#$&*
Phy 231
Your 'cq_1_02.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_02.2_labelMessages **
The problem:
A graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds. The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).
What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
midpoint=(initial time + final time)/2
=(13s+5s)/2
=(18s)/
=9sec
#$&*
What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
velocity at midpoint=(initial velocity + final velocity)/2
=(16cm/s+40cm/sec)/2
=(56cm/sec)/2
=27cm/sec
@&
Careful. 56/2 = 28, not 27.
*@
#$&*
How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Velocity at midpoint is the average velocity in this case (because the line is straight, we are to assume a steady increase in velocity over the course of the interval).
So, average velocity is 27cm/sec.
Time interval is (13sec-5sec)= 8sec.
VAve=`ds/`dt
(vAve)(`dt)=`ds
(27cm/sec)(8sec)=`ds
216cm=`ds
#$&*
By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Change in clock time=final clock time-initial clock time
Change in clock time = 13sec-5sec
Change in clock time = 8 sec.
#$&*
By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Change in velocity = (final velocity-initial velocity)
Change in velocity = 40cm/sec - 16 cm/sec
Change in velocity = 24 cm/sec
#$&*
What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Ave. rate of change in velocity = (change in velocity)/(change in clock time)
Ave. rate of change in velocity = (24cm/sec)/(8sec)
Ave. rate of change in velocity = 3cm/sec^2
Velocity changes, on average, 3cm/sec for every second of this interval.
#$&*
What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Rise=(change in y coordinate)
=(40-16)
=24
@&
Rise and run both have units.
The units of the slope would thus be (cm/s) / s = cm/s^2.
*@
#$&*
What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Run=(change in x coordinate)
=(13-5)
=8
#$&*
What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Slope=rise/run
=24/8
=3
#$&*
What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The slope of this graph tells us how much velocity is changing over the course of this time interval. It starts out at 16cm/sec and increases to 40cm/sec over the course of an 8-sec span. So the graph is increasing as velocity increases.
#$&*
What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Ave. rate of change in velocity = (change in velocity)/(change in clock time)
=(24cm/sec)/(8sec)
=3cm/sec^2
On a graph of velocity vs. time, the slope represents the change in velocity. So we calculate average rate of of change in the object's velocity the same way we calculate the slope, rise/run, or in this case, velocity/time.
#$&*
@&
The slope is 3 cm/s^2.
This isn't the change in velocity, which is 24 cm/s.
The slope represents the rate of change of the velocity with respect to clock time.
*@
*#&!*#&!
&#Good responses. See my notes and let me know if you have questions. &#