course Phy 231 gՈݕ|{yeԁassignment #007
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19:00:12 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 --> You would have a line connecting v0 and vf and pointing to dv and vAve. You then connect dv and dv to find ave acc.You could also have a line from vAve and dt to find ds. confidence assessment: 3
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19:17:31 ** 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 self critique assessment: 3
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19:22:09 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 --> We would have a line connecting dt, a, and v0 to find ds. Then a line from dt, a, and v0 to find vf. Then vf to v0 to find vAve and dv. confidence assessment: 3
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19:23:17 ** 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 --> I used the more complicated eqns to find the missing variable, I hope this is okay although there diagram is less cluttered then mine. self critique assessment: 2
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19:27:32 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 --> We would have a line from v0 and vf to find dv. The a line from dv and dt to find ave acc. This eqn would be ave acc = dv/dt. Then we would have a line from v0 and vf to find vAve. Then we would have a line from vAve and dt to find ds. This eqn would be ds = vAve*dt. These are the most used eqns used in this course so far. confidence assessment: 3
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19:27:52 **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 --> OK self critique assessment: 3
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19:42:21 qaExplain 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 --> Knowing v0 and a gives you the eqn: vf = adt + v0 because of ave acc = (vf-v0)/dt. ds = v0 `dt + .5 a `dt^2. Knowing a and dt, you get the eqn a = dv/dt, then dv = a*dt. You then have the eqn: ds = v0dt + .5adt. For the previous question concerning the first two eqns of motion, I misunderstood the question. The two eqns are ds = (v0+vf)/2 * dt which (v0+vf)/2 is the average velocity on that time interval. And the second eqn is: vf = v0 + a *dt. This is found by realizing a*dt is just the dv for that time interval. Then you find the final velocity by adding the first velocity to dv. Sorry I missed this question before. self critique assessment: 3
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19:42:38 ** 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 self critique assessment: 3
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19:51:17 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 --> If we know ds, dt, v0, vf, and a, we can easily find vAve and dv but vAve and dv are not necessary to analyze uniformly accelerated motion. confidence assessment: 3
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19:52:01 ** 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 self critique assessment: 2
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20:24:39 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 --> My guess is that a constant distance has little to do with uniform acceleration. For example ave acc cannot be found directly with use of ds. I have no idea, this is just a guess. confidence assessment: 0
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20:27:32 ** 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 --> Okay so the change in velocity is dependent on the dt. So as dt increases, dv increases. If you start with a higher velocity there will be less time on the ramp so the dt would be less so dv would be less. This helps. self critique assessment: 2
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