ast 2

course phy 201

ʢy𬧖T雛理

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assignment #002

002. Velocity

Physics I

01-27-2008

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01:47:07

`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 -->

12meters/4secs=3meters/s

confidence assessment: 2

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01:47:43

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 I'l be more detailed in my decriptions

self critique assessment: 2

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01:48:33

`q002. How is this problem related to the concept of a rate?

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RESPONSE -->

because a rate is an action and in this equation there is an action

confidence assessment: 0

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01:48:59

01-27-2008 01:48:59

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 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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01:50:44

`q003. Is object position dependent on time or is time dependent on object position?

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RESPONSE -->

position is dependant on time

confidence assessment:

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01:51:01

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 -->

YES

self critique assessment: 2

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01:52:38

`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 was a bit skepical that i knew what i was talking about. but time is consistantly changing which means that position is dependant on time because no matter what you are measuring its an object is here for this many mins ect. not this many mins is position here.

confidence assessment: 0

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01:53:04

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

self critique assessment: 0

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01:57:35

`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 -->

velocity=displacement/time = -6/3= -2meters/sec

speed = distance/time

since teh equation gives us displacement instead of an actual distance I am speculating that i can not solve for speed too many variables.

confidence assessment: 1

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01:58:13

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 -->

oops guess i wasnt supose to solve for that one

self critique assessment: 1

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02:00:00

`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

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

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.

confidence assessment: 0

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02:01:46

`q007. How do you write the expressions `ds and `dt on your paper?

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RESPONSE -->

triangle s and triangle t

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

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.

confidence assessment: 3

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02:02:17

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 -->

delta is a triangle

self critique assessment: 2

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02:03:30

`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 -->

5 times 10 equals 50 meters

its the action of moving so its a rate of change

confidence assessment: 1

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02:04:18

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 -->

self critique assessment: 0

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02:05:29

`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 -->

vave='ds/'dt

confidence assessment: 2

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02:08:12

01-27-2008 02:08:12

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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NOTES -------> 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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02:09:33

`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 -->

we would multiply bth sides by time to isolate s and solve.

vave*'dt='ds

confidence assessment: 3

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02:09:53

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 -->

got it

self critique assessment: 2

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02:10:35

`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 -->

i dont understand the intuition reference

confidence assessment: 0

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02:12:18

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 -->

i am trying to save notes here and im hoping everything looks right on the submission page so i can print it because it keeps bringing up teh last coped page in other words its not working for me tonight

self critique assessment: 0

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02:13:37

`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 divide both sides by 'ds to isolate d't and solve so its going to look like this=

vAve/'ds='dt

confidence assessment: 2

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02:14:37

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 -->

ok so i left it in the denominator location ill bring it up next time

self critique assessment: 1

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02:15:13

`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 -->

i still have no idea what to do with this intuition

confidence assessment: 0

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02:15:38

01-27-2008 02:15:38

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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NOTES -------> f 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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Good responses. Let me know if you have questions. &#