traveled and the change in its position or how far it is from its starting point. self critique assessment: 2
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20:09:09 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> displacement (change in position)/ time elapsed confidence assessment: 1
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20:10:22 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 --> So really when you say Delta, you talking about change. The change in position over the change in time. self critique assessment: 2
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20:14:44 `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 could potentially travel 50 meters. The object is moving at an average rate of 5m/s over an interval of 10 seconds. So if 5m/s is the objects average, then in 10 seconds the object should have traveled 50 meters. Rate its showing the ratio between distance traveled over time elapsed. confidence assessment: 1
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20:19:57 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 --> Subtract the initial point from the ending point to get the resulting change in the first. ???? self critique assessment: 0
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20:21:42 `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 --> final positon-initial positon = the change 'ds confidence assessment: 1
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20:23:47 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 --> This makes sense in hourly wage to dollars and miles per hour. I can visualize the results a little better with that scenario. self critique assessment: 1
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20:27:00 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 --> Average velocity is the amount or rate the position changes. The displacement (`ds) is the change in position. The time elapsed is (`dt). self critique assessment: 1
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20:29:26 `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 figure out displacement : you must look at it in terms of final position - initial position. or change in x=X2 - X1 confidence assessment: 1
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20:30:52 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 --> Okay, I understand what you wanted there. Just to factor out in order to solve for change in position. So you must know the Average velocity already and time elapsed. self critique assessment: 2
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20:33:00 `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 --> If you know two of them, then you can figure the other one out. To me again average velocity=displacement/time elapsed=final position-initial postion/time elapsed confidence assessment: 1
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20:35:28 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 --> Thats a good way to word it I believe. If i'm going 60 miles and hour and go for 1 hour, then i've traveled 60 miles. self critique assessment: 3
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20:36:59 `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 --> If you're solving for `dt....then you could multiply both side of the equation by `dt/1 to factor the `dt out. confidence assessment: 1
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20:37:42 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 --> I believe i understood the concept of that question fairly easy and answer it correctly. self critique assessment: 3
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20:39:17 `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 --> Most people when they see average velocity, displacment, and clock time are thinking how fast something is going over what distance and for how long. confidence assessment: 0
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20:40:26 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 --> I understand this concept. Thats a good illustration of the idea behind how to get the duration of the time interval. self critique assessment: 3
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