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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, 12 cm/s) to the point (13 sec, 36 cm/s).
What is the clock time at the midpoint of this interval?
answer/question/discussion:
13 - 5 = 8 sec interval
½ of 8 = 4 sec
5+4 = 9 secs is the midpoint clock time
This midpoint clock time can be calculated in either of two ways.
You can determine that the interval lasts for 13 s - 9 s = 4 s. Half of this is 2 s so the midpoint of the interval occurs 2 s after it begins (and 2 s before it ends) at clock time t = 9 s + 2 s = 11 s.
Or you can use the fact that the number halfway between two numbers is the mean of the two numbers. so the midpoint lies at clock time (9 s + 13 s) / 2 = 22 s / 2 = 11 s.
Neither method is better than the other, but the latter will be the standard method in this course. You may, however, choose to use either.
What is the velocity at the midpoint of this interval?
answer/question/discussion:
36 cm/s - 12 cm/s = 24 cm/s
½ of a difference of 24 cm/s = 12
12 + 12 = 24 cm/s is the midpoint velocity
How far do you think the object travels during this interval?
answer/question/discussion:
Acceleration = 3 cm/sec/sec ( see the last question of this problem set - I had to come back to this one).
There is an 8 sec interval.
The object starts out going 12 cm/sec and increases by 3 cm/sec each sec
Seconds distance traveled
1st 15 cm
2nd 18 cm
3 21 cm
4 24 cm
5 27 cm
6 30 cm
7 33 cm
8th 36 cm
Add up all of distances traveled in each second (calculator used) = 204 cm
This answer may not be right, but I think it is pretty creative!
Very good, in fact. This estimate is a little high. It applied the velocity at the end of each interval to the entire interval; since the velocity at the end of each interval is the highest velocity attained during that interval, the object won't go quite as far as you estimate.
If you use the average velocity on the interval you will get the correct result.
Since the v vs. t graph is a straight line the average velocity on the interval occurs at the midpoint of the interval, so the average velocity is
ave vel = (12 cm/s + 36 cm/s) / 2 = 24 cm/s.
Since average velocity is the average rate at which position changes with respect to clock time, we have
ave velocity = (change in position) / (change in clock time) so that
change in position = average velocity * change in clock time = 24 cm/s * 8 s = 192 cm.
This is close to, but a little lower than your estimate.
By how much does the clock time change during this interval?
answer/question/discussion:
13 sec - 5 sec = 8 secs
The clock time changes by 8 seconds in the interval.
By how much does velocity change during this interval?
answer/question/discussion:
36 cm/s - 12 cm/s = 24 cm/s
The velocity changes by 24 cm/s
What is the rise of the graph between these points?
answer/question/discussion:
Rise is the change in y (or velocity) so the rise is 24
What is the run of the graph between these points?
answer/question/discussion:
Run is the change in x (or clock time) so the run is 8
What is the slope of the graph between these points?
answer/question/discussion:
Slope = rise / run
Slope = 24 / 8
Slope = 3
What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion:
The slope is positive so it tells us that the motion of the object is increasing at a constant rate.
What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion:
Another way to say rate of change of the object’s velocity is to ask what is it’s acceleration
A = dv/dt
a=24(cm/s)/8s
a=3 cm/s/s
The rate of change is 3 cm/sec/sec
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20 mins
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&#Your work looks good. See my notes. Let me know if you have any questions. &#