Qa 05

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course Phy 121

7/1 5

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

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:

To find out how much this object changes you would subtract 5 m/s from 25 m/s to get 20 m/s which is the first answer.

aAve or the average acceleration = change in quantity/duration of time interval

So, we know that the change is 20 m/s and the time interval is 4 seconds.

So, (20 m/s ) / (4 seconds) = 5 m/s^2

The average velocity would be equal to the average of the initial and final velocities. So, 5 m/s + 25 m/s = (30 m/s) / 2 = 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):

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Self-critique rating: OK

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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:

To find how far the object travels in a certain time interval you would need to multiply the average velocity by the time interval

So, 15 m/s * 4 seconds, which would be 60 m/s

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):

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Self-critique rating: OK

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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:

First of all, we can find the acceleration by subtracting the initial velocity from the final velocity and dividing the answer by the time interval.

Once we have the average velocity we can multiply it by the time interval to find how far it traveled.

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):

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Self-critique rating: OK

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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:

The average acceleration is equal to the change in quantity divided by the time interval or a = vf - v0 / `dt

vAve = average of initial and final velocities/ 2 or vAve = (v0 + vf) /2

`dt = final velocity - initial velocity or `dt = vf - v0

`ds = (v0 + vf) /2 * `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):

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Self-critique rating: OK

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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:

I am unsure of how to do this problem. I understand that these graphs are straight lines and remain constant but I don’t really understand how to construct one given just these coordinates.

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 wasn’t sure on how to do this, but now I understand that you had to put the initial and final velocities in to make the coordinates.

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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:

This graph seems to be increasing when you sketch out the coordinates. Because the x-axis is going from 0 to 4 and the y coordinates is increasing from 5 to 25. Also, it’s increasing at a constant rate, as all of these graphs do since it is uniform.

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):

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Self-critique rating: OK

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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:

Since the coordinates are (0, 5 m/s) and (4 s, 25 m/s) and the slope is rise/run, you need to divide the two v coordinates over the t coordinates. So you need to subtract 5 m/s from 25 m/s which is 20 m/s. Then you subtract 0 from 4 s which is 4 sec. So, (20 m/s) / (4 sec) = 5 m/s^2 for the slope.

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):

I didn’t really know what to put for what it represents, which is the change in `dv in the velocity divide by the change `dt in the clock time which represents the acceleration.

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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:

To find the average altitude you need to multiply the two attitudes and divide by 2. So 25 m/s and 5 m/s = 30 m/s. Then you divide it by 2 which is 15 m/s.

And, the area is the graph width * the average altitude. So, 4 * 15 m/s = 60 m/s^2

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):

I understand how to find the average altitude and multiply it by the amount of seconds. I also understand how to find the area of the trapezoid. But, again I don’t understand what it repreents, which is the product of the average velocity and the time interval, or the displacement.

@&

If you multiply the average velocity on a time interval by the duration of the interval, you get the displacement.

Since the average altitude represents the average velocity and the width represents the duration of the time interval, the product therefore represents the displacement.

Since the product of average altitude and width is area, it follows that this product represents the displacement.

*@

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Self-critique rating: OK

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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 see an assignment with 9 problems, similar to this assignment to which I can answer the question and then look at the given solution. I think I understand these quantities pretty well and I have written them down in my notebook and I keep going over them and looking back to help me with my problems. If I have any trouble or want to practice these problems, I have bookmarked this site so I can go back and have more practice.

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):

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Self-critique rating: OK

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:

To figure out how far an object travels you would need to multiply the average velocity by the time interval.

So, to find the average velocity you need to add the initial and final velocities and divide it by 2. So 25 m/s + 5 m/s = 30 m/s. Then (30 m/s) / 2 = 15 m/s.

Then, 15 m/s* 6 sec = 90 m/s^2

@&

Good reasoning, but m/s * s does not give you m/s^2. If you divide m by s, then multiply by s, what do you get?

*@

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:

To find the average graph altitude you need to add the 2 altitudes, 10 m/s + 20 m/s = 30 m/s and divide it by 2, so that gives you 15 m/s.

The average graph altitude is the average of the two altitudes.

The area of the trapezoid is the graph width multiplied by the average altitude.

So, to find the graph width you subtract the x coordinates, 10 s - 5 s = 5 s. Then, you multiply 5 s and 30 m/s to get 150 m/s^2 for the area of the trapezoid.

@&

Your units are not correct; the rest of your reasoning is excellent.

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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 = (v0 + vf) / 2: in order to find the average velocity you must add your initial and final velocities and divide by 2.

`dv = vf - v0: to find the change in velocity you have to subtract the initial velocity from the final.

`ds = (v0 + vf) /2 * `dt: you need find the average velocity and multiply it by the time interval.

aAve = `dv/ `dt: you need to divide the change in quantity over the time duration you find the answer.

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating: OK

@&

You're doing very well with everything but the units.

See my notes, including an answer to your one question.

Let's be sure to get a handle on this so it doesn't cause you problems:

Please use a submit work form or a question form to submit just the following, along with your answer to my questions:

m/s * s does not give you m/s^2. If you divide m by s, then multiply by s, what do you get?

What therefore are the units of your (correct) calculation

15 m/s* 6 sec

and why are those units appropriate to the quantity you were finding?

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