Phy 231
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A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
answer/question/discussion: I drew the graph, we the clock time (t) on the x-axis and velocity (v) on the y axis
Sketch a straight line segment between these points.
answer/question/discussion: I sketched the line. It is sloping upward and to the right. Because it is from one point to another, it is a straight line, thereby indicating constant acceleration.
What are the rise, run and slope of this segment?
answer/question/discussion: The rise of this segment is = 40 cm/s - 10 cm/s = 30 cm/s. The run is = 9 s - 4 s = 5 s. The slope is therefore 30 cm/s / 5 s = 6 cm/s^2
What is the area of the graph beneath this segment?
answer/question/discussion: The area = h * [(b1 + b2) / 2], which = 5 s * [(10 cm/s + 40 cm/s) / 2] = 5 s * 25 cm/s = 125 cm
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10 minutes
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&#This looks very good. Let me know if you have any questions. &#