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course PHY 201

2/09 9

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:

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

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

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

confidence rating #$&*:

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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:

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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:

`ds = 15m/s * 4sec = 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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Self-critique rating:

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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:

Acceleration is found by taking the difference in vF and v0 and dividing that figure by the amount of time transpired, or 'dt.

Distance traveled is found by multiplying the known time interval by the average velocity which is vf and v0 divided by 2.

confidence rating #$&*:

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Given Solution:

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

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

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

we know the time interval we can now determine the distance traveled.

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Self-critique (if necessary):

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Self-critique rating:

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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 = (v_f -v_0) / `dt

vAve = (v_f + v_0) / 2

`dt = aAve * vAve

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:

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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 graph has velocity in m/s on the vertical axis and clocktime represented on the horizontal axis. v_0 = (0,5) and v_f = (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):

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Self-critique rating:

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

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

decreasing rate?

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Your solution:

Increasing at constant rate between (0s, 5m/s) and (4s, 25m/s).

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:

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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:

m = (25m/s - 5m/s) / (4s - 0s) = 20m/s / 4s = 5m/s/s

The slope, 5m/s/s is the average acceleration between the two intervals.

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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Self-critique rating:

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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:

ave. altitude is equal to (5m/s + 25m/s) /2 =15 m/s, which represents the vAve.

Area= 4sec * 15m/s = 60m which represents the total displacement during the interval of 4sec.

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):

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Self-critique rating:

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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:

The URL brought up a document with 9 questions related to uniformly accelerated motion. I have save this URL in my notebook, and will refer back to it if I feel I need more practice.

confidence rating #$&*:

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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:

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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 find the vAve by adding 5m/s and 25m/s to calculate 30m/s. Divide this by 2 to get 15m/s.

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

confidence rating #$&*:

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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 points (5s, 10m/s) and (10s, 20m/s) produce a graph whose trapezoidal shape has graph altitudes of 20m/s and 10m/s which when added and divided by 2 find 15m/s, the average altitude of the graph. This is also the vAve of the object for the given interval between 5seconds and 10 seconds.

The area of the trapezoid is equal to the graph altitude multiplied by the graph latitude between the given points, which is: 15m/s * 5sec = 75meters, or the total displacement for this interval.

confidence rating #$&*:

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

'dv= v_f - v_0

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

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

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating:

Asst_5_QA5

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