Assignment 5 qa

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course Phy 121

10/9 2

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 velocity of the object changes by 20 m/s. (You find this by taking the vf - v0, which in this case is 25 m/s - 5 m/s)

aAve = (25 m/s - 5 m/s) / 4s = 20 m/s / 4s = 5 m/s/s = 5 m/s^2.

The average velocity of the object can be found by taking the mean or average of the two velocities. vAve = (25 m/s + 5 m/s) / 2 = 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):OK

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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 the distance traveled by the object in the preceding problem, we need to solve for `ds. The formula needed for this is the average velocity formula, and luckily we have been given the appropriate factors to solve for this.

`ds = `dt * vAve

`ds = 4 s * 15 m/s

`ds = 60 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):OK

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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:

To find the acceleration if we know the v0 and vf, we simply set up the formula like this. aAve = (vf - v0) / `dt The expression (vf - v0) represents the change in velocity (`dv).

To find the distance traveled we must have the average velocity. Therefore we have to incorporate the average velocity (or mean of the velocities) in the formula. `ds = vAve * `dt this can be written in terms of v0 and vf here: `ds = [(vf + v0) / 2] * `dt

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):

I think I took my explanation a little farther than what the question was asking, but I think I explained it right.

@&

You had a good explanation.

*@

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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

`ds = [(vf + v0) / 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):OK

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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 don't know where I am getting these numbers from, unless it is from questions 1 and 2. I will go by that, since I am not sure.

The graph of velocity vs clock time for this problem would have points at the coordinates of (0, 5) and (4, 25). The midpoint for the vertical quantities is at 15 m/s (average velocity as found in problems 1 and 2). The line is straight (assuming it has a uniform acceleration) and has a slope (which is found the same as the acceleration in problems one and two) of 5 m/s^2.

The clock time where the final velocity is attained is at 4 sec.

When t = 0, v = 5

The coordinates of the point corresponding to the final velocity is (4, 25).

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 did not include my units of measure in my coordinates. I did not know if that was needed or if the standard way of writing out coordinates includes the unit of measure with them.

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Self-critique rating:3

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Question: `q007. This situation continues the preceding.

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.

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: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:

The slope of the graph between these two points is 5 m/s^2. This is found by taking the difference in the v values of the coordinates and dividing it by the difference in the corresponding t values of the coordinates. Essentially it is the exact same values for finding the average acceleration of the object.

m = (25 m/s - 5 m/s) / 4 s = 20 m/s / 4s = 5 m/s/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:OK

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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 is 15 m/s. This number represents the average velocity. Both were found by taking (25 m/s - 5 m/s) / 2 = 30 m/s / 2 = 15 m/s.

The area of the trapezoid is found by taking the new altitude of 15 m/s and multiplying it by 4 s (this is treating the trapezoid as a rectangle with points at (0, 0), (0, 15), (4, 15), and (4, 0). Therefore the Area of this trapezoid is 15 m/s * 4s = 60 m/s * s = 60 m (I don't know if that unit is right. It could supposed to be squared, because when you find the area of something, the units are always squared.)

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.

STUDENT COMMENT

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.

INSTRUCTOR RESPONSE

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 (if necessary):OK

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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:

In this document, there are 9 problems that ask you various questions about average velocity, average acceleration, initial velocity, final velocity, change in distance, and change in clock time. I feel like I understand these questions and aspects of the various aspects of velocity and acceleration, but if I do feel like I need a brushing up on the material, I will know where it is and how to find it again.

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: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:

We know that to find how far (change in distance) an object travels, we need to have a formula that has change in distance somewhere in it. We know that the formula to find average velocity has change in distance in it, therefore with the given information we can alter the formula to solve for change in distance.

vAve = `ds / `dt

`ds = vAve * `dt

Now that we have a formula to use, we need to take what we have been given in the problem and make it fit our needs. We know that we need a change in time, which is given as 6 sec, and we also need the vAve which we can find by knowing that (vf + v0 / 2) will give us the average velocity of the object.

`ds = [(vf + v0) / 2] * `dt

`ds = [(25 m/s + 5 m/s) / 2] * 6 s = ( 30 m/s / 2) * 6 s = 15 m/s * 6 s = 90 m/s * s = 90 m

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 average 'graph altitude' for this trapezoid is the sum of the two altitudes of 20 and 10 divided by 2. This is worked out like (20 m/s + 10 m/s) / 2 = 30 m/s / 2 = 15 m/s

The average 'graph altitude' here shows where our coordinates are in order to find the area, but it also represents (in this case) the average velocity of the moving object.

The area of this trapezoid is found just like a rectangle, after we have found the average 'graph altitude' (which we already have). We treat the new altitude as the other 2 points that make a rectangle. Those points are (5 s, 0 m/s), (5 s, 15 m/s), (10 s, 15 m/s), and (10 s, 0 m/s).

Therefore we now have 2 sides that are 5 s wide and 2 sides that are 15 m/s high. A = 5 s * 15 m/s = 75 m/s *s = 75 m In the last step we see the units m/s *s, this ends up making the 2 s cancel out (or equal one) leaving just the m.

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 = `ds / `dt If we are not given a `ds, but we are given v_0 and v_f, we can also find vAve by using this: vAve = (v_f + v_0) / 2 This is essentially just taking the mean of the initail and final velocities

`dv is found by finding the difference between the final velocity and the initail velocity. `dv = v_f - v_0

`ds is found by altering the formula for vAve. vAve = `ds / `dt We need to get the `ds by itself and after that is done algebraically, we have `ds = vAve * `dt. In this form, only knowing the initial and final velocities, and not being given the average velocity beforehand, we need to solve for vAve using the information we do have. It can be written in one step as: `ds = [(v_f + v_0) / 2] * `dt

aAve = `dv / `dt To find the aAve in terms of only knowing the initial and final velocities as well as the time, we need to allocate for `dv, this is simply done by taking v_f minus v_0. This all written out will look like: aAve = (v_t - v_0) / `dt

confidence rating #$&*:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q014. (University Physics) Given the velocity function

v(t) = 0.1 m/s*3 * t^2 + 2 m/s^2 * t + 4 m/s

evaluate v for t = 5 seconds and for t = 10 second, being sure to verify that the units of our calculation are consistent and that they work out to units of velocity.

On a graph of v vs. t, what would be the coordinates of the points corresponding to t = 5 sec and t = 10 sec?

What would be the area of the 'graph trapezoid' defined by these coordinates, and what does that area mean?

Would this area be the same as, more than, or less than the area under the actual graph of v vs. t.?

What would be the actual area, and what is its meaning?

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Your solution:

confidence rating #$&*:

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Self-critique (if necessary):

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