Asst_2

course Phy 201

…Ç„ˆª¬Îäl²r™ßæ¾×wŒ´Â«Úassignment #002

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

002. Velocity

Physics I

06-21-2007

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

......!!!!!!!!............................

.......

15:17:10

`q002. How is this problem related to the

concept of a rate?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

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)

..........................................

.............!!!!!!!!.....................

..............

15:20:14

`q003. Is object position dependent on

time or is time dependent on object

position?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

NOTES -------> speed = average rate at

which distance changes (never negative)

velocity = average rate at which position

changes (pos or neg)

..........................................

.............!!!!!!!!.....................

..............

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?

......!!!!!!!!............................

.......

RESPONSE -->

vAve = 'ds/'dt

confidence assessment: 2

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

......!!!!!!!!............................

.......

11:32:37

`q007. How do you write the expressions

`ds and `dt on your paper?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

NOTES -------> 'd is the symbol for Delta.

..........................................

.............!!!!!!!!.....................

..............

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?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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?

......!!!!!!!!............................

.......

RESPONSE -->

'ds stands for the change in position of

the object. It could be found by dividing

the vAve by 'dt.

confidence assessment: 2

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

......!!!!!!!!............................

.......

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?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

......!!!!!!!!............................

.......

22:05:53

`q012. How is this result related to our

intuition about the meanings of the terms

average velocity, displacement and clock

time?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

......!!!!!!!!............................

.......

22:09:13

`q013. What algebraic steps do we use to

solve the equation vAve = `ds / `dt for

`dt, and what is our result?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

......!!!!!!!!............................

.......

22:12:24

`q014. How is this result related to our

intuition about the meanings of the terms

average velocity, displacement and clock

time?

......!!!!!!!!............................

.......

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

..........................................

.......

......!!!!!!!!............................

.......

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.

......!!!!!!!!............................

.......

RESPONSE -->

self critique assessment: 3

..........................................

.......

"

Your answers and your self-critiques look very good. Keep up the good work.

The copy you are submitting of the SEND file is compressed into narrow columns. This makes it a little more difficult for me to read, and will likely make it significantly more difficult for you to read when it is posted.

When you open the file to copy it, I suggest you maximize the window in which you open it before copying the contents.