áÙ‹ŒCØØ•ŒhˆéÉÚå„G Ư°Ý–ŸÅà«M¨ Student Name: assignment #002
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18:35:36 `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.......!!!!!!!!...................................
RESPONSE --> 3m /s -- If the object is moving 12m in 4 seconds then the rate at which the object moves each second would be 12 divided by 4.
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18:35:44 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 --> ok
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18:39:49 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> A rate is the relationship of two given quantites..examples: distance in respect to time, qantity in respect to weight, etc. The problem gives the relationship of an objects dispacement in respect to time
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18:39:57 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
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18:43:02 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> I think that both go hand in hand in the respect of an object changes position over time. If an object is moving at a rate then the position of the object is given at a specific time, and also the specific timing of the object moving at specific rate will result in a distance.
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18:43:50 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 --> I certainly see how this is true....the distance of the object is very much dependant upon the time it travels
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18:45:18 `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 logic I put an object in motion and said that both the distance and time were equally important...however in a rate the distance is changing in repect to time...thus the distance is dependant upon the time the object travles.
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18:45:28 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
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18:47:15 `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 of the object is 2m/s and the average velocity would also be -2m/s
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18:48:42 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 --> OK -- I assumed that a velocity generally has a direction associated with it that the velocity would be -2 since the dispacement is in the negative direction.
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18:49:49 `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 --> 'ds/'dt=vAve
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18:49:57 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
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18:51:13 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> I would write the symbol delta and the letter s or x for the distance, and the symbol delta and the letter t for the time interval
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18:51:32 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
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18:53:14 `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?......!!!!!!!!...................................
RESPONSE --> 5m/s*10s=50m
The object would move 50m since the rate of the object moving at 5m/s would reach a distance of 50m after 10s.................................................
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18:55:09 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 --> The rate of the first quantity is based on the rate of the object. If given the second quantity this is incorporated into the rate of the first quantity to change it. All values must be associtated with the inital given quanitiy
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18:58:07 `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 --> the change in position 'ds is calculated by the following:
'ds = vAve*'dt.................................................
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18:58:16 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
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19:02:11 `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 formulas to calculate both the velocity and rate of an object are very similar. The rate of an object deals with the movement of an object over a given time interval and is always positive. The velocity of an object deals not only with the rate an object moves but also includes a displacement factor. velocity can be both positive and negative since the displacement of an object can be directional
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19:02:22 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
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19:04:11 `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 --> The rearrance the formula for 'ds we would get:
'ds=vAve*'dt where we are looking for the displacement of the object when we know the average velocity and the time interval it travels.................................................
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19:04:57 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.......!!!!!!!!...................................
RESPONSE --> Sorry...I did not show the simple calc. rearrangements of the formula..but I do understand the concept.
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19:05:45 `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 average velocity is the average change in displacement in relationship to the average time for a given interval.
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19:05:52 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 --> ok
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19:48:45 `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 --> 'ds/vAve ='dt
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19:50:22 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.......!!!!!!!!...................................
RESPONSE --> Again...got in too big a hurry....didnt show the calc.
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19:51:49 `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 --> The fomula ties together the velocity, displacement and clock time where by simple rearrangement of the formula using calculus will yeild results for the values we are looking for.
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19:51:56 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 --> ok
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