course Phy 201 ÙñôÛ¦òžñ˜¶~¡–~»³ÒãÑÔoà…šîÛ¢assignment #004
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RESPONSE --> Given those terms, I would say that the equation would have to be v0 + a = vf and a * 'dt = 'ds because if you add the initial velocity and the acceleration, you would have the final velocity. If you multiply the acceleration by the change in time you would get the displacement. confidence assessment: 1
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15:12:38 **You would use accel. and `dt to find `dv: a * `dt = `dv. Adding `dv to initial vel. vo you get final vel. Then average initial vel. and final vel. to get ave. vel.: (v0 + vf) / 2 = ave. vel. You would then multiply ave. vel. and `dt together to get the distance. For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s: 3 m/s^2 * 5 s = 15 m/s = `dv 15 m/s + 3 m/s = 18 m/s = fin. vel. (18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve 10.5 m/s * 5 s = 52.5 m = dist. In more abbreviated form: a * `dt = `dv v0 + `dv = vf (vf + v0) /2 = vAve vAve * `dt = `ds so `ds = (vf + v0) / 2 * `dt. **
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RESPONSE --> So, the acceleration multiplied by the change in time gives us the change in velocity. Then if you add the change in velocity to the initial velocity you get the final velocity, and then the average velocity is the inital velocity plus the final then divided by 2. For the distance you multiply the ave. velocity and the change in time to get the distance. self critique assessment: 3
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15:17:56 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> The displacement associated with the uniform acceleration from v0 to vf in clock time is the distance an object moves at the specified velocity within a specified time. So 'ds = (vf - v0) * 'dt confidence assessment: 1
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15:20:42 ** Since accel is uniform vAve = (v0 + vf) / 2. Thus displacement is `ds = vAve * `dt = (v0 + vf) / 2 * `dt, which is the first equation of uniformly accelerated motion. **
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RESPONSE --> ok, because the acceration is uniform, the displacement is equal to the average velocity multiplied by the change in time. self critique assessment: 2
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15:24:09 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> if we know the inital velocity, the final velocity and the change in time then we are able to determine the displacement as well as the average velocity confidence assessment: 0
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15:25:38 ** The first level in the diagram would contain `dt, v0 and vf. Then v0 and vf would connect to `dv in the second level. The second level would also contain vAve, connected from vf in the first level to v0 in the first level. The third level would contain an a, connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, connected to vAve in the fourth level and `dt in the first level. **
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RESPONSE --> the diagram would show the components listed in levels and lines connecting them as they would be used together to find another component. self critique assessment: 2
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15:27:04 Query Add any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I am finally beginning to realize that if you take an objective point of view from the parts of the equations/experiments, the easier physics is to understand. confidence assessment: 3
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15:28:05 ** Student Response: I think I really confused what information stood for what in the Force and Pendulum Experiment. However, I enjoy doing the flow diagrams. They make you think in a different way than you are used to. INSTRUCTOR NOTE: These diagrams are valuable for most people. Not all--it depends on learning style--but most. **
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RESPONSE --> I can learn from a flow diagram, but I have a very hard time trying to draw one up on my own. self critique assessment: 3
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course Phy 201 ÙñôÛ¦òžñ˜¶~¡–~»³ÒãÑÔoà…šîÛ¢assignment #004
......!!!!!!!!...................................
RESPONSE --> Given those terms, I would say that the equation would have to be v0 + a = vf and a * 'dt = 'ds because if you add the initial velocity and the acceleration, you would have the final velocity. If you multiply the acceleration by the change in time you would get the displacement. confidence assessment: 1
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15:12:38 **You would use accel. and `dt to find `dv: a * `dt = `dv. Adding `dv to initial vel. vo you get final vel. Then average initial vel. and final vel. to get ave. vel.: (v0 + vf) / 2 = ave. vel. You would then multiply ave. vel. and `dt together to get the distance. For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s: 3 m/s^2 * 5 s = 15 m/s = `dv 15 m/s + 3 m/s = 18 m/s = fin. vel. (18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve 10.5 m/s * 5 s = 52.5 m = dist. In more abbreviated form: a * `dt = `dv v0 + `dv = vf (vf + v0) /2 = vAve vAve * `dt = `ds so `ds = (vf + v0) / 2 * `dt. **
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RESPONSE --> So, the acceleration multiplied by the change in time gives us the change in velocity. Then if you add the change in velocity to the initial velocity you get the final velocity, and then the average velocity is the inital velocity plus the final then divided by 2. For the distance you multiply the ave. velocity and the change in time to get the distance. self critique assessment: 3
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15:17:56 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> The displacement associated with the uniform acceleration from v0 to vf in clock time is the distance an object moves at the specified velocity within a specified time. So 'ds = (vf - v0) * 'dt confidence assessment: 1
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15:20:42 ** Since accel is uniform vAve = (v0 + vf) / 2. Thus displacement is `ds = vAve * `dt = (v0 + vf) / 2 * `dt, which is the first equation of uniformly accelerated motion. **
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RESPONSE --> ok, because the acceration is uniform, the displacement is equal to the average velocity multiplied by the change in time. self critique assessment: 2
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15:24:09 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> if we know the inital velocity, the final velocity and the change in time then we are able to determine the displacement as well as the average velocity confidence assessment: 0
.................................................
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15:25:38 ** The first level in the diagram would contain `dt, v0 and vf. Then v0 and vf would connect to `dv in the second level. The second level would also contain vAve, connected from vf in the first level to v0 in the first level. The third level would contain an a, connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, connected to vAve in the fourth level and `dt in the first level. **
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RESPONSE --> the diagram would show the components listed in levels and lines connecting them as they would be used together to find another component. self critique assessment: 2
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15:27:04 Query Add any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I am finally beginning to realize that if you take an objective point of view from the parts of the equations/experiments, the easier physics is to understand. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:28:05 ** Student Response: I think I really confused what information stood for what in the Force and Pendulum Experiment. However, I enjoy doing the flow diagrams. They make you think in a different way than you are used to. INSTRUCTOR NOTE: These diagrams are valuable for most people. Not all--it depends on learning style--but most. **
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RESPONSE --> I can learn from a flow diagram, but I have a very hard time trying to draw one up on my own. self critique assessment: 3
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