course Phy 201 Rx????s?a???N??assignment #004004. `Query 4
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19:35:41 Intro Prob 6 given init vel, accel, `dt find final vel, dist If initial velocity is v0, acceleration is a and time interval is `dt, then in symbols what are the final velocity vf and the displacement `ds?
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RESPONSE --> I don't understand what this question is asking. In symbols what are the final velocity it would be vf and the displacement symbol is 'ds i mean the d stands for delta but dont know what this question is asking for. confidence assessment: 0
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19:45:56 **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 --> I completly misunderstood what this problem was asking for now i understand that a * 'dt = 'dv and (v0 + vf)/2 = vAve. and that you multiply vAve with 'dt to get distance. self critique assessment: 3
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19:58:13 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> 'ds = (vf - v0)t confidence assessment: 1
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20:25:55 ** 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 --> I messed up on this equation it should have been 'ds = vAve * 'dt = v0 + vf) /2 * 'dt self critique assessment: 3
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20:56:33 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> on this graph we will travel from an initial velocity to the final velocity to in a certain amount of time.the graph would flow from the right up. the rise between two graph points repersents the change in velocity and the run between the two graph points represents change in position. the slope between two graph points represents average acceleration for the corresponding time interval. confidence assessment: 3
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21:14:26 ** 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 --> Ok i didnt understand this problem i didnt see anything in the book about the flow diagram but it makes some sense not alot but some. level 1 'dt,v0 vf. then v0 and vf connect to 'dv in 2nd levelalso vAve connected from vf n 1st level to v0 in 1st level. 3rd level would contain a, connected to 'dv in 2nd level and 'dt in the first level the 3rd level contains 'ds connected to vAve in the 4th level and 'dt in the 1st level. self critique assessment: 3
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21:17:22 Query Add any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> i understand a lot more about the wording of these types of problems. I didnt completly understand any of these problems because of the way they were worded.. confidence assessment: 3
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21:33:01 ** 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 dont like the diagrams very much dont understand them. self critique assessment: 3
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