vidclips_09-10_wk3quiz2v11

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01:36:13 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 --> We can calculate the displacement over the time interval by looking at the average velocity. If we have an initila velocity of 0 and a final of 8m/s, and the change in time was 4 s. We can figure out that the average velocity is 4m/s (8+0/2), since the rate at which the velocity changes is constant. We can then multiply this average by 4 seconds to get the displacment of 16 m.

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01:36:23 ** 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 --> Ok.

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01:36:32 Physics video clip 10 continuing 09: calculation of area

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RESPONSE --> Ok.

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01:38:07 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 --> We can also calculate the displacement by finding the area of the trapezoid formed from the velocity versus time graph. The area of a trapezoid is its altitude times its width. Since we find the average altitude of the trapezoid, the midpoint between initial and final velocity,we can then multiply that by the width, or the change in time. This average altitude also corresponds to the average velocity.

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01:38:34 ** 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 --> Ok.

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01:38:56 What aspect of the graph gives the displacement during the time interval?

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RESPONSE --> The area under the line, the area of the trapezoid.

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01:39:29 ** 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 --> The area of the trapezoid, avg. altitude * width. Average velocity * change in time.

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Randomized Problems: Week 3, quiz 2, version 11 Determine the acceleration of an object which, after accelerating through a distance of 85 cm if 4.6 sec, is moving at 12.95652 cm/s. We have the following information: `ds = 85cm `dt = 4.6s vf=12.95652 cm/s Using `ds=(v0+vf)/2*`dt, we solve for v0=2`ds/`dt-vf v0=2(85cm)/4.6s-12.95652cm/s 170cm/4.6s-12.95652cm/s = 36.95652cm/s-12.95652cm/s v0=24cm/s Since a = `dv/`dt, we now have (12.95652cm/s-24cm/s/4.6s a = -2.4cm/s^2 Since we went from a higher initial velocity to a lower, it makes sense that its accerlation is negative. "

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