#$&*
PHY 201
Your 'cq_1_04.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_04.1_labelMessages **
The problem:
A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
• Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I sketched the graph. The x-axis represents the time domain and the y-axis represents the velocity range.
#$&*
• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
A straight line was sketched between the points (4 s, 10 cm/s) and (9 s, 40 cm/s)
#$&*
• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Rise = (y2 - y1) = (v2 - v1) = (40 cm/s - 10 cm/s) = 30 cm/s
Run = (x2 - x1) = (t2 - t1) = 9 s - 4 s = 5 s
Slope = 30 cm/s / 5 s = 6 cm/s/s
#$&*
• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Area is a trapezoid.
A = (h/2)(b1 + b2)
A = (9s / 2)(10 cm/s + 40 cm/s)
A = 4.5 s * 50 cm/s = 225 cm
#$&*
@&
This is fine, but
9 s * (10 cm/s + 40 cm/s) / 2 = 9 s * 25 cm/s = 225 cm
arranges the calculation as time interval * average velocity, which makes more physical sense.
*@
@&
The result is correct either way.
*@
*#&!*#&!
&#Good responses. See my notes and let me know if you have questions. &#