course Phy 201 z˾yƿ`|assignment #007
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23:20:55 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 --> If you know v0 and vf, you can get both the average velocity and the change in velocity. From the change in velocity divided by the `dt, you can arrive at the average acceleration. confidence assessment: 2
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23:24:05 ** 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 --> Put each known on the first line of the diagram. Use vo and vf to find the average velocity by drawing lines from v0 and vf to average velocity. Also use v0 and vf to get the change in velocity, `dv. From `dv and `dt, you can find average acceleration. Draw lines from `dv to `dt and to acceleration. self critique assessment: 2
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23:28:12 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 --> From v0, a, and `dt, we can find the final velocity. You would draw a line between a and `dt, and multiply them together. Then add a line to v0 to get vf. confidence assessment: 1
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23:30:51 ** 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 --> To find `dv: Find vf from v0 and a and `dt. Then subtract v0 from vf to get `dv. Also vf and v0 can be used to get vave. vAve can be used with `dt to get aAve. From the change in time and the average velocity, we can get the change in position. self critique assessment: 2
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23:33:05 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 --> In order to calculate both the velocity and the acceleration, v0, vf and `dt are crucial. From (vf+v0)/2 you get the average velocity. From the change in velocity, vf-vo, you can calculate average acceleration when the change in velocity is divided by the time interval. confidence assessment: 2
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23:35:41 **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 --> aAve=(vf-v0)/`dt vf=v0 + a*`dt vAve=`ds/`dt vAve*`dt=`ds self critique assessment: 2
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23:41:15 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 --> aAve=`dv/`dt aAve*`dt=`dv `dv=vf-v0 aAve*`dt=vf-v0 aAve*`dt+v0=vf vf=v0 + a*`dt From vf and v0 we can get the average velocity and the change in velocity. From the acceleration, v0 and `dt, we can calculate vf. vf = v0 + a * `dt. self critique assessment: 2
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23:51:28 ** 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 --> I numbered the laws differently and solved the wrong one. I used 7.3. `dv = a * `dt gives us vf = v0 + a * `dt. Should have solved for `ds = v0 `dt + .5 a `dt^2 I wasn't sure whether to used them as written in the book or on the memorize this. From aAve and `dt you can get `dv. `dv=vf-v0 v0 and vf give vAve. vAve and `dt give you `ds. `dv=a*`dt vf-v0=a*`dt vf=v0+`dv=v0 +a`dt vAve=(vf +v0)/2=(v0 +(v0 + a`dt))/2=v0 + .5 a`dt `ds=vAve*`dt=(v0`dt +.5a`dt)*`dt=v0 `dt +.5 a `dt^2 self critique assessment: 2
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23:52:50 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 --> Because `dv and and average velocity can be calculated if you know the time interval and the initial and final velocities. confidence assessment: 2
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23:54:11 ** 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 --> Five fundamental quantities- v0, vf, a `dt, and `ds vf and v0 help us to find change in velocity and along with `dt, we can get average velocity also. self critique assessment: 2
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23:59:49 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 --> When calculating average acceleration, you calculated the change in velocity, divided by the time interval. The change in velocity is the same no matter what you started at. To go from a distance, you use the formula vAve=`ds/`dt vAve*`dt=`ds From a distance, we calculate the average velocity, not the change in velocity. Below you would have different values for v0. `ds = v0 `dt + .5 a `dt^2 confidence assessment: 1
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00:01:25 ** 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 --> Greater initial velocity results in a greater average velocity. From a greater average velocity, there is less time on the incline, and less time for its velocity to change or accelerate. self critique assessment: 2
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