Asst__5_QA

#$&*

course Phy 121

11/2 800To speed my progress through the course, I followed your advice and noted 'OK' where I felt confident I worked the problem out correctly. I then checked the given solution and if different submitted my answer in the self-critique.

005. Uniformly Accelerated Motion

Preliminary notes:

On any interval there are seven essential quantities in terms of which we analyze the

motion of a nonrotating object:

the time interval `dt between the beginning and the end of the interval

the displacement `ds of the object during the interval

the initial velocity v0, the velocity at the beginning of the interval

the final velocity vf, the velocity at the end of the interval

the average velocity vAve of the object during the interval

the change `dv in the velocity of the object during the interval

the average acceleration a_Ave of the object during the interval

You should remember these symbols and their meanings. You will be using them

repeatedly, and you will soon get used to them.

You should at any time be able to list these seven quantities and explain the meaning of

each.

In any question or problem that involves motion, you should identify the interval of

interest, think about what each of these quantities means for the object, and identify

which quantities can be directly determined from the given information.

You will of course improve your understanding and appreciation of these quantities as

you work through the qa and the associated questions and problems.

Note also that `dt = t_f - t_0, where t_f represents the final clock time and t_0 the initial

clock time on the interval, and that `ds = s_f - s_0, where s_f represents the final position

and t_0 the initial position of the object on the interval.

Further discussion of symbols (you can just scan this for the moment, then refer to it when

and if you later run into confusion with notation)

the symbol x is often used instead of s for the position of an object moving along a

straight line, so that `dx might be used instead of `ds, where `dx = x_f - x_0

some authors use either s or x, rather that s_f or x_f, for the quantity that would

represent final position on the interval; in particular the quantity we express as `dx might

be represented by x - x_0, rather than x_f - x_0

some authors use t instead of `dt; there are good reasons for doing so but at this point in

the course it is important to distinguish between clock time t and time interval `dt; this

distinction tends to be lost if we allow t to represent a time interval

the quantity we refer to as `dt is often referred to as 'elapsed time', to distinguish it from

'clock time'; once more we choose here to use different symbols to avoid confusion at this

critical point in the course)

If the acceleration of an object is uniform, then the following statements apply. These are

important statements. You will need to answer a number of questions and solve a

number of problems in order to 'internalize' their meanings and their important. Until you

do, you should always have them handy for reference. It is recommended that you write

a brief version of each statement in your notebook for easy reference:

1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a

constant rate or decreasing at a constant rate.

2. The average velocity of the object over any time interval is equal to the average of

its velocity at the beginning of the time interval (called its initial velocity) and its velocity at

the end of the time interval (called its final velocity).

3. The velocity of the object changes at a constant rate (this third statement being

obvious since the rate at which the velocity changes is the acceleration, which is assumed

here to be constant).

4. The acceleration of the object at every instant is equal to the average acceleration of

the object.

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Question: `q001. Note that there are 13 questions in this assignment.

Suppose that an object increases its velocity at a uniform rate, from an initial velocity of 5

m/s to a final velocity of 25 m/s during a time interval of 4 seconds.

By how much does the velocity of the object change?

What is the average acceleration of the object?

What is the average velocity of the object?

(keep your notes on this problem, which is continued through next few questions)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

'dv= 25m/s - 5m/s = 20m/s

a_Ave= 20m/s / 4 sec = 5m/s^2

vAve= (25m/s + 5m/s)/2 = 30m/s / 2 = 15m/s

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

The velocity of the object changes from 5 meters/second to 25 meters/second so the

change in velocity is 20 meters/second. The average acceleration is therefore (20

meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the

average of its initial and final velocities, as asserted above, and is therefore equal to (5

meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers

are averaged by adding them and dividing by 2).

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Self-critique (if necessary):

ok

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Self-critique rating:

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Question: `q002. How far does the object of the preceding problem travel in the 4

seconds?

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Your solution:

ok

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

The displacement `ds of the object is the product vAve `dt of its average velocity and the

time interval, so this object travels 15 m/s * 4 s = 60 meters during the 4-second time

interval.

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Self-critique (if necessary):

ok

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Self-critique rating:

*********************************************

Question: `q003. Explain in commonsense terms how we determine the acceleration and

distance traveled if we know the initial velocity v0, and final velocity vf and the time

interval `dt.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Acceleration is found by taking the difference in vF and v0 and dividing that figure by

the amount of time transpired, or 'dt.

Distance traveled is found by multiplying the known time interval by the average velocity

which is vf and v0 divided by 2.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

In commonsense terms, we find the change in velocity since we know the initial and final

velocities, and we know the time interval, so we can easily calculate the acceleration.

Again since we know initial and final velocities we can easily calculate the average

velocity, and since we know the time interval we can now determine the distance

traveled.

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Self-critique (if necessary):

ok

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Self-critique rating:

*********************************************

Question: `q004. Symbolize the situation by first giving the expression for the

acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of

v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf

and `dt.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

ok

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

The acceleration is equal to the change in velocity divided by the time interval; since the

change in velocity is vf - v0 we see that the acceleration is a = ( vf - v0 ) / `dt.

The average velocity is the average of the initial and final velocities, which is expressed

as (vf + v0) / 2.

When this average velocity is multiplied by `dt we get the displacement, which is `ds =

(v0 + vf) / 2 * `dt.

STUDENT SOLUTION (mostly but not completely correct)

vAve = (vf + v0) / 2

aAve = (vf-v0) / dt

displacement = (vf + v0)/dt

INSTRUCTOR RESPONSE

Displacement = (vf + v0)/dt is clearly not correct, since greater `dt implies greater

displacement. Dividing by `dt would give you a smaller result for larger `dt.

From the definition vAve = `ds / `dt, so the displacement must be `ds = vAve * `dt. Using

your correct expression for vAve you get the correct expression for `ds.

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Self-critique (if necessary):

ok

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Self-critique rating:3

*********************************************

Question: `q006. This situation is identical to the previous, and the conditions implied by

uniformly accelerated motion are repeated here for your review: If the acceleration of

an object is uniform, then the following statements apply:

1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a

constant rate or decreasing at a constant rate.

2. The average velocity of the object over any time interval is equal to the average of

its velocity at the beginning of the time interval (called its initial velocity) and its velocity at

the end of the time interval (called its final velocity).

3. The velocity of the object changes at a constant rate (this third statement being

obvious since the rate at which the velocity changes is the acceleration, which is assumed

here to be constant).

4. The acceleration of the object at every instant is equal to the average acceleration of

the object.

Describe a graph of velocity vs. clock time, assuming that the initial velocity occurs at clock

time t = 0.

At what clock time is the final velocity then attained?

What are the coordinates of the point on the graph corresponding to the initial velocity

(hint: the t coordinate is 0, as specified here; what is the v coordinate at this clock time?

i.e., what is the velocity when t = 0?).

What are the coordinates of the point corresponding to the final velocity?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

v0 = (0, 5)

vf= (4, 25)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5

m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4

seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the

corresponding graph point is ( 4 s, 25 m/s).

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Self-critique (if necessary):

ok

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Self-critique rating:3

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Question: `q007. Is the v vs. t graph increasing, decreasing or level between the two

points, and if increasing or decreasing is the increase or decrease at a constant,

increasing or decreasing rate?

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Your solution:

Increasing at constant rate

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

Since the acceleration is uniform, the graph is a straight line. The graph therefore

increases at a constant rate from the point (0, 5 m/s) to the point (4 s, 25 m/s).

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Self-critique (if necessary):

ok

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Self-critique rating:

*********************************************

Question: `q008. What is the slope of the graph between the two given points, and

what is the meaning of this slope?

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Your solution:

ok

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which

represents the change in the velocity of the object. The run of the graph is from 0

seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval

during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s)

= 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in

the clock time and therefore represents the acceleration of the object.

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Self-critique (if necessary):

ok

------------------------------------------------

Self-critique rating:3

*********************************************

Question: `q009. The graph forms a trapezoid, starting from the point (0,0), rising to

the point (0,5 m/s), then sloping upward to (4 s, 25 m/s), then descending to the point (4

s, 0) and returning to the origin (0,0). This trapezoid has two altitudes, 5 m/s on the left

and 25 m/s on the right, and a base which represents a width of 4 seconds. What is the

average altitude of the trapezoid and what does it represent, and what is the area of

the trapezoid and what does it represent?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

ok

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

The two altitudes are 5 meters/second and 25 meters/second, and their average is 15

meters/second. This represents the average velocity of the object on the time interval.

The area of the trapezoid is equal to the product of the average altitude and the base,

which is 15 m/s * 4 s = 60 meters. This represents the product of the average velocity

and the time interval, which is the displacement during the time interval.

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Self-critique (if necessary):

ok

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Self-critique rating:3

*********************************************

Question: `q010. Students at this point often need more practice identifying which of the

quantities v0, vf, vAve, `dv, a, `ds and `dt are known in a situation or problem. You

should consider running through the optional supplemental exercise

ph1_qa_identifying_quantities.htm . The detailed URL is

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_identifying_quantit

ies.htm If you are able to quickly identify all the quantities correctly 'in your head', the

exercise won't take long and it won't be necessary to type in any responses or submit

anything. If you aren't sure of some of the answers, you can submit the document,

answer and/or asking questions on only the problems of which you are unsure.

You should take a quick look at this document. Answer below by describing what you

see and indicating whether or not you think you already understand how to identify the

quantities. If you are not very sure you are able to do this reliably, indicate how you

have noted this link for future reference. If you intend to submit all or part of the

document, indicate this as well.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The URL brought up a document with 9 questions related to uniformly accelerated

motion. I have save this URL in my notebook, and will refer back to it if I feel I need

more practice.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

You should have responded in such a way that the instructor understands that you are

aware of this document, have taken appropriate steps to note its potential usefulness,

and know where to find it if you need it.

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Self-critique (if necessary):

ok

------------------------------------------------

Self-critique rating:

*********************************************

Question: `q011. The velocity of a car changes uniformly from 5 m/s to 25 m/s during

an interval that lasts 6 seconds. Show in detail how to reason out how far it travels.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First find the vAve by adding 5m/s and 25m/s to calculate 30m/s. Divide this by 2 to

get 15m/s.

Next calculate 'ds by multiply the interval 6 by the vAve, 15m/s to find 'ds=90meters.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q012. The points (5 s, 10 m/s) and (10 s, 20 m/s) define a 'graph

trapezoid' on a graph of velocity vs. clock time.

What is the average 'graph altitude' for this trapezoid?

Explain what the average 'graph altitude' means and why it has this meaning.

What is the area of this trapezoid? Explain thoroughly how you reason out this result,

and be sure to include and explain your units.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The points (5s, 10m/s) and (10s, 20m/s) produce a graph whose trapezoidal shape has

graph altitudes of 20m/s and 10m/s which when added and divided by 2 find 15m/s,

the average altitude of the graph. This is also the vAve of the object for the given

interval between 5seconds and 10 seconds.

The area of the trapezoid is equal to the graph altitude multiplied by the graph latitude

between the given points, which is: 15m/s * 5sec = 75meters, or the total displacement

for this interval.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q013. On a certain interval of duration `dt an object has initial velocity v_0

and final velocity v_f. In terms of the symbols v_0, v_f and `dt, what are the values of

the following?

vAve

`dv

`ds

aAve

Be sure to explain your reasoning.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

vAve=ave. graph altitude = (v_0 + v_f)/2

'dv= v_f - v_0

'ds= [(v_0 + v_f)/2] * 'dt ; b/c: 'ds = vAve*'dt

aAve= (v_f - v_0) / 'dt ; b/c: aAve='dv / 'dt

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

Question: `q011. The velocity of a car changes uniformly from 5 m/s to 25 m/s during

an interval that lasts 6 seconds. Show in detail how to reason out how far it travels.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First find the vAve by adding 5m/s and 25m/s to calculate 30m/s. Divide this by 2 to

get 15m/s.

Next calculate 'ds by multiply the interval 6 by the vAve, 15m/s to find 'ds=90meters.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q012. The points (5 s, 10 m/s) and (10 s, 20 m/s) define a 'graph

trapezoid' on a graph of velocity vs. clock time.

What is the average 'graph altitude' for this trapezoid?

Explain what the average 'graph altitude' means and why it has this meaning.

What is the area of this trapezoid? Explain thoroughly how you reason out this result,

and be sure to include and explain your units.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The points (5s, 10m/s) and (10s, 20m/s) produce a graph whose trapezoidal shape has

graph altitudes of 20m/s and 10m/s which when added and divided by 2 find 15m/s,

the average altitude of the graph. This is also the vAve of the object for the given

interval between 5seconds and 10 seconds.

The area of the trapezoid is equal to the graph altitude multiplied by the graph latitude

between the given points, which is: 15m/s * 5sec = 75meters, or the total displacement

for this interval.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q013. On a certain interval of duration `dt an object has initial velocity v_0

and final velocity v_f. In terms of the symbols v_0, v_f and `dt, what are the values of

the following?

vAve

`dv

`ds

aAve

Be sure to explain your reasoning.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

vAve=ave. graph altitude = (v_0 + v_f)/2

'dv= v_f - v_0

'ds= [(v_0 + v_f)/2] * 'dt ; b/c: 'ds = vAve*'dt

aAve= (v_f - v_0) / 'dt ; b/c: aAve='dv / 'dt

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

*********************************************

Question: `q011. The velocity of a car changes uniformly from 5 m/s to 25 m/s during

an interval that lasts 6 seconds. Show in detail how to reason out how far it travels.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First find the vAve by adding 5m/s and 25m/s to calculate 30m/s. Divide this by 2 to

get 15m/s.

Next calculate 'ds by multiply the interval 6 by the vAve, 15m/s to find 'ds=90meters.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q012. The points (5 s, 10 m/s) and (10 s, 20 m/s) define a 'graph

trapezoid' on a graph of velocity vs. clock time.

What is the average 'graph altitude' for this trapezoid?

Explain what the average 'graph altitude' means and why it has this meaning.

What is the area of this trapezoid? Explain thoroughly how you reason out this result,

and be sure to include and explain your units.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The points (5s, 10m/s) and (10s, 20m/s) produce a graph whose trapezoidal shape has

graph altitudes of 20m/s and 10m/s which when added and divided by 2 find 15m/s,

the average altitude of the graph. This is also the vAve of the object for the given

interval between 5seconds and 10 seconds.

The area of the trapezoid is equal to the graph altitude multiplied by the graph latitude

between the given points, which is: 15m/s * 5sec = 75meters, or the total displacement

for this interval.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q013. On a certain interval of duration `dt an object has initial velocity v_0

and final velocity v_f. In terms of the symbols v_0, v_f and `dt, what are the values of

the following?

vAve

`dv

`ds

aAve

Be sure to explain your reasoning.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

vAve=ave. graph altitude = (v_0 + v_f)/2

'dv= v_f - v_0

'ds= [(v_0 + v_f)/2] * 'dt ; b/c: 'ds = vAve*'dt

aAve= (v_f - v_0) / 'dt ; b/c: aAve='dv / 'dt

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!#*&!

@&

Excellent work. You clearly understand this very well.

*@