qa 05

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course Phy 231

2/2 9:20pm

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:

change in velocity = `dv = vf-v0 = 25m/s-5m/s = 20m/s. The velocity changes by 20m/s.

average acceleration = change in velocity/change in time =`dv/`dt = (20m/s)/(4s)= 5m/s^2. Average acceleration is 5m/s^2.

Average velocity = vAve =(v0+vF)/2 =(5m/s+25m/s)2 =(30m/s)/2 =15m/s. Average velocity is 15m/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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Question: `q002. How far does the object of the preceding problem travel in the 4 seconds?

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Your solution:

Average velocity is the rate of change in position with respect to time.

vAve=`ds/`dt

15m/s=(`ds)/4s

(15m/s)(4s)=`ds

60m=`ds

The object travels 60m.

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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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:

We know that the object increases in speed from a velocity of v0 to a velocity of vf, and we assume that it is increasing at a consistent rate. So, we look at the time interval and see how much the velocity would have to change per unit of time within that interval in order to increase to vf in the given time. This means we need to divide the change in velocity, or (vf-v0), by the time interval. That will tell us the rate the velocity is changing per unit of time; that is the acceleration.

To find the distance travelled, we first need to find the average speed during that interval by adding up the initial and final velocities and dividing by 2. This tells us the average distance per unit of time the object is moving. So now we can multiply by the number of units of time, which well tells us the distance the object travels over the entire interval.

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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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:

If given v0 and vf, we find their difference to calculate `dv. Now we can evaluate aAve=`dv/`dt.

We find the sum of v0 and vf and divide by 2 to find vAve. vAve=(v0+vf)/2

Finally, we know that vAve=`ds/`dt, which can be rearranged as (vAve)(`dt)=`ds. Therefore, `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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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:

Since the time interval of 4 seconds begins at t=0, the interval would end at t=4. So the clock time for final velocity would be t=4.

The coordinate corresponding to initial velocity would occur at t=0, initial clock time. Initial velocity is given as 5m/s, so the coordinate would occur at point (0,5), with time on the x-axis and velocity on the y-axis.

The coordinate corresponding to final velocity would occur at t=4, final clock time. Final velocity is given as 25m/s, so the ccoordinate would occur at point (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):

??? In general, should we use units when naming coordinates on a graph ???

@&

Yes. There are many good reasons for this, one being that the units help us keep the meanings of the quantities in mind.

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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 graph is increasing because velocity is increasing during that interval. The increase is at a constant rate (acceleration is uniform), so the line is straight.

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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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:

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.

Slope is calculated as rise/run. In this case, the rise is the change in velocity, which begins at 5m/s and increases to 25m/s for a rise of 20m/s. The run is the change in time, which begins at 0s and ends at 4s for a run of 4s. So, the rise/run is 20m/s/4s, or 5m/s^2. The slope of a velocity v. time graph shows the rate at which velocity is changing with respect to time, or the rate at which the object is accelerating.

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):

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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 can be found by calculating the sum of the lower and higher altitudes and dividing by 2. In this case, the altitudes are initial velocity and final velocity, so our calculation here will lead us to the average velocity once again. (v0+vf)/2=(30m/s)2=15m/s.

To find the area of the trapezoid, we multiply base by average height. Here, that's 4s*15m/s, the same as time interval * average velocity. We end up with a result of 60m, the total displacement of the object. This shows us that on a velocity v. time graph, the area under the graph equals the total displacement of the object.

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):

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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 feel good about my ability to solve these problems. But, I went about one of them completely differently. The formula you suggest for question 5, 'ds=v0`dt + .5a `dt^2, is totally unfamiliar to me. So I didn't feel good plugging into that formula, but I reasoned through using a combination of formulas (which I'm assuming this is). I came up with the right answer.

Is it important that I know this formula? Or is knowing all the basic relationships ok?

@&

There are four formulas that can be used to solve any problem with uniformly accelerated motion, provided you know at least three of the relevant quantities (three of the quantities v0, vf, `ds, `dt, a) without the need to combine formulas.

If you haven't yet seen them yet you will very soon.

This one is among them. It can be derived from the expressions for average velocity and acceleration

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

and

a = (vf - v0) / `dt

treating these as simultaneous linear equations and eliminating vf.

You'll see all of this shortly.

*@

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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The formula you quote here can also be derived very easily by integrating a constant acceleration twice with respect to time.

We have

a = constant

so that

v(t) integral(a dt) = a t + c

and c = v(0). Thus

v(t) = a t + v(0),

and

s(t) = integral(v(t) dt) = integral ( ( a t + v(0) ) dt) = 1/2 a t^2 + v(0) * t + c

where now c = s(0), so

s(t) = 1/2 a t^2 + v(0) * t + s(0),

or

s(t) - s(0) = 1/2 a t^2 + v(0) * t

s(t) - s(0) is the change in position s so we have

`ds = 1/2 a t^2 + v(0) * t.

This is easily identified with the equation

`ds = 1/2 a `dt^2 + v_0 * `dt.

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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:

vAve=(v0+vf)/2 =(5m/s+25m/s)/2 =(30m/s)2 = 15m/s.

`ds=(vAve)(`dt)

`ds=(15m/s)(6s)

`ds=90m

The car travels 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:

Average graph altitude is the average of the lowest height and the highest height of the trapezoid formed under the graph.

We calculate it by adding the two heights, in this case the lowest velocity and the highest velocity, and dividing by two. On a graph of velocity v. time, average graph altitude is the same as average velocity.

Here, average graph altitude is (10m/s+20m/s)/2, or 15m/s. So the average velocity represented in this graph is 15m/s.

The area is calculated by multiplying the base by the average graph altitude.

Here, that's (10s-5s)*(15m/s) or 5s*15m/s or 75m. On a graph of velocity v. time, the area under the graph represents displacement. So this graph tells us that an object is displaced 75m.

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=(v_0+v_f)/2, or the average of the initial and final velocities.

`dv=(vf-v0), or the difference between the initial and final velocities; how much the velocity changes over the interval

`ds=[(v_0+v_f)/2](`dt), or the average distance travelled per unit of time, multiplied by the number of units of time to give us total distance travelled

aAve=(vf-v0)/(`dt), or the change in velocity with respect to time

confidence rating #$&*:

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Self-critique (if necessary):

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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:

v(5)= 0.1 m/s*3 * 5^2 + 2 m/s^2 * 5 + 4 m/s

@&

t is not 5. t is 5 seconds. So

v(5 s)= 0.1 m/s*3 * (5 s)^2 + 2 m/s^2 * 5 s + 4 m/s

The units of every term now work out to m/s.

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=0.3m/s * 25 + 2m/s^2 * 5 + 4 m/s

=0.75m/s + 10m/s^2 + 4m/s

=4.75m/s + 10m/s^2

v(10)= 0.1 m/s*3 * 10^2 + 2 m/s^2 * 10 + 4m/s

=0.3m/s * 100 + 20m/s^2 + 4 m/s

=3m/s + 20m/s^2 + 4m/s

=7m/s + 20m/s^2

(Stuck here! Units!!)

The presence of a power of 2 leads me to believe that the actual shape of the graph might be a curve.

The actual area signifies total displacement over the time interval fom 5 sec to 10 sec. We find it by integrating the velocity function.

confidence rating #$&*:

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Self-critique (if necessary):

I cannot figure out how this can come out in terms of units of velocity or m/s. Unless I am messing something up with order of operations, but I can't see how the s^2 is ever going to go away.

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Self-critique rating:

@&

I'm very glad to see you trying to reconcile the units here. This is very important, and your confusion is easy enough to correct, per my more detailed note above.

You're just leaving the units off of the value of t. That's easy to fix. Check my note.

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#