cq_1_041

Your 'cq_1_04.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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Good, but 30 cm/s / (5 s) = 6 (cm/s) / s = 6 cm/s * (1/s) = 6 cm/s^2, not 6 cm/s.

`dt is not equal to `ds so it is not the case that vAve ‘ds = vAve * ‘dt.

The average rate of change of velocity with respect to clock time is 6 cm/s^2, but this is not the average velocity. In fact 6 cm/s^2 is not a velocity at all; it doesn't have the units of velocity.

Multiplying 6 cm/s^2 by 5 s would result in the velocity chnage, which is 30 cm/s.

When acceleration is uniform the v vs. t graph is linear; the average value of v is therefore the average of the initial and final velocities. This average is the velocity at the midpoint of the trapezoid.

The region below the graph forms a trapezoid with altitudes 10 cm/s and 40 cm/s, and width 5 cm.

The average of these altitudes is 25 cm/s, and the area of the trapezoid is ave altitude * width = 25 cm/s * 5 s = 125 cm.

The altitudes represent initial and final velocities, so the average of the altitudes represents the midpoint velocity, which since the graph is linear does in fact represent the average velocity. So when you multiply average altitude by width, you are multiplying average velocity by time interval and the result is displacement.

The area of the trapezoid represents the 125 cm displacement corresponding to the motion of the object on this interval.

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See my notes and be sure you understand. Let me know if you have questions.