#$&*
Phy 241
Your 'cq_1_02.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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The problem:
A graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds. The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).
What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&* The clock time at the midpoint = (13 sec + 5 sec)/2 = 9 sec.
What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&*The velocity at the midpoint is = (40cm/sec + 16cm/sec)/2 = 28cm/sec.
How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&* We know that velocity = dx/dt, dx = velocity * dt = 28 cm/sec*4 sec =7cm.
@& 28 cm/s * 4 s is correct, but the product is 28 * 4 (cm / s) * s = 108 cm, not 7 cm. You appear to have divided where you intended to multiply.*@
By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&* The clock time has changed from the difference of 9 sec - 5 sec = 4 sec.
By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&* The velocity has changed by 28cm/sec - 16cm/sec = 12 cm/sec
@& Change in velocity is final velocity - initial velocity, not average velocity - initial velocity.*@
What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&* The average rate of change of velocity with respect to clock time is = (28cm/sec - 16cm/sec)/(9sec -5sec)= 3cm/sec^2
What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&*The rise is =40cm/sec - 16cm/sec =24cm/sec
What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&*The run = 13sec - 5sec = 8sec
What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&*The slope is = rise/run = (24cm/sec)/8sec = 3cm/sec^2
What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&*The slope is showing how the acceleration in increasing.
@& The rise is the change in velocity, the run is the change in clock time, and the rise / run is the acceleraiton, which is constant.
Your result is 3 cm/s^2, and this is the correct acceleration.
Compare this with your preceding answers and reasoning to see what you did correctly and what you did incorrectly.
*@
What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&*The average rate of change is: (40cm/sec - 16cm/sec)/(13sec-5sec) =3cm/sec^2
*#&!
@& This is the correct average rate of change of velocity with respect to clock time, which is therefore the correct acceleration.
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*@