course Phy 201 ?????????g?? X?assignment #005N????????J???Physics I Vid Clips
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22:14:05 `questionNumber 50000 Physics video clip 11 `ds on single time interval: ave. ht * width = vAve * `dt = `ds How do we calculate the displacement during a single specific time interval, given a linear v vs. t graph?
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RESPONSE --> (v1 + v2)/2 * (t1/ t2)
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22:14:09 `questionNumber 50000 ** STUDENT SOLUTION: ((v1+v2)/2)*(t2-t1) or area of the trapezoid created by the given time interval **
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RESPONSE --> Correct
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22:23:45 `questionNumber 50000 If we know the initial and final velocities over some time interval, and if the rate at which velocity changes is constant, then how do we calculate the displacement of the object during that time interval?
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RESPONSE --> ((vf + v0)/2) * (t2 -t1) or we could also use ((vf + v0)/2 * 'dt
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22:23:49 `questionNumber 50000 ** ((vf + v0)/2) * `dt or ((vf + v0)/2) * (t2 - t1) **
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RESPONSE --> Correct
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22:34:14 `questionNumber 50000 In how many different ways can we represent the calculation of the displacement over a constant-acceleration time interval?
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RESPONSE --> We could graph this as a velocity vs. time graph or as ((vf+ v0)/2) * (t2 - t1)
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22:34:19 `questionNumber 50000 ** At least two: graphically as the area under the v vs. t graph and algebraically as ((vf + v0)/2) * (t2 - t1) **
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RESPONSE --> Correct
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22:38:53 `questionNumber 50000 Physics video clip 12 continuing ph11 How does the graphical calculation connect with our common sense about velocity, displacement and time?
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RESPONSE --> For example if we graphed velocity and time we could find discplacent. As long as we have two of the values on the planes then we can solve the third value.
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22:38:56 `questionNumber 50000 ** STUDENT SOLUTION: First of all, the graph illustrates the fact that the three are related. We can graph for two of the values on the x-y plane, in order to solve for the third one. We usually graph for velocity vs. time, to find the displacement. } STUDENT QUESTION: Is it possible to find the time by setting up a graph of distance vs. velocity or velocity vs. distance? ANSWER (UNIVERSITY PHYSICS LEVEL): Excellent question. Basically if you know the velocity at every position you have completely defined the motion; however depending on the situation it might be that the best you can do is an approximation to find x vs. t (and therefore v vs. t, again in approximation). If you know the velocity at a certain position then you can predict how long it will take to get to a nearby position; knowing the velocity at that position you can predict how long it will take to get to a new nearby position. These predictions would be approximations, the accuracy of which would depend on the magnitude of the second derivative of the velocity vs. position function. They would effectively give you an estimate of position vs. clock time, from which you could also estimate velocity vs. clock time. The situation would be that you would be able to calculate for a given position the value of the rate of position change, which is the derivative of the position. If the position function is x(t) then a graph of v vs. x would be a graph of dx/dt vs. x, so that for every x you could find dx/dt. So what you would effectively have is the differential equation dx / dt = v(x). The solution of the differential equation would be your position function x(t). Depending of v(x) there might or might not be a closed-form solution of the differential equation. If not the type of numerical approximation in the paragraph before last (which is effectively Euler's Method; there are other more sophisticated and more accurate methods of approximaton) would be your only resort, and this could be done graphically. If there is a closed-form solution to the diff eq then you might or might not be able to solve the equation x = x(t) for t in terms of x, but in any case t would be given implicitly as a function of x. **
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RESPONSE --> Ok
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