q_avideo9-10_0602010

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17:29:33 Physics video clip 09 displacement for linear v vs. t graph: common sense, formula, area If we know the initial and final velocities over some time interval, and if the rate which velocity changes is constant, then how do we calculate the displacement over a the time interval?

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RESPONSE --> You would multiply the average velocity by the time interval and then get the displacement.

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17:29:42 ** Displacement is the product of average velocity and time interval. Since acceleration is uniform average velocity is average of initial and final velocities. Displacement could therefore be calculated from the final and initial velocities => `ds =[(vf + v0)/2] * `dt. **

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RESPONSE --> right

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17:31:02 Physics video clip 10 continuing 09: calculation of area

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RESPONSE --> is the average height multiplied by the average width so it would be the average velocity multiplied by the time interval.

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17:31:07 How do we use a graph of v vs. t to depict the calculation of the displacement over a time interval?

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RESPONSE --> right

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17:33:25 ** Looking at the graph, we notice a trapezoid created by the slope line across the top, and an imaginary line drawn from the y value down to and perpendicular to the x axis. The area of this trapezoid represents the displacement or 'signed distance' the object travels. The displacement for any time interval can be found finding the average of the two 'altitudes' of the trapezoid, which represent initial and final velocities. Multiplying the average 'altitude' by the width is therefore equivalent to multiplying the approximate average velocity by the time interval, giving us the area of the trapezoid, which represents the approximate displacement. In the case where the graph is linear (which corresponds to uniform acceleration) the average of the two altitudes in fact represents the average velocity, and the result is the displacement, not the approximate displacement. **

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RESPONSE --> You would take the max and min velocity and add them and divide by two to find the average velocity then you would take the time on the x axis. then you would multiply the time interval by the velocity.

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17:33:40 What aspect of the graph gives the displacement during the time interval?

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RESPONSE --> the line

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17:34:15 ** STUDENT ANSWER: It is the average of the two sides that are as high as 'y' in each case, multiplied by the width-units of 'x'. **

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RESPONSE --> right it s the velcoity side by the time side

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Good. Let me know if you have questions.