course phy 201 ¢ìñÌ—ZÌíÇÂì䧜®ñŒyÆÆÀ„P{³£assignment #002
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18:40:01 `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 --> The object is moving 3 meters per second. Divide the number of meters by the number of seconds. confidence assessment: 3
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18:41:39 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> Rate is the concept of change. confidence assessment: 3
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18:42:04 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 --> correct self critique assessment:
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18:42:54 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> object poition is only dependent on time if the object is in motion. confidence assessment: 3
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18:44:09 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 --> The change in the objects position for a given time interval. self critique assessment: 3
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18:45:27 `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 --> none confidence assessment: 3
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18:46:20 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 --> The concept of position change due to motion in a given amount of time. self critique assessment: 3
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18:47:22 `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 2 meters per second as is the velocity. confidence assessment: 3
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18:48:32 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 --> The average speed would still be 2 meters per second if the object was moving weither if it is moving forward or backwards. self critique assessment: 3
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18:49:38 `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 confidence assessment: 2
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18:50:07 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 now understand it was asking for the equation. self critique assessment: 3
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18:50:40 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> Change of distance and Change of time confidence assessment: 3
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18:51:03 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 --> Change and Delta is the same self critique assessment: 3
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18:51:54 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 --> 50 meters. The problem is asking to compute using the rate given. self critique assessment: 3
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18:52:29 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 --> i understand the solution given self critique assessment: 3
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18:53:22 `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 'ds / 'dt to give you the average velocity. confidence assessment: 2
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18:53:45 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 --> correct given solution self critique assessment: 2
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18:54:57 `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 --> vAve = 'ds/'dt to solve for 'ds take the vAve * 'dt to give you 'ds. confidence assessment: 3
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18:55:14 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 --> given solution was correct self critique assessment: 3
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18:56:11 `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 --> by subsitution of given you can solve for part of an equation that is not given. confidence assessment: 3
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18:56:48 `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 --> I understand the given solution mine was basicly the same confidence assessment: 2
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18:57:23 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 --> correct self critique assessment: 3
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18:58:33 `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 --> you are using given information to solve for an unknown to have all the information such as velocity, time and distance confidence assessment: 3
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18:59:54 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 --> This is true if you continuely travel at the constant rate of speed or if you maintain the required average rate of speed. self critique assessment: 3
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