course phy121
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02:05:54 Describe the flow diagram you would use for the uniform acceleration situation in which you are given v0, vf, and `dt.
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RESPONSE --> From this first layer, v0, vf and `dt, you can then form the 2nd layer a few ways. From v0 and vf you can find `dv, or vAve, from `dv and `dt you can find aAve, from `dt and vAve you can find `ds.
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02:06:09 ** We start with v0, vf and `dt on the first line of the diagram. We use vO 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. **
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RESPONSE --> ok.
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02:07:46 Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0
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RESPONSE --> First layer is `dt, aAve, and v0 From `dt and aAve we can find `dv From `dv and v0, we can find vf From v0 and vf we get vAve From `dt and vAve we find `ds
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02:08:01 ** Student Solution: Using 'dt and a, find 'dv. Using 'dv and v0, find vf. Using vf and vO, find vave. Using 'dt and Vave, find 'ds. **
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RESPONSE --> ok
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02:09:59 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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RESPONSE --> From these three variables we are able to find `ds as well as aAve. v0 and Vf provide us with both `dv and also vAve for uniform acceleration. From `dv and `dt, we can therefore find aAve. From vAve and `dt, we can then find `ds. So these 3 provide us with the information needed to gather the rest, for uniform acceleration.
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02:10:31 **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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RESPONSE --> I did not go into the equations, but more by reasoning and common sense.
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02:12:31 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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RESPONSE --> From v0, a and `dt, we are able to calculate `ds. a and `dt provide you with `dv, which then provides you with vf = v0 + `dv = v0 + a `dt. Since you have vf and v0, you can find vAve = (vf + v0)/2 or replace it in the previous equation to be (v0 + (v0 + a `dt)) / 2) = v0 + 1/2 a `dt. Since `ds is found from vAVe * `dt, we can also replace this to be (v0`dt + 1/2 a `dt )*`dt, or the third equation `ds = v0 `dt + .5a dt^2.
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02:18:16 ** 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. **
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RESPONSE --> ok.
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02:20:33 Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five?
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RESPONSE --> From these five that we model, we can find the other two. Without these five however, or combination of, we cannot find the vAve or `dv.
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02:24:58 ** 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. **
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RESPONSE --> Ok, the answer provided gives me pretty good insight into what you were asking. You are correct. We can fit numbers into equations all day but unless we can tie some meaning to them, we will never really understand uniform acceleration. How does one interpret a negative acceleration, what does that mean? How does that look on a moving object? I think average velocity and change in velocity provide us with the picture behind the numbers. I just had a problem whose final velocity was 12.9cm/s, and ended up having a negative acceleration, and show its initial velocity was 24cm/s. So what does that mean? As in the example of the car with the parachute to slow itself down, its velocity is still propelling it forward, but it's decelerating with the release of the parachute.
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02:31:52 Accelerating down an incline through a given distance vs. accelerating for a given time Why does a given change in initial velocity result in the same change in final velocity when we accelerated down a constant incline for the same time, but not when we accelerated down the same incline for a constant distance?
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RESPONSE --> I am not sure I am understanding this problem. The first case states there is the same change in velocity when an object goes down an incline for the same time? I am not sure what the same time means? My take on this is that an object moving down an incline, even when it is constant, is accelerating at a constant rate, so acceleration is the same throughout, so its intial and final velocities will be affected by this constant acceleration. I don't think I am understanding constant distance. If an object is going down the same incline, for a constant distance, it does not have change in velocity?
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02:33:19 ** 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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RESPONSE --> Ok. I think I did not understand the constant distance. The distance was the same, but the times were different, hence, the velocities were different. A greater initial velocity, hence less time to cover the same distance so less change from initial to final velocity.
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