Q-007

course PHY 231

7/13 2305

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

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.

007. `query 7

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Question: `qDescribe the flow diagram you would use for the uniform acceleration situation in which you are given v0, vf, and `dt.

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Your solution:

You can use v0 and vf to find vavg and 'dv. You can combine vavg and 'dt to determine 'ds, finally you can combine 'dv and 'dt to find a.

confidence rating #$&* 3

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Given Solution:

We start with v0, vf and `dt on the first line of the diagram.

We use v0 and vf to find Vave, indicated by lines from v0 and vf to vAve.

Use Vave and 'dt to find 'ds, indicated by lines from vAve and `dt to `ds.

Then use `dv and 'dt to find acceleration, indicated by lines from vAve and `dt to a. **

STUDENT COMMENT i dont understand how you answer matches up with the question

INSTRUCTOR RESPONSE All quantities are found from basic definitions where possible; where this is possible each new quantity will be the result of two other quantities

whose value was either given or has already been determined.

Using 'dt and a, find 'dv (since a = `dv / `dt, we have `dv = a `dt).

Using 'dv and v0, find vf, indicated by lines from `dv and v0 to vf (vf = v0 + `dv).

Using vf and v0, find vAve, indicated by lines from vf and v0 to vAve ( (vf + v0) / 2 = vAve, for uniform acceleration).

Using 'dt and vAve, find 'ds, indicated by lines from `dt and vAve to `ds (vAve = `ds / `dt so `ds = vAve * `dt).

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Self-critique (if necessary):

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Question: Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0

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Your solution:

'dt and a can be combined to find 'dv. 'dv and v0 can be used to find vf which in turn can be used to find vavg. vavg and 'dt can be used to find 'ds.

confidence rating #$&* 3

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Given Solution:

Student Solution: Using 'dt and a, find 'dv.

Using 'dv and v0, find vf, indicated by lines from `dv and v0 to vf.

Using vf and v0, find vAve, indicated by lines from vf and v0 to vAve

Using 'dt and vAve, find 'ds, indicated by lines from `dt and vAve to `ds.

STUDENT QUESTION

Can you only have two lines that connect to one variable because i utilized the formula vf=v0 +a `dt and connected all three

to find vf? I do see how it could be done using two in the above solution.

INSTRUCTOR RESPONSE

The idea is to use the definitions of velocity and acceleration whenever possible. This is possible in this case:

If you know `dt and a you can use the definition of acceleration to find `dv (which is equal to a `dt).

Then you can use v0 and `dv to get vf (which is equal to v0 + `dv; from this you could conclude that vf = v0 + a `dv).

You end up with the same result you would have gotten from the formula, but you are using insight into the nature of velocity and acceleration by using the

definitions, as opposed to a memorized formula that can be applied whether or not you understand its meaning.

The only exceptional cases are when you know v0 or vf (but not both), acceleration a and displacement `ds. In that case you need to start with the third or fourth

equation, where I recommend that you start with the fourth.

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Self-critique (if necessary):

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Question: Check out the link flow_diagrams and give a synopsis of what you see there.

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Your solution:

It shows how to construct the flow diagram that we have been using.

confidence rating #$&* 2

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Given Solution: You should have seen a detailed explanation of a flow diagram, and your 'solution' should have described the page.

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Question: Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion.

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Your solution:

V_avg*'dt = 'ds

v_avg can be rewritten using v0 and vf to form the equation

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

We can then solve for 'ds

a = 'dv/'dt

'dv can be rewritten using v0 and vf thus we have:

a = (vf-v0)/'dt

We can then solve for a

confidence rating #$&* 2

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Given Solution:

Student Solution:

v0 and vf give you `dv = vf - v0 and vAve = (vf + v0) / 2.

`dv is divided by `dt to give accel. So we have a = (vf - v0) / `dt.

Rearranging this we have a `dt = vf - v0, which rearranges again to give vf = v0 + a `dt.

This is the second equation of motion.

vAve is multiplied by `dt to give `ds. So we have `ds = (vf + v0) / 2 * `dt.

This is the first equation of motion

Acceleration is found by dividing the change in velocity by the change in time. v0 is the starting velocity, if it is from rest it is 0. Change in time is the ending

beginning time subtracted by the ending time. **

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Question: Explain in detail how the flow diagram for the situation in which v0, a and `dt are known gives us the third fundamental equations of motion.

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Your solution:

`ds = v0 `dt + (a `dt^2)/2

a multiplied by 'dt gives us 'dv from which we can then calculate vf because we know v0.

with v0 and vf we can get vavg which is also equal to 'ds/'dt

vavg = 'ds/'dt

(v0 + vf)/2 = 'ds/'dt

a = 'dv/'dt

a*'dt = 'dv = vf - v0

vf = 'dv+v0 = a*'dt + v0

Plug this into the first equation

[v0+(a*'dt+v0)]/2 = 'ds/'dt

(v0*2/2) + (a*'dt)/2 = 'ds/'dt

Finally:

'dt*v0 + (a*'dt^2)/2 = 'ds

confidence rating #$&* 2

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Given Solution:

a and `dt give you `dv. `dv and v0 give you vf. v0 and vf give you vAve. vAve and `dt give you `ds.

In symbols, `dv = a `dt.

Then vf = v0 + `dv = v0 + a `dt.

Then vAve = (vf + v0)/2 = (v0 + (v0 + a `dt)) / 2) = v0 + 1/2 a `dt.

Then `ds = vAve * `dt = [ v0 `dt + 1/2 a `dt ] * `dt = v0 `dt + 1/2 a `dt^2. **

STUDENT COMMENT:

I do not understand how to get the equation out of the flow diagram or calculations.

INSTRUCTOR RESPONSE:

Presumably the flow diagram was the basis for your responses

'You can get the `dv from the `dt and a by: a * `dt = `dv

Then, you can get the vf by: `dv + v0 = vf

Next, you can get the vAve by: (v0 + vf) / 2 = vAve

Then, you can get the `ds by: vAve * `dt = `ds

The change in position is what is being solved for in the equation: `ds = v0 * `dt + .5 a `dt^2.'

Using your responses as a basis:

You can get the `dv from the `dt and a by: a * `dt = `dv

Then, you can get the vf by: `dv + v0 = vf.

Since `dv = a * `dt, we have a * `dt + v0 = vf

Next, you can get the vAve by: (v0 + vf) / 2 = vAve

Then, you can get the `ds by: vAve * `dt = `ds

v0 is considered to be one of the given quantities, and vf = v0 + a `dt from the line before the preceding line. So

vAve * `dt

= (v0 + vf) / 2 * `dt

= (v0 + (v0 + a `dt) ) / 2 * `dt

= (2 v0 + a `dt) / 2 * `dt

= (v0 + 1/2 a `dt) * `dt

= v0 `dt + 1/2 a `dt^2.

The change in position is what is being solved for in the equation: `ds = v0 * `dt + .5 a `dt^2.

the preceding showed that

`ds = v0 `dt + 1/2 a `dt^2

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Question: Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five?

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Your solution:

Because each of those seven terms are all involved somehow, it is useful to know in case some of them are not known.

confidence rating #$&* 1

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Given Solution:

ONE WAY OF PUTTING IT:

The four equations are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. However to think in terms of meanings we have to be able to think

not only in terms of these quantities but also in terms of average velocity vAve and change in velocity `dv, which aren't among these five quantities. Without the

ideas of average velocity and change in velocity we might be able to use the equations and get some correct answers but we'll never understand motion.

ANOTHER WAY:

The four equations of unif accelerated motion are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds.

The idea here is that to intuitively understand uniformly accelerated motion, we must often think in terms of average velocity vAve and change in velocity `dv as well

as the five quantities involved in the four fundamental equations.

one important point is that we can use the five quantities without any real conceptual understanding; to reason things out rather than plugging just numbers into

equations we need the concepts of average velocity and change in velocity, which also help us make sense of the equations. **

STUDENT QUESTION

I understand how to make flow diagrams and use all of the concepts to figure out the missing variable from the equation. I even understand `dv and vAve are intuitive

but don't these still show up in the flow diagrams?

Aren't they still in a sense being modeled?

Good question.
They show up in the diagrams but not in the four equations of uniformly accelerated motion.
The point is that in the process of reasoning out a situation, we must always use `dv and vAve, both of which are part of our definitions of velocity and acceleration.
However we can write a set of equations that do not include vAve and `dv as variables. These equations involve only v0, vf, a, `ds and `dt. Given any three of these
five we can use the equations to find the other two, and we never have to think about `dv and vAve to do so. We reduce the physics to a mechanical process involving
only simple algebra, unconnected to the basic definitions.
The five-variable formulation is very nice and easy to use. We can use it to solve problems in fewer steps than the direct-reasoning-from-definitions approach, and
this is something we very much want to be able to do.
The trick in a first-semester physics course is to achieve a very basic understanding of uniformly accelerated motion, eventually learning to use the equations without
using them as a crutch to bypass understanding.
So we learn to reason using the seven quantities, then we learn to use the four-equation model.
There is an additional approach for University Physics students, which involves calculus and is not relevant (and not accessible) to anyone who doesn't know calculus.
We first understand how the derivative is an instantaneous rate-of-change function, so that the velocity function is the derivative of the position function, and the
acceleration function the derivative of the velocity function. Then, understanding how the integral is the change-in-quantity function, we integrate the acceleration
function with respect to clock time to get the velocity function, and finally integrate the velocity function to get the position function.

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Self-critique (if necessary):

Interesting.

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Question: Accelerating down an incline through a given distance vs. accelerating for a given time

If we accelerate down a constant incline for `dt seconds, starting at some initial velocity, then repeat the process, accelerating for `dt second but with another

initial velocity, the change `dv in velocity will be the same for both trials.

If we accelerate through displacement `ds on a constant incline, starting at some initial velocity, then repeat the process, accelerating through displacement `ds but

with another initial velocity, the change `dv in velocity will be different for the two trials.

Why does a given change in initial velocity result in the same change in final velocity when we accelerate down a constant incline for the same time, but not when we

accelerate down the same incline for a constant distance?

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Your solution:

It is just a displacement of velocity. If you picture a linear graph you have v0 and vf spaced so far apart from eachother. The distance is dependent on the

accelertation and 'dt. If you start with a different initial velocity, 'dv remains the same, but the graph shifts.

confidence rating #$&* 3

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Given Solution:

If we accelerate down a constant incline our rate of change of velocity is the same whatever our initial velocity.

So the change in velocity is determined only by how long we spend coasting on the incline. Greater `dt, greater `dv.

If you travel the same distance but start with a greater speed there is less time for the acceleration to have its effect and therefore the change in velocity will be

less.

You might also think back to that introductory problem set about the car on the incline and the lamppost. Greater initial velocity results in greater average velocity

and hence less time on the incline, which gives less time for the car to accelerate. **

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Question: Explain how the v vs. t trapezoid for given quantities v0, vf and `dt leads us to the first two equations of linearly accelerated motion.

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Your solution:

vf and v0 are the legs of the trapezoid while 'dt is the width of the base. Vavg can be inferred, which is the average height. From Here we can get the change in

distance by multiplying the base by the average height. You can infer 'dv from vf and v0 and thus find a.

confidence rating #$&* 2

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Given Solution:

If acceleration is uniform then the v vs. t graph is linear. So the average velocity on the interval is vAve = (vf + v0) / 2.

From the definition of average velocity we conclude that `ds = vAve * `dt.

Thus `ds = (vf + v0) / 2 * `dt. This is the first equation of uniformly accelerated motion.

Note that the trapezoid can be rearranged to form a rectangle with 'graph altitude' vAve and 'graph width' equal to `dt. The area of a rectangle is the product of its

altitude and its width. Thus the product vAve * `dt represents the area of the trapezoid.

More generally the area beneath a v vs. t graph, for an interval, represents the displacement during that interval.

For University Physics, this generalizes into the notion that the displacement during a time interval is equal to the definite integral of the velocity function on

that interval.

The definition of average acceleration, and the fact that acceleration is assumed constant, leads us to a = `dv / `dt.

`dv = vf - v0, i.e., the change in the velocity is found by subtracting the initial velocity from the final

Thus a = (vf - v0) / `dt.

`dv = vf - v0 represents the 'rise' of the trapezoid, while `dt represents the 'run', so that a = `dv / `dt represents the slope of the line segment which forms the

top of the trapezoid.

For University Physics, this generalizes into the notion that the acceleration of an object at an instant is the derivative of its velocity function, evaluated at

that instant.

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Question:

(required only of University Physics students): If s(t) = .3 m/s^3 * t^3 - 2 m/s^2 * t^2 + 5 m/s * t + 12 m then what are the velocity and acceleration functions?

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Your solution:

v(t) = 0.9mt^2/s^3 - 4mt/s^2 + 5m/s

a(t) = 1.8mt/s^3 - 4m/s^2

confidence rating #$&* 3

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Given Solution:

If s(t) = .3 m/s^3 * t^3 - 2 m/s^2 * t^2 + 5 m/s * t + 12 m then:

The derivative of .3 m/s^3 * t^3 is

(.3 m/s^3 * t^3 ) ' = (.3 m/s^3) * (t^3) ' = (.3 m/s^3) * (3 t^2) = .9 m/s^3 * t^2.

Note that .3 m/s^2 is a constant, and also that if t is in seconds the units of the result are m/s^3 * (s)^2 = m/s, which is the unit of velocity.

Similarly the derivatives for the other terms are

(-2 m/s^2 * t^2 ) ' = -4 m/s^2 * t

(5 m/s * t) ' = 5 m/s and

(12 m) ' = 0

Thus the derivative of s(t) is

v(t) = s ' (t) = .9 m/s^3 * t^2 - 4 m/s^2 * t + 5 m/s

The acceleration function is the derivative of v(t):

a(t) = v ' (t) = 1.8 m/s^3 * t - 4 m/s^2

You should check to be sure you understand that the units of each of these terms are m/s^2, which agrees with the unit for acceleration.

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