course Phy 121
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11:38:17
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RESPONSE -->
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11:38:59 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 --> The displacement is given by the area of the trapezoid located under the graph of the given data.
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11:39:03 ** STUDENT SOLUTION: ((v1+v2)/2)*(t2-t1) or area of the trapezoid created by the given time interval **
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RESPONSE --> ok
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11:44:34 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 --> 'ds = avg height (avg. velocity) * width ('dt)
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11:44:45 ** ((vf + v0)/2) * `dt or ((vf + v0)/2) * (t2 - t1) **
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RESPONSE --> ok
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11:45:42 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 can manipulate one of the four basic equations and plug in values or we can take the area of the trapezoid of the graph
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11:45:50 ** 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 --> ok
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11:48:16 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 --> Both of these connect by one of the equations will work out to give us specifically displacement, time, and velocity...given at least two of the three variables
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11:48:24 ** 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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