Open Query 7

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course Phy 121

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:

Start with what you know, and work toward what you don’t. The known data, v0, vf, and dt go toward the top. Drawing a line connecting the v0 and vf will give you the change in velocity. From there, you can connect change in velocity and dt to get acceleration.

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):x

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Self-critique rating #$&*3

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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:

A is the change in velocity over the time, so you can figure out the change invelocity by multiplying a by dt. Then given the change, and v0, you can get vf. From vf and v0, you can get vAve.

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

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Your solution:

The information here is actually pretty useful. It explains how to draw, read and reason through a flow diagram. I’m more of a visual person, so seeing the different steps of the flow diagram drawn out and the diagram get bigger with different colors connecting the entities needed to give the answer to the third, is pretty cool.

confidence rating #$&* 3

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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:

Vf and v0 go on the first level, as well as dt. Connecting them, you get vAve. Connect vAve back to dt, and you get distance. From the distance, you can get you cn set up the equation.Also connecting v0 and vf, you get the change in velocity. Connecting the change to dt, you get accelerastion.

The equation is ds = (vf +v0)/ds * dt

confidence rating #$&* 3

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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:

Putting v0, a, and dt in the top line, we connect dt and a to get the change in velocity, from the change in velocity, we figure out vAve, and then manipulate the numbers to get the third equation of motion, which is ds = v0 * dt + ½ a dt^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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Self-critique (if necessary):X

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Self-critique rating #$&*3

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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?

Your solution:

The five fundamental qualities are vf, v0, ds, dt, and a. In order to work with these guys though, we also need vAve, and dv.

confidence rating #$&* 2

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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):X

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Self-critique rating #$&*3

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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?

Your solution:

For the acceleration, the ratio of initial velocity to final velocity and therefore change in velocity is the same, whether we start at 0/ms, or 10m/s, since it is for the dame period of time.

However, if we have the same distance, but different velocities, then the change in velocities will differ.

In other words, when the distance is fixed, the velocities will differ, but when the time is fixed, and the distance is not, the difference in distance will make for fixed time and account for the same velocity.

confidence rating #$&*: 2

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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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Self-critique (if necessary):X

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Self-critique rating #$&*3

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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:

The graph v vs t, the slope would be the acceleration, while the area under the trapezoid represents the distance traveled. The first equation is ds = vAve *dt, this we can get from just multiplying the definition of vAve by dt. The second equation is ds = (vf + v0)/2 *dt, this we can also get from the definition of vAve, but we need to find dv first.

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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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Self-critique (if necessary):

I think I got this, not sure if I explained it the way it needed to be….

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Self-critique rating #$&*2

&#Good work. &#

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