course Phy 201 …Ç„ˆª¬Îäl²r™ßæ¾×wŒ´Â«Úassignment #002
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15:14:05 `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 --> If an object moves 12 meters in 4 seconds, it is moving at an average rate of 3meters per second. 12 divided by 4 is 3, so it moves 3 meters every one second. confidence assessment: 2
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15:14:44 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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15:17:10 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> In the previous problem, we were moving at an average rate of 3 meters per second. The concept of a rate is the speed at which something is traveling or happening. confidence assessment: 2
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15:18:56 06-21-2007 15:18:56 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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NOTES -------> A rate is found by dividing the change in a quantity by the change in another quantity on which is dependent. (we previously divided the change in position byb the time which that change occured)
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15:20:14 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> The object position is dependent on time. Time will always go on, reguardless of the position of the object, therefore it is independent. confidence assessment: 2
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15:20:33 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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15:24:06 `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 --> In my explanations, I didn't specify that the rate, in this case, is the average rate at which the position is changing with respect to clock time. The clock time is the independent variable and the object's change in position is the dependent variable. confidence assessment: 2
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15:24:25 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 --> self critique assessment: 3
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15:37:37 `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 found by dividing distance by time. (6meters)/(3seconds) =2meters per second. The average velocity is found by dividing displacement by time. (-6meters)/ (3seconds)=-2 meters per second. If 6 meters are divided up into 3 parts (one second intervals), the objct is traveling at an average speed of 2 meters every one second. Also, if the object starts at zero meters, and moves backward 6 meters in 3 seconds, it would be traveling at an average velocity of -2meters per second. confidence assessment: 2
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15:39:23 06-21-2007 15:39:23 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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NOTES -------> speed = average rate at which distance changes (never negative) velocity = average rate at which position changes (pos or neg)
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15:42:09 `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: 2
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15:42:32 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 --> self critique assessment: 3
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11:32:37 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> 'ds is the change is position 'dt is the time interval I write it on paper with a triangle in front of a lower case d and a triangle preceeding the t,also. confidence assessment: 2
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11:33:52 06-22-2007 11:33:52 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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NOTES -------> 'd is the symbol for Delta.
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01:33:06 `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 --> If an object changes position of 5 meters every one second, then in 10 seconds, it should have moves 50 meters. confidence assessment: 2
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21:34:55 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 --> We obtain the resulting change in the first by multiplying the rate at which one quantity changes with respect to another (5m/s) with the change in the second quantity (10m). self critique assessment: 2
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21:44:26 `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 stands for the change in position of the object. It could be found by dividing the vAve by 'dt. confidence assessment: 2
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21:47:44 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 made a mistake when finding the formula to calculate 'ds. I said vAve/'dt, when actually you would MULTIPLY vAve by 'dt. Also, the units will help me not to make that mistake again because the will be in the units of distance (cm, meters, etc.) self critique assessment: 2
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21:57:09 `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 --> vAve is the rate at which the position of the object changes vAve = 'ds / 'dt 'ds =change in position 'dt =time interval You could use a similar formula to calculate almost any rate. confidence assessment: 2
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21:57:30 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 --> self critique assessment: 3
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22:01:04 `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, vAve = `ds / `dt, for 'ds: Simply multiply vAve by 'dt. The new equation would read: 'ds = vAve * 'dt The units would be in the same units as our distance. confidence assessment: 2
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22:01:16 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 --> self critique assessment: 3
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22:05:53 `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 equation shows us that the change in position is equal to the average rate at which the position changes times the time interval. displacement = velocity * clock time confidence assessment: 2
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22:06:36 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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22:09:13 `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 --> We multiply both sides by 'dt, giving us 'ds = vAve * 'dt. Next we divide both sides by vAve, giving us 'dt = 'ds / vAve. confidence assessment: 2
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22:09:22 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 --> self critique assessment: 3
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22:12:24 `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 --> clock time = displacement / velocity The time interval is equal to the change in position divided by the average rate at which the position of the object changes. confidence assessment: 2
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22:12:59 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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