Lesson 2

course phy 121

Tӟ̆s˗assignment #002

002. Velocity

Physics I

09-14-2008

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assignment #002

002. Velocity

Physics I

09-14-2008

VⱓܩTyNJ

assignment #002

002. Velocity

Physics I

09-14-2008

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22:25:36

`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.

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RESPONSE -->

If you laid out 12 meters in a straigt line, rolled a toy car 3 meters along it and total time was only one second , then you could assume that for every second that you rolled a toy car you could add 3 more meters. If you rolled the car a total of 4 seconeds it would have traveled a total of 12 meters.(3m+3m+3m+3m)= 12meters

confidence assessment: 3

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22:26:00

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

self critique assessment: 3

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22:32:20

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

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RESPONSE -->

Giving these numbers you could determin the time or speed needed to travel a certin distance.

In order to travel 12 meters in four seconds you would need to be going at least

3 meters/second to reach 12 meters in 4sec.

At a rate of 3 meters/second.(3m/s)

confidence assessment: 3

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22:32:52

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

self critique assessment: 3

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22:40:07

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

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RESPONSE -->

As long as you have one cordanat the other can be obtianed giving slope or y intrval.

Both

confidence assessment: 2

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22:44:02

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 -->

Cool I hadn't thought of it like that, guess you cant realy stop time; only measure it in intravels.

self critique assessment: 2

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22:47:50

`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 -->

Clock time dominates.

###But is clock time always indicaited by y or can it be x as well?###

confidence assessment: 2

When clock time is the independent variable, it is traditionally graphed along the horizontal axis. There are exceptions, but this will be the general rule.

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22:51:55

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 -->

I think I've been pretty though in my self-critqueing and addressed any troubles I have had.

self critique assessment: 2

I agree.

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23:03:09

`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 -->

If an object is moving at a rate of 6 meters every three seconds you would divide the total distance traveled by the amount of time it took to travel it, then you would have the distance -6 meters/time 3sec would result in a rate of -2 meters/second.

confidence assessment: 2

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23:07:11

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 -->

hmmm. I think I get it.

So no matter what if you are moveing in any direction you are still moveing in a positive motion.

self critique assessment: 2

It can be put a little simpler than this. If you're moving, then you're moving and if you're moving then you have speed. Speed is always positive. However you can be moving in the positive or negative direction, and when direction is considered we call it velocity instead of speed.

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23:10:41

`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) or

`ds = (vAve/`dt) or

`dt = (vAve/`ds) .

confidence assessment: 3

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23:12:45

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 -->

So `dt = (vAve*`ds)?

I might have gotten confused.

self critique assessment: 1

vAve is the average rate of change of position with respect to clock time, which is then defined as (change in position)/(change in clock time). Thus vAve = `ds / `dt. Your equation vAve = `ds * `dt is not correct and if you think about it does not make sense.

Your algebra steps were correct, but they were applied to an incorrect equation. If you solve the correct equation for `dt you get `dt = `ds / vAve.

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23:20:40

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

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RESPONSE -->

ds = position

dt = time

vAve = avarage rate of speed

dv = velocity increase.

confidence assessment: 2

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23:21:28

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 -->

yes delta

self critique assessment: 2

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23:25:47

`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 -->

Well for every second that the object changes position it moves 5meters you could add 5 meters/second 10 times (seconds) or you could multiply 5 meters/second by 10 meters and get 50 meters.

confidence assessment:

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23:25:56

`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 -->

confidence assessment: 2

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23:30:32

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 -->

By forming a graph and using y axis for time and x axis for position;plotting points on the graph coenciding with the (x,y) form to create a slope of incresing, decreasing, or moving at a constant rate.

self critique assessment: 2

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23:33:45

`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 = dt * vAve

confidence assessment: 2

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23:34:28

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

self critique assessment: 3

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23:39:17

`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 rate can be determined by the the distance traveled divided by the time it took to move from one interval to the next .

confidence assessment: 2

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23:39:46

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: 2

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23:43:59

`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 = dt * ds

100miles = 10hrs * 10mph

confidence assessment: 2

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23:44:49

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 -->

I wasn't sure how much detail to go into here.

self critique assessment: 2

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23:46:54

`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 understan that velocity is the rate of travel and

displacement is the distance travled and clock time is the measurment of rate increase.

confidence assessment: 2

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23:47:27

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 -->

Exactly

self critique assessment: 3

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23:48:12

`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 -->

`dt = vAve * `dt

confidence assessment: 2

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23:48:39

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 -->

I did it again. I knew this.

self critique assessment: 1

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23:52:07

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 -->

The average velocity is determined by the distance travled or `ds = vAve/ dt

self critique assessment: 2

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You're still making some errors, but you have the general idea and you are approaching this work correctly. See my notes and be sure to let me know if you have questions.