#$&*
Phy 121
Your 'cq_1_04.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_04.1_labelMessages **
The problem:
A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
•Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
•Sketch a straight line segment between these points.
•What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise 30 cm/s
The run is 5 sec
The slope is 6 cm/s^2
#$&*
•What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
If I recall correctly, the correct way to find the area of such a graph would be to find the average of the two points on the vertical axis and consider that the new height of a rectangle and solve accordingly.
Therefore I would say that the area of the graph beneath this segment is [(40 + 10) / 2] * 5 = 25 * 5 = 125 cm^2
@&
Your calculation should include the units with every quantity that has units. A bare number, lacking units, does not correctly represent any quantity which does have units.
*@
@&
Except for the units, this is correct.
Think of the area of the trapezoid. You multiply its 5 second width by the average of its two 'graph altitudes', which is 25 cm/sec.
However you don't get 125 cm^2. The 125 is right, but the cm^2 is not.
*@
@&
Except for the units, you've done everything very well.
However we do need to get the issue of the units out of the way.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@