course Phy 201 ??????o?????m?assignment #007
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15:17:30 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 --> vo, vf & `dt on top row, vo & vf give `dv on second row, and `dt & `dv give a on third row.
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15:18:36 ** 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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15:20:35 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 Line: a, dt & vo Second Line: a & `dt give `dv Third Line: `dv & vo give vf Fourth Line: vf & vo give vAve Fifth Line: `dt & vo give `ds
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15:21:02 ** 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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15:36:39 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 --> Flow diagram: line 1: vo, vf, `dt line 2: vo & vf give vAve; vo & vf also give `dv line 3: vAve & `dt give `ds; `dv & `dt give a To derive vAve from vo & vf, we average them: vAve = (vo +vf) / 2 To then derive `ds from vAve and `dt, we multiply: 'ds = vAve * `dt Therefore, the first most fundamental equation can be written from the flow diagram as follows: 'ds = ( (vo +vf) / 2 ) * `dt To derive `dv from vo & vf, we subtract: vf - vo = `dv To then derive a from `dv & `dt, we divide: a = `dv / `dt Therefore, we can substitute from the flow chart to get `dv in the second most fundamental equation as follows, then rearrange the equation so we're solving for vf: a = (vf - vo) / `dt -- multiply both sides by `dt to get: a * `dt = vf - vo -- then add vo to both sides to get: vo + a * `dt = vf
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15:37:14 **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
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16:09:38 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 --> Flow diagram: line 1: vo, a & `dt line 2: a & `dt give `dv line 3: vo & `dv give vf line 4: vo & vf give vAve line 5: vAve & `dt give `ds since we know from the flow diagram that vAve & `dt give us `ds, we can substitute for vAve as follows: `ds = ( (vo + vf) / 2 ) * `dt since we know from the diagram that vf is derived from vo & `dv and that `dv is derived from a & `dt, we can substitute for vf to include the substitute for `dv as follows: vf = vo + `dv `dv = a * `dt, therefore, vf = vo + a * `dt, therefore, `ds = ( (vo + (vo + a*`dt)) / 2 ) * 'dt This equation can be simplified algebraically as follows: `ds = (vo + (a`dt / 2)) * `dt, or `ds = (vo + .5a`dt) * dt, and factoring gives: `ds = vo`dt + .5a`dt^2 which is the third fundamental equation of motion.
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16:10:21 ** 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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16:17:42 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 think in terms of 7 fundamental quantities while modeling uniformly accelerated motion only in 5 because with uniform acceleration as long as 2 of the quantities: vo, vf, vAve & `dv are known, the other 2 can easily be derived and that derivative equation used in ultimate calculations instead.
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16:18:41 07-13-2006 16:18:41 ** 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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NOTES -------> ok
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16:46:45 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 --> At a constant acceleration, final velocity is directly related to both initial velocity and time interval so if initial velocity is increased but time remains the same, final velocity is increased only by initial velocity. However, displacement is directly related to acceleration, so if initial velocity were to increase but displacement and acceleration were to remain the same, final velocity would not be the increased by the initial velocity amount.
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16:48:46 07-13-2006 16:48:46 ** 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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NOTES -------> ok
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