#$&*
Phy 201
Your 'cq_1_04.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_04.1_labelMessages **
The problem:
A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
Sketch a straight line segment between these points.
What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise or +30cm/s is the change in velocity. The run is 5 seconds or the duration of time from one clock time to another (time passed in this interval). The slope is
the change in velocity divided by the change in time, or acceleration.
@&
You've described (correctly) meaning of the slope (very good), but you also need to calculate it.
*@
#$&*
What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The area here is 1/2bh or 1/2 * 5s * 30cm/s. Cross out the secons and you get 75cm or the distance traveled. Or more generally speaking, since Velocity = 'ds/'dt, 'ds
@&
You could easily form a triangle with this area, but it wouldn't be the total area under the graph. The area represented by this triangle would indeed be 75 cm.
However the region under the graph goes all the way to the horizontal axis, and is a trapezoid.
Can you calculate the area of this trapezoid?
*@
= velocity * 'dt, making the area displacement or distance traveled.
#$&*
@&
Good work, but you neglected part of the area under the graph, and you didn't actually calculate the slope.
Hopefully this will be easy and quick for you to correct.
&#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. &#
*@