course phy 201 Z?zz????????z?assignment #007007. `query 7
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11:39:17 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 --> a = (vf-v0)/'dt confidence assessment: 3
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11:39:27 ** 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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11:40:18 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 --> a = (vf-v0)/'dt a('dt) + v0 = vf confidence assessment: 3
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11:40:28 ** 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 self critique assessment: 3
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11:42:04 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 --> vf = v0 + a('dt) 'ds = (vf + v0)/2 * ('dt) confidence assessment: 3
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11:42:25 **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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11:44:24 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 --> 'ds = v0('dt) + .5a('dt^2) self critique assessment: 3
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11:44:28 ** 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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11:46:12 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 --> we need the other two quantities sometimes to find the 5 main quantities used in uniformly accelerated motion. confidence assessment: 3
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11:46:32 ** 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: 3
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11:54:57 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 --> it will cover the same distance faster when the initial velocity increased, but if measured in time, the final will be the same for that given amount of time since the acceleration is the constant in both situations. confidence assessment: 3
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11:55:09 ** 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 self critique assessment: 3
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