course PHY 201 Mr. Smith,
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12:24:51 `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 --> Commonsense tells us that if an object is moving 12 meters every 4 seconds, then the average rate or speed of this object is 3 m/s. I got this by taking the 12 meters and dividing the 4 seconds into it. confidence assessment: 2
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12:25:05 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 --> self critique assessment: 3
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12:27:09 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> This problem is related to the concept of a rate because the student is able to find the rate with the information provided. The object of that problem was to find the rate of change (or speed) of the object. confidence assessment: 3
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12:27:23 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 --> self critique assessment: 3
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12:28:26 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> When speaking of actual position, the object's position is dependent on both time and the initial position of the object. When speaking of relative position, the object's position is dependent on the time and speed of the object. confidence assessment: 2
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12:28:53 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 --> self critique assessment: 3
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12:29:36 `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 left out the fact that whether the object is moving or not, the clock is still running. I left that out as an ""understood"" point. confidence assessment: 3
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12:30:00 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 --> okie dokie self critique assessment: 3
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12:33:47 `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 --> By dividing the displacement of -6 meters by the time it takes to move those 6 meters, 3 seconds, I found the average speed to be -2 m/s. I foudn this to be a negative 2, simply because the object had been displaced 6 meters backwards on a ""number line"" or similar circumstance. confidence assessment: 3
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12:34:33 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 --> So when talking about speed, the number is always positive, but when talking about velocity, the number can be either positive or negative. Gotcha.. self critique assessment: 2
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12:35:52 `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 The average velocity is equal to the change in position divided by the change in time. confidence assessment: 3
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12:36:06 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 --> Correct! self critique assessment: 3
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12:37:40 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> I dont really understand what is being asked here.. but 'ds stands for the change in position, and i usually just use a ""d"" for ""distance"" while the 'dt stands for the change in time, where i usually just use a ""t"" for ""time."" I guess this is because thats what we were always told to use in high school, and its just a habit. confidence assessment: 2
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12:38:37 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 use Delta as another way of communicating the ""change in"" self critique assessment: 2
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12:41:24 `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 in this 10 seconds of moving at a velocity of 5 m/s. This problem is related to the concept of a rate because it is ultimately the same problem, only looking for a different variable. The same formula is still being used, just in a different order, with a different variable as the solution. confidence assessment: 3
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12:42:01 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 --> in the first we dividng the position by the time. in this one we multiply the velocity and the time, in order to figure otu the distance. self critique assessment: 3
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12:43:46 `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 the change of position equals the average velocity multiplied by the change in time. confidence assessment: 3
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12:43:58 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 --> mmkay self critique assessment: 3
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12:45:39 `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 --> the quantities, average velocity, time interval, and displacement are all related by the definition of a rate in which the displacement is equal to the average velocity multiplied by the time interval (this equation can be written many different ways, depending on the variable which is being looked for). confidence assessment: 3
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12:45:56 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 --> another way to write the same equation. self critique assessment: 3
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12:47:28 `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 so to solve for 'ds, the student must take the 'dt and multiply it ot the other side of the equation, in order to solve for 'ds. vAve * 'dt = 'ds then the student can simply solve the equation for the missing factor. confidence assessment: 3
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12:47:42 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 --> same thing self critique assessment: 3
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12:49:07 `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 --> this result relates to our intuition about the meanings of the terms average velocity, displacement, and clock time because we now know how each of these variables are related, and how to quickly find the missing variable if the other two are present. confidence assessment: 2
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12:49:14 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 --> self critique assessment: 3
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12:50:40 `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 --> in order to solve the equation for 'dt, the student must multiply the 'dt to the other side, and then turn around and divide the vAve back to the side with the 'ds. the final equation would be: 'dt = 'ds / vAve which is the change in clock time equal to the displacement of an object divided by the average velocity of that object. confidence assessment: 3
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12:50:49 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 --> yup self critique assessment: 3
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12:51:34 `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 --> again, the result helps the student to better understand how each of these three variables are related to each other, and how to easily solve the equation for any of these different variables, as long as the other two are present. confidence assessment: 3
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12:51:40 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 --> self critique assessment: 3
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