QA Assignment 5

#$&*

course PHY 121

6/15 11

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)

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Your solution:

The change in velocity is 25 m/s - 5 m/s or 20 m/s

The average acceleration is (25 m/s - 5 m/s) / 4 s = 20 m/s / 4 s or 5 ms^2

The average velocity of the object is (5 m/s + 25 m/s) / 2 = 30 m/s / 2 or 15 m/s

confidence rating #$&*:

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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: 3

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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:

Every second it is traveling 5 m/s, so for 4 seconds, it would be 5 m/s * 4 s = 20 m.

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):

I knew that I was off when I had the m/s^2, but didn't stop to think it through. To

find out how far it traveled, I needed the average speed times the seconds not the

acceleration times the seconds.

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Self-critique rating: 3

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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.

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Your solution:

If you have a certain interval of time where you know how fast an object is moving at

the beginning and how fast the object is moving at the end, you can find the amount of

distance traveled by averaging those 2 speeds and multiplying by the amount of time that

has elapsed. To find the acceleration, you would take the first velocity and subtract

it from the second speed and then divide that difference by the amount of time that

elapsed.

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: 3

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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.

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Your solution:

aAve = (vf-v0)/'dt

vAve = (v0+vf)/2

'ds=(vf-v0) * 'dt

confidence rating #$&*:

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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):

On the final one, I forgot to divide by 2 to get the average before I multiplied by the

time. I do better when I use actual numbers at this point. I think if I'd done that

with numbers, I would have said, ""That answer can't be right."" I'm still working on

these formulas and how they relate.

@&

You're making excellent progress.

*@

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Self-critique rating: 3

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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?

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Your solution:

The x axis represents seconds. The y axis represents meters/second. The initial time,

as given in this problem is 0. The velocity as given in problem 1 is 5 m/s. So the

first point is plotted at (0,5). The elapsed time, as given in problem 1 is 4 seconds

and the final velocity is given as 25 m/s, so the second point is plotted at (4,25).

The graph has a very steep slope of 5 m/s for every 1 second.

confidence rating #$&*:

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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):

I guess I should also put the units of measure in the ordered pairs. I did not think

that was usually done.

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Self-critique rating: 3

@&

If you include the units you avoid a lot of errors, and you also provide yourself with a quick reference to meanings.

*@

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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:

The v vs. t graph is increasing at a constant rate of 5 m/s for every second.

confidence rating #$&*:

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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: 3

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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:

The slope is 4 m/s / 1 s. You get this from the rise of 25 m/s - 5 m/s = 20 m/s, and

the run of 4 s - 0 s = 4 s. 20 m/s / 4 s = 5 m/s^2

confidence rating #$&*:

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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

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Self-critique rating: 3

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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?

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Your solution:

The average altitude of the trapezoid (5 m/s + 25 m/s)/2 = 30 m/s / 2 = 15 m/s. This

represents the average velocity of the object during this time interval.

The area of the trapezoid is base times average height or 15 m/s * 4 s = 60 m, which is

the amount of displacement of the object during this 4 second time frame.

confidence rating #$&*:

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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

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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_quantities.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.

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Your solution:

I was able to access the first link but not the second. I went through and wrote out

what each formula meant so that I could associate words with the symbols. I substituted

the information that was given into the formula to solve. At first, I just substituted

the numbers and then worked from there without rearranging the formulas. After the

first few I tried rearranging the formulas first before I plugged in the numbers. I

generally got a good start and then had some minor difficulty with my math or I was on

the right track but lacked self confidence in my solution. I copied and pasted this

document onto a Word Document so that I can refer back to it. I did all of my work on

the paper and marked on the ""Given Solution"" where I was on track and where I had

difficulty.

confidence rating #$&*:

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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

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Self-critique rating: 2

If you understand the assignment and were able to solve the previously given problems

from your worksheets, you should be able to complete most of the following problems

quickly and easily. If you experience difficulty with some of these problems, you will

be given notes and we will work to resolve difficulties.

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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.

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Your solution:

First you have to find the average velocity of the car. You do this by averaging the

initial and final velocities. 5 m/s + 25 m/s = 30 m/s 30 m/s / 2 = 15 m/s.

Then you have to multiply the average velocity by the elapsed time. 15 m/s * 6 s = 90

meters.

confidence rating #$&*:

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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.

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Your solution:

The graph altitude is the average of the two vertical sides of the trapezoid. 10 m/s +

20 m/s = 30 m/s; 30 m/s / 2 = 15 m/s. This represents the average velocity of the

object during this time interval. You obtain the average for the sides of the trapezoid

in the same way you would mathematically obtain the average velocity by taking the

initial velocity and the final velocity, adding them and dividing that sum by 2.

In order to find the area, you have to multiply the average height that you just found

by the elapsed time. In this case the elapsed time is 10 s - 5 s = 5 s. When you

multiply 15 m/s * 5 s, you will get 75 meters. This represents the distance the object

moved during these 5 seconds at this average velocity.

confidence rating #$&*:

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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.

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Your solution:

vAve means average velocity. To find average velocity, you add the initial velocity (v_0) to the final velocity (v_f) and divide the sum by 2

vAve=(v_0 + v_f)/2

'dv means the change in velocity. To find the change in velocity, you subtract the initial velocity (v_0) from the final velocity (v_f)

'dv=v_f-v_0

'ds means change in distance. To find the change in distance, you find the average of the velocity (above) and multiply by the elapsed time ('dt)

'ds=(v_0+v_f)/2 * 'dt

aAve means the average acceleration. To find the average acceleration, you find the change in velocity (above) and divide by the elapsed time ('dt)

aAve=(v_f-v_0)/'dt

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating:

&#Very good responses. Let me know if you have questions. &#