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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15 minutes
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The problem:
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).
o Ok
• Sketch a straight line segment between these points.
o Ok
• What are the rise, run and slope of this segment?
o Slope = 6/1, or 6 (reduced from 30/5)
The slope has units, and units should be included in your calculation. Be sure you see why the units shoud be cm/s^2.
• What is the area of the graph beneath this segment?
o A= ½ bh
o A = ½ 5 * 30
o A= ½ 150
o A= 75
The region beneath the line segment is a trapezoid, not a triangle. However you could break the trapezoid into a triangle and a rectangle, and if you did the altitude and base would be 5 sec and 30 cm/s. The area beneath the segment would also include a rectangle with dimensions 5 cm by 10 cm/s.
When you multiply sec by cm/sec you get cm, not cm/s/s. The area of your triangle would represent 75 cm, not 75 cm/s/s.
The area of that rectangle would represent 50 cm., so the total area represents 75 cm + 50 cm = 125 cm.
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.
&#Your work looks good. See my notes. Let me know if you have any questions. &#