.
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
062. General rate definition.
Question: `q001.
A 'graph rectangle' is a rectangle, one of whose sides is part of the horizontal axis.
On a graph of speed in miles / hour vs. clock time in hours, we find a graph rectangle with base 3 and altitude 40.
Your solution:
Confidence Assessment:
Given Solution:
A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of speed in miles / hour vs. clock time in hours represents speed as the 'vertical' quantity and clock time as the 'horizontal' coordinate.
On such a graph, the altitude of the rectangle therefore represents the speed.
The base rests on the horizontal axis, so it runs from one clock time to another.
We will call the first clock time the 'initial clock time' and the second the 'final clock time'.
The base therefore represents the change from the initial clock time to the final. We call this quantity the 'change in clock time'.
On the present graph, the speed is said to be in miles/hour and the clock time is said to be in hours.
Thus the altitude of the graph represents the speed in miles / hour, and the base represents the change in clock time in hours.
In the process of solving this problem you should have sketched a graph, with the vertical axis labeled 'speed in miles / hour' and the horizontal axis labeled 'clock time in hours'.
Your graph would have included a rectangle resting on the x axis, with its 'graph altitude' labeled 40 (representing the speed of 40 miles/hour) and its base labeled 3 (representing 3 hours).
When you multiply the base of the rectangle by the altitude to get the area, your result will represent the product of 40 mile / hour and 3 hours.
area = 40 * 3 = 120, representing 40 miles / hour * 3 hours = 120 miles / hour * hours.
The units of the product are miles / hour * hours.
miles / hour * hours = miles / hour * hours / 1 = miles * hours / (hour * 1) = miles * hour / hour = mile * (hours / hour) = miles
According to the order of operations this could be written in standard mathematical notation as
.
Thus the area represents the quantity
area = 120 miles / hour * hours = 120 miles.
This makes perfect sense. Recall that 40 miles/hour is the speed and 3 hours the time interval. Moving at 40 miles/hour for 3 hours, an object would move 120 miles.
Any 'graph rectangle' which represents speed vs. time interval will have an area which represents the product of the speed and the time interval. This product is the distance an object will travel at that speed, during that time interval.
Self-critique (if necessary):
Self-critique rating:
Question: `q002.
On a graph of income stream in dollars per month vs. clock time in months, we find a graph rectangle with base 36 and altitude 1000.
Your solution:
Confidence Assessment:
Given Solution:
A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of income stream in dollars / month vs. clock time in months represents income stream as the 'vertical' quantity and clock time as the 'horizontal' coordinate.
On such a graph, the altitude of the rectangle therefore represents the income stream.
The base rests on the horizontal axis, so it runs from one clock time to another.
We will call the first clock time the 'initial clock time' and the second the 'final clock time'.
The base therefore represents the change from the initial clock time to the final. We call this quantity the 'change in clock time'.
On the present graph, the income stream is said to be in dollars / month and the clock time is said to be in months.
Thus the altitude of the graph represents the income stream in dollars / month, and the base represents the change in clock time in months.
In the process of solving this problem you should have sketched a graph, with the vertical axis labeled 'income stream in dollars / month' and the horizontal axis labeled 'clock time in months'.
Your graph would have included a rectangle resting on the x axis, with its 'graph altitude' labeled 1000 (representing the income stream of 1000 dollars/month) and its base labeled 36 (representing 36 months).
When you multiply the base by the altitude to get the area, your result will represent the product of 1000 dollars / month and 36 months.
area = 1000 * 36 = 36 000, representing 1000 dollars / month * 36 months = 36 000 dollars / month * months.
The units of the product are dollars / month * months.
dollars / month * months = dollars / month * months / 1 = dollars * months / (month * 1) = dollars * month / month = mile * (months / month) = dollars
Thus the area represents the quantity
area = 36 000 dollars / month * months = 36 000 dollars.
This makes perfect sense. Recall that 1000 dollars/month is the income stream and 36 months the time interval. Earning money at a rate of 1000 dollars/month for 36 months, we would earn 36 000 dollars.
Any 'graph rectangle' which represents income stream vs. time interval will have an area which represents the product of the income stream and the time interval.
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Question: `q003. On a graph of force in pounds vs. position in feet, we find a graph rectangle with base 200 and altitude 30.
Your solution:
Confidence Assessment:
Given Solution:
A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of force in lb vs. position in ft represents force as the 'vertical' quantity and position as the 'horizontal' coordinate.
On such a graph, the altitude of the rectangle therefore represents the force.
The base rests on the horizontal axis, so it runs from one position to another.
We will call the first position the 'initial position' and the second the 'final position'.
The base therefore represents the change from the initial position to the final. We call this quantity the 'change in position'.
On the present graph, the force is said to be in lb and the position is said to be in ft.
Thus the altitude of the graph represents the force in lb, and the base represents the change in position in ft.
In the process of solving this problem you should have sketched a graph, with the vertical axis labeled 'force in lb' and the horizontal axis labeled 'position in ft'.
Your graph would have included a rectangle resting on the x axis, with its 'graph altitude' labeled 30 (representing the force of 30 lb) and its base labeled 200 (representing 200 ft).
When you multiply the base of the rectangle by the altitude to get the area, your result will represent the product of 30 lb and 200 ft.
area = 30 * 200 = 6000, representing 30 lb * 200 ft = 6000 lb * ft.
The units of the product are lb * ft. These units do not simplify.
Thus the area represents the quantity
graph area = 6000 lb * ft.
This quantity might not make sense in terms of what you know right now. However it does make physical sense, and actually does so with a couple of different interpretations. You aren't expected to know these interpretations at this point, but it would be your advantage to 'get the wheels turning' by thinking a little bit about them now:
In one interpretation you can think of pushing a box down a long hall.
You have to exert certain number of pounds to keep it moving, and to get all the way down the hall you have a certain distance.
If, for example, you have to an object twice as far, then you have to do twice as much work. And if you have to exert more force to keep it moving, you have to do more work.
If you multiply the number of pounds you have to exert by the number of feet you have to move the object, you get the work done in lb * ft. Since multiplication of two quantities can take place in either order, this unit can also be written as ft * lb.
In the present problem if you exert 30 lb of force to move the box 200 ft down the hall, you do 6000 ft * lb of work.
In another interpretation you can think of two people trying to raise a heavy rock with levers.
The weight of the rock they can each lift is determined by how much torque each can exert with their respective levers. The longer the lever, the more torque can be exerted for a given force. Also, the more force exerted, the greater the torque.
To get the torque each person can produce you multiply the length of his or her lever by the force he or she can exert. If the force is measured in lb and the length of the lever in ft then the product is in lb * ft.
(Note that for simplicity this discussion isn't completely precise regarding the meaning of the length of the lever. We will clarify that issue later.).
In the present problem if you exert 30 lb of force on a 200 ft lever, then you are exerting a torque of 6000 lb * ft. Of course a lever 200 ft long might not be practical, since it would probably require more that 6000 lb * ft of torque just to lift one end; but that's another issue we'll deal with later.
A graph of force vs. position will usually have the 'work' interpretation. In this case any 'graph rectangle' will have an area which represents the product of the force and the displacement, which represents work. The topic of work and energy is developed elsewhere in the course.
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Question: `q004. On a graph of density in grams / centimeter vs. position in centimeters, we find a graph rectangle with base 16 and altitude 50.
Your solution:
Confidence Assessment:
Given Solution:
A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of density in g / cm vs. position in cm represents density as the 'vertical' quantity and position as the 'horizontal' coordinate.
On such a graph, the altitude of the rectangle therefore represents the density.
The base rests on the horizontal axis, so it runs from one position to another.
We will call the first position the 'initial position' and the second the 'final position'.
The base therefore represents the change from the initial position to the final. We call this quantity the 'change in position'.
On the present graph, the density is said to be in g / cm and the position is said to be in cm.
Thus the altitude of the graph represents the density in g / cm, and the base represents the change in position in cm.
In the process of solving this problem you should have sketched a graph, with the vertical axis labeled 'density in g / cm' and the horizontal axis labeled 'position in cm'.
Your graph would have included a rectangle resting on the x axis, with its 'graph altitude' labeled 50 (representing the density of 50 g / cm) and its base labeled 16 (representing 16 cm).
When you multiply the base of the rectangle by the altitude to get the area, your result will represent the product of 50 g / cm and 16 cm.
area = 50 * 16 = 800, representing 50 g / cm * 800 cm = 800 g / cm * cm.
The units of the product are g / cm * cm. These units simplify to g * cm / cm = g * (cm/cm) = g * 1 = g.
Thus the area represents the quantity
graph area = 800 g.
This quantity has a fairly straightforward meaning.
The quantity grams / centimeter tells us how many grams of mass are in a centimeter of length (this might apply, for example, to a board; for example a 2 x 4 made of a certain wood might contain about 20 grams of mass for each centimeter of its length).
If we multiply the number of grams in a centimeter of length by the number of centimeters of length, we get the total number of grams in that length.
The area of the rectangle therefore represents the mass of the given length of a material.
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Question: `q005. A 'graph trapezoid' is defined by two points on a graph, as follows:
The 'graph slope' between two points is the slope of the 'slope segment' of the graph trapezoid defined by the two points.
On a graph of speed in miles / hour vs. clock time in hours, we find graph points (2, 50) and (7, 60)
A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of speed in miles / hour vs. clock time in hours represents speed as the 'vertical' quantity and clock time as the 'horizontal' coordinate.
On such a graph, an altitude of the rectangle therefore represents the speed.
The base rests on the horizontal axis, so it runs from one clock time to another.
We will call the first clock time the 'initial clock time' and the second the 'final clock time'.
The base therefore represents the change from the initial clock time to the final. We call this quantity the 'change in clock time'.
On the present graph, the speed is said to be in miles/hour and the clock time is said to be in hours.
Your solution:
Confidence Assessment:
Given Solution:
A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of speed in miles / hour vs. clock time in hours represents speed as the 'vertical' quantity and clock time as the 'horizontal' coordinate.
On such a graph, an altitude therefore represents a speed.
The base rests on the horizontal axis, so it runs from one clock time to another.
We will call the first clock time the 'initial clock time' and the second the 'final clock time'.
The base therefore represents the change from the initial clock time to the final. We call this quantity the 'change in clock time'.
On the present graph, the speed is said to be in miles/hour and the clock time is said to be in hours.
Thus an altitude of the graph represents a velocity in miles / hour, and the base represents the change in clock time in hours.
In the process of solving this problem you should have sketched a graph, with the vertical axis labeled 'speed in miles / hour' and the horizontal axis labeled 'clock time in hours'.
Your graph would have included a trapezoid resting on the x axis, with its left and right left and right 'graph altitudes' labeled 50 and 60 (representing speeds of 50 miles/hour and 60 miles/hour) and its base running from 2 to 7 (representing 2 hours and 7 hours), so that its base is 7 - 2 = 5, representing 7 hr - 2 hr = 5 hr.
The 'slope segment' of this trapezoid rises from its left altitude 50 to its right altitude 60, representing the velocities 50 miles/hr and 60 miles/hr.
The rise of this trapezoid is the rise of its 'slope segment', which is the change in its 'graph altitude'.
From left to right this change is 60 - 50 = 10 and represents 60 miles/hr - 50 miles/hr = 10 miles/hr.
Since the vertical coordinate represents the velocity, this is the change in velocity on the interval represented by this trapezoid.
The run of this trapezoid is the run of its 'slope segment', which is the change in the horizontal coordinate on the corresponding interval.
From left to right this change is 7 - 2 = 5 and represents 7 hr - 2 hr = 5 hr.
Since the horizontal coordinate represents the clock time, this is the change in clock time on the interval represented by this trapezoid.
The slope of the trapezoid is 'rise / run', the rise of the slope segment divided by its run.
The rise is 10 and the run is 5 so the slope is 2.
This slope represents rise / run = (10 miles/hr) / (2 hr) = 5 miles / hr^2.
Details of the calculation of units:
(miles / hr) / (hr) = (miles / hr) * (1 / hr) = (miles * 1) / (hr * hr) = miles / hr^2
The order of operations can be used to represent this series of calculations as follows:
To interpret the meaning of this slope:
The rise represents the change in the vertical quantity, which in this case is the velocity in mile / hr.
The run represents the change in the horizontal quantity, which in this case is the clock time in hr.
Thus the slope = rise / run represents (change in vertical quantity) / (change in horizontal quantity), which by the definition of average rate is the average rate of change of the vertical quantity with respect to the horizontal quantity.
So the slope represents the average rate of change of velocity with respect to clock time, in units of miles / hr^2.
The average rate of change of velocity with respect to clock time is by definition the average acceleration on the interval defined by the trapezoid.
So for a graph of velocity vs. clock time, a slope between two points represents the average acceleration on the corresponding interval.
The base of the equal-area rectangle represents the change in the 'horizontal' quantity. The 'horizontal quantity' in this case the clock time, so the base represents the change in clock time.
Strictly speaking the base is a positive quantity.
However the change in the 'horizontal' quantity can be either positive or negative, depending on whether you think of moving from left to right or from right to left, and also on whether your horizontal axis happens to be pointing to the right or the left.
For now agree on the convention that the horizontal axis points toward the right, and both rise and run will be calculated from the left to the right (so that the change will be the rightmost quantity minus the leftmost). Once we are used to this convention we will have a point of reference that will allow us to easily adapt our thinking to situations where the axis might run in the opposite direction, or the quantities are considered in reverse order.
The equal-area rectangle is formed by cutting the trapezoid along a horizontal line originating from the midpoint of the slope segment. This cuts the trapezoid into two pieces, one being a triangle which can be rotated 180 deg and 'pasted' to the other piece to form a rectangle. Since the rectangle is formed by the two pieces of the original trapezoid, it has the same area as that trapezoid.
The 'graph altitude' of this rectangle is halfway between the 'graph altitudes' of the original trapezoid, and can therefore be calculated by averaging the two 'altitudes':
'graph altitude' of equal-area rectangle = 'average altitude' = (left altitude + right altitude) / 2.
In this case the 'graph altitudes' are 50 and 60, representing 50 miles/hr and 60 miles/hr, so the altitude of the 'equal-area rectangle' is
'graph altitude' of equal-area rectangle = (50 + 60) / 2 = 55,
and represents
(50 miles / hr + 60 miles / hr) / 2 = 55 miles / hr,
which is the average of the initial and final velocities on the interval.
This quantity makes perfect sense as the approximate average velocity for this interval, which based only on the given information we expect (but are not assured) will be between the initial and final velocity and will lie somewhere around halfway between the two.
When you multiply the base of the rectangle by the altitude to get the area, your result will represent the product of the average altitude and the base of the original trapezoid.
In this case we obtain
area = 55 * 5 = 275, representing 55 miles / hour * 5 hours = 275 miles / hour * hours.
The units of the product are miles / hour * hours = miles
Thus the area represents the quantity
area = 275 miles / hour * hours = 275 miles.
This makes perfect sense.
Recall that 55 miles/hour is the approximate average velocity and 5 hours the time interval.
Moving at an approximate average velocity of 55 miles/hour for 5 hours, an object would move approximately 275 miles.
We can generalize this:
Any 'graph trapezoid' which represents velocity vs. clock time will have an area equal to that of an equal-area rectangle which represents the product of the approximate average velocity and the time interval.
This product is the displacement of an object traveling at that average velocity, during the time interval defined by the trapezoid.
The slope of this trapezoid will represent the change in velocity / change in clock time, which is the average acceleration on the interval defined by the trapezoid.
Self-critique (if necessary):
Self-critique rating:
Question: `q006. On a graph of income stream in dollars per month vs. clock time in months, we find the two points (2, 50) and (7, 60).
Your solution:
Confidence Assessment:
Given Solution:
A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of income stream in dollars/month vs. clock time in month(s) represents income stream as the 'vertical' quantity and clock time as the 'horizontal' coordinate.
On such a graph, an altitude therefore represents a income stream.
The base rests on the horizontal axis, so it runs from one clock time to another.
We will call the first clock time the 'initial clock time' and the second the 'final clock time'.
The base therefore represents the change from the initial clock time to the final. We call this quantity the 'change in clock time'.
On the present graph, the income stream is said to be in dollars/month and the clock time is said to be in month(s).
Thus an altitude of the graph represents a income stream in dollars/month, and the base represents the change in clock time in month(s).
In the process of solving this problem you should have sketched a graph, with the vertical axis labeled 'income stream in dollars/month' and the horizontal axis labeled 'clock time in month(s)'.
Your graph would have included a trapezoid resting on the x axis, with its left and right left and right 'graph altitudes' labeled 50 and 60 (representing income stream of 50 dollars/month and 60 dollars/month) and its base running from 2 to 7 (representing 2 month(s) and 7 month(s)), so that its base is 7 - 2 = 5, representing 7 month(s) - 2 month(s) = 5 month(s).
The 'slope segment' of this trapezoid rises from its left altitude 50 to its right altitude 60, representing the income streams 50 dollars/month and 60 dollars/month.
The rise of this trapezoid is the rise of its 'slope segment', which is the change in its 'graph altitude'.
From left to right this change is 60 - 50 = 10 and represents 60 dollars/month - 50 dollars/month = 10 dollars/month.
Since the vertical coordinate represents the income stream, this is the change in income stream on the interval represented by this trapezoid.
The run of this trapezoid is the run of its 'slope segment', which is the change in the horizontal coordinate on the corresponding interval.
From left to right this change is 7 - 2 = 5 and represents 7 month(s) - 2 month(s) = 5 month(s).
Since the horizontal coordinate represents the clock time, this is the change in clock time on the interval represented by this trapezoid.
The slope of the trapezoid is 'rise / run', the rise of the slope segment divided by its run.
The rise is 10 and the run is 5 so the slope is 2.
This slope represents rise / run = (10 dollars/month) / (2 month(s)) = 5 dollars / month^2.
Details of the calculation of units:
(dollars/month) / (month(s)) = (dollars/month) * (1 / month(s)) = dollars / month^2
The order of operations can be used to represent this series of calculations as follows:
To interpret the meaning of this slope:
The rise represents the change in the vertical quantity, which in this case is the income stream in dollars/month.
The run represents the change in the horizontal quantity, which in this case is the clock time in month(s).
Thus the slope = rise / run represents (change in vertical quantity) / (change in horizontal quantity), which by the definition of average rate is the average rate of change of the vertical quantity with respect to the horizontal quantity.
So the slope represents the average rate of change of income stream with respect to clock time, in units of dollars / month^2.
The average rate of change of income stream with respect to clock time is by definition the average acceleration on the interval defined by the trapezoid.
So for a graph of income stream vs. clock time, a slope between two points represents the average acceleration on the corresponding interval.
The base of the equal-area rectangle represents the change in the 'horizontal' quantity. The 'horizontal quantity' in this case the clock time, so the base represents the change in clock time.
Strictly speaking the base is a positive quantity.
However the change in the 'horizontal' quantity can be either positive or negative, depending on whether you think of moving from left to right or from right to left, and also on whether your horizontal axis happens to be pointing to the right or the left.
For now agree on the convention that the horizontal axis points toward the right, and both rise and run will be calculated from the left to the right (so that the change will be the rightmost quantity minus the leftmost). Once we are used to this convention we will have a point of reference that will allow us to easily adapt our thinking to situations where the axis might run in the opposite direction, or the quantities are considered in reverse order.
The equal-area rectangle is formed by cutting the trapezoid along a horizontal line originating from the midpoint of the slope segment. This cuts the trapezoid into two pieces, one being a triangle which can be rotated 180 deg and 'pasted' to the other piece to form a rectangle. Since the rectangle is formed by the two pieces of the original trapezoid, it has the same area as that trapezoid.
The 'graph altitude' of this rectangle is halfway between the 'graph altitudes' of the original trapezoid, and can therefore be calculated by averaging the two 'altitudes':
'graph altitude' of equal-area rectangle = 'average altitude' = (left altitude + right altitude) / 2.
In this case the 'graph altitudes' are 50 and 60, representing 50 dollars/month and 60 dollars/month, so the altitude of the 'equal-area rectangle' is
'graph altitude' of equal-area rectangle = (50 + 60) / 2 = 55,
and represents
(50 dollars/month + 60 dollars/month) / 2 = 55 dollars/month,
which is the average of the initial and final velocities on the interval.
This quantity makes perfect sense as the approximate average income stream for this interval, which based only on the given information we expect (but are not assured) will be between the initial and final income stream and will lie somewhere around halfway between the two.
When you multiply the base of the rectangle by the altitude to get the area, your result will represent the product of the average altitude and the base of the original trapezoid.
In this case we obtain
area = 55 * 5 = 275, representing 55 dollars/month * 5 month(s) = 275 dollars/month * month(s).
The units of the product are dollars/month * month(s) = dollars
Thus the area represents the quantity
area = 275 dollars/month * month(s) = 275 dollars.
This makes perfect sense.
Recall that 55 dollars/month is the approximate average income stream and 5 month(s) the time interval.
With an approximate average income stream of 55 dollars/month for 5 month(s), the total income would be approximately 275 dollars.
We can generalize this:
Any 'graph trapezoid' which represents income stream vs. clock time will have an area equal to that of an equal-area rectangle which represents the product of the approximate average income stream and the time interval.
This product is the total income corresponding to that average income stream, during the time interval defined by the trapezoid.
The slope of this trapezoid will represent the change in income stream / change in clock time, which is the average rate of change of the income stream on the interval defined by the trapezoid. This quantity basically tells us how quickly the income stream is changing, and whether it is increasing or decreasing.
Self-critique (if necessary):
Self-critique rating:
Question: `q007. On a graph of force in pounds vs. position in feet, we find a graph rectangle with base 200 and altitude 30.
Your solution:
Confidence Assessment:
Given Solution:
On a graph of force in pounds vs. position in feet, we find a graph rectangle with base 200 and altitude 30.
A 'graph trapezoid' is defined by two points on a graph, as follows:
The 'graph slope' between two points is the slope of the 'slope segment' of the graph trapezoid defined by the two points.
On a graph of force in lb vs. position time in ft, we find graph points (70, 20) and (190, 50)
Your solution:
Confidence Assessment:
Given Solution:
A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of force in lb vs. position in ft represents force as the 'vertical' quantity and position as the 'horizontal' coordinate.
On such a graph, an altitude therefore represents a force.
The base rests on the horizontal axis, so it runs from one position to another.
We will call the first position the 'initial position' and the second the 'final position'.
The base therefore represents the change from the initial position to the final. We call this quantity the 'change in position'.
On the present graph, the force is said to be in lb and the position is said to be in ft.
Thus an altitude of the graph represents a force in lb, and the base represents the change in position in ft.
In the process of solving this problem you should have sketched a graph, with the vertical axis labeled 'force in lb' and the horizontal axis labeled 'position in ft'.
Your graph would have included a trapezoid resting on the x axis, with its left and right left and right 'graph altitudes' labeled 20 and 50 (representing force of 20 lb and 50 lb) and its base running from 70 to 190 (representing 70 ft and 190 ft), so that its base is 190 - 70 = 120, representing 190 ft - 70 ft = 120 ft.
The 'slope segment' of this trapezoid rises from its left altitude 20 to its right altitude 50, representing the forces 20 lb and 50 lb.
The rise of this trapezoid is the rise of its 'slope segment', which is the change in its 'graph altitude'.
From left to right this change is 50 - 20 = 30 and represents 50 lb - 20 lb = 30 lb.
Since the vertical coordinate represents the force, this is the change in force on the interval represented by this trapezoid.
The run of this trapezoid is the run of its 'slope segment', which is the change in the horizontal coordinate on the corresponding interval.
From left to right this change is 190 - 70 = 120 and represents 190 ft - 70 ft = 120 ft.
Since the horizontal coordinate represents the position, this is the change in position on the interval represented by this trapezoid.
The slope of the trapezoid is 'rise / run', the rise of the slope segment divided by its run.
The rise is 30 and the run is 120 so the slope is .25.
This slope represents rise / run = (30 lb) / (70 ft) = .25 lb / ft.
Details of the calculation of units:
(lb) / (ft) = (lb) * (1 / ft) = lb / ft
To interpret the meaning of this slope:
The rise represents the change in the vertical quantity, which in this case is the force in lb.
The run represents the change in the horizontal quantity, which in this case is the position in ft.
Thus the slope = rise / run represents (change in vertical quantity) / (change in horizontal quantity), which by the definition of average rate is the average rate of change of the vertical quantity with respect to the horizontal quantity.
So the slope represents the average rate of change of force with respect to position, in units of lb / ft.
The average rate of change of force with respect to position is by definition the average acceleration on the interval defined by the trapezoid.
So for a graph of force vs. position, a slope between two points represents the average acceleration on the corresponding interval.
The base of the equal-area rectangle represents the change in the 'horizontal' quantity. The 'horizontal quantity' in this case the position, so the base represents the change in position.
Strictly speaking the base is a positive quantity.
However the change in the 'horizontal' quantity can be either positive or negative, depending on whether you think of moving from left to right or from right to left, and also on whether your horizontal axis happens to be pointing to the right or the left.
For now agree on the convention that the horizontal axis points toward the right, and both rise and run will be calculated from the left to the right (so that the change will be the rightmost quantity minus the leftmost). Once we are used to this convention we will have a point of reference that will allow us to easily adapt our thinking to situations where the axis might run in the opposite direction, or the quantities are considered in reverse order.
The equal-area rectangle is formed by cutting the trapezoid along a horizontal line originating from the midpoint of the slope segment. This cuts the trapezoid into two pieces, one being a triangle which can be rotated 180 deg and 'pasted' to the other piece to form a rectangle. Since the rectangle is formed by the two pieces of the original trapezoid, it has the same area as that trapezoid.
The 'graph altitude' of this rectangle is halfway between the 'graph altitudes' of the original trapezoid, and can therefore be calculated by averaging the two 'altitudes':
'graph altitude' of equal-area rectangle = 'average altitude' = (left altitude + right altitude) / 2.
In this case the 'graph altitudes' are 20 and 50, representing 20 lb and 50 lb, so the altitude of the 'equal-area rectangle' is
'graph altitude' of equal-area rectangle = (20 + 50) / 2 = 35,
and represents
(20 lb + 50 lb) / 2 = 35 lb,
which is the average of the initial and final forces on the interval.
This quantity makes perfect sense as the approximate average force for this interval, which based only on the given information we expect (but are not assured) will be between the initial and final force and will lie somewhere around halfway between the two.
When you multiply the base of the rectangle by the altitude to get the area, your result will represent the product of the average altitude and the base of the original trapezoid.
In this case we obtain
area = 35 * 120 = 4200, representing 35 lb * 120 ft = 4200 lb * ft.
The units of the product are lb * ft = lb * ft
Thus the area represents the quantity
area = 4200 lb * ft = 4200 lb * ft.
This makes perfect sense.
Recall that 35 lb is the approximate average force and 120 ft the time interval.
Moving at an approximate average force of 35 lb for 120 ft, an object would move approximately 4200 lb * ft.
We can generalize this:
Any 'graph trapezoid' which represents force vs. position will have an area equal to that of an equal-area rectangle which represents the product of the approximate average force and the time interval.
This product is the displacement of an object traveling at that average force, during the time interval defined by the trapezoid.
The slope of this trapezoid will represent the change in force / change in position, which is the average acceleration on the interval defined by the trapezoid.
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Question: `q008. On a graph of density in grams / centimeter vs. position in centimeters, we find the points (5, 12) and (20, 10).
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A 'graph trapezoid' is defined by two points on a graph, as follows:
The 'graph slope' between two points is the slope of the 'slope segment' of the graph trapezoid defined by the two points.
On a graph of density in grams / cm vs. position time in cm, we find graph points (5, 12) and (20, 10)
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A graph of y vs. x represents y on the vertical axis and x on the horizontal axis. So a graph of density in grams / cm vs. position in cm represents density as the 'vertical' quantity and position as the 'horizontal' coordinate.
On such a graph, an altitude therefore represents a density.
The base rests on the horizontal axis, so it runs from one position to another.
We will call the first position the 'initial position' and the second the 'final position'.
The base therefore represents the change from the initial position to the final. We call this quantity the 'change in position'.
On the present graph, the density is said to be in grams / cm and the position is said to be in cm.
Thus an altitude of the graph represents a density in grams / cm, and the base represents the change in position in cm.
In the process of solving this problem you should have sketched a graph, with the vertical axis labeled 'density in grams / cm' and the horizontal axis labeled 'position in cm'.
Your graph would have included a trapezoid resting on the x axis, with its left and right left and right 'graph altitudes' labeled 12 and 10 (representing density of 12 grams / cm and 10 grams / cm) and its base running from 5 to 20 (representing 5 cm and 20 cm), so that its base is 20 - 5 = 15, representing 20 cm - 5 cm = 15 cm.
The 'slope segment' of this trapezoid rises from its left altitude 12 to its right altitude 10, representing the densityPlural 12 grams / cm and 10 grams / cm.
The rise of this trapezoid is the rise of its 'slope segment', which is the change in its 'graph altitude'.
From left to right this change is 10 - 12 = -2 and represents 10 grams / cm - 12 grams / cm = -2 grams / cm.
Since the vertical coordinate represents the density, this is the change in density on the interval represented by this trapezoid.
The run of this trapezoid is the run of its 'slope segment', which is the change in the horizontal coordinate on the corresponding interval.
From left to right this change is 20 - 5 = 15 and represents 20 cm - 5 cm = 15 cm.
Since the horizontal coordinate represents the position, this is the change in position on the interval represented by this trapezoid.
The slope of the trapezoid is 'rise / run', the rise of the slope segment divided by its run.
The rise is -2 and the run is 15 so the slope is -7.5.
This slope represents rise / run = (-2 grams / cm) / (5 cm) = -7.5 grams / cm^2.
Details of the calculation of units:
(grams / cm) / (cm) = (grams / cm) * (1 / cm) = grams / cm^2
The order of operations can be used to represent this series of calculations as follows:
To interpret the meaning of this slope:
The rise represents the change in the vertical quantity, which in this case is the density in grams / cm.
The run represents the change in the horizontal quantity, which in this case is the position in cm.
Thus the slope = rise / run represents (change in vertical quantity) / (change in horizontal quantity), which by the definition of average rate is the average rate of change of the vertical quantity with respect to the horizontal quantity.
So the slope represents the average rate of change of density with respect to position, in units of grams / cm^2.
The average rate of change of density with respect to position is by definition the average acceleration on the interval defined by the trapezoid.
So for a graph of density vs. position, a slope between two points represents the average acceleration on the corresponding interval.
The base of the equal-area rectangle represents the change in the 'horizontal' quantity. The 'horizontal quantity' in this case the position, so the base represents the change in position.
Strictly speaking the base is a positive quantity.
However the change in the 'horizontal' quantity can be either positive or negative, depending on whether you think of moving from left to right or from right to left, and also on whether your horizontal axis happens to be pointing to the right or the left.
For now agree on the convention that the horizontal axis points toward the right, and both rise and run will be calculated from the left to the right (so that the change will be the rightmost quantity minus the leftmost). Once we are used to this convention we will have a point of reference that will allow us to easily adapt our thinking to situations where the axis might run in the opposite direction, or the quantities are considered in reverse order.
The equal-area rectangle is formed by cutting the trapezoid along a horizontal line originating from the midpoint of the slope segment. This cuts the trapezoid into two pieces, one being a triangle which can be rotated 180 deg and 'pasted' to the other piece to form a rectangle. Since the rectangle is formed by the two pieces of the original trapezoid, it has the same area as that trapezoid.
The 'graph altitude' of this rectangle is halfway between the 'graph altitudes' of the original trapezoid, and can therefore be calculated by averaging the two 'altitudes':
'graph altitude' of equal-area rectangle = 'average altitude' = (left altitude + right altitude) / 2.
In this case the 'graph altitudes' are 12 and 10, representing 12 grams / cm and 10 grams / cm, so the altitude of the 'equal-area rectangle' is
'graph altitude' of equal-area rectangle = (12 + 10) / 2 = 11,
and represents
(12 grams / cm + 10 grams / cm) / 2 = 11 grams / cm,
which is the average of the initial and final densities on the interval.
This quantity makes perfect sense as the approximate average density for this interval, which based only on the given information we expect (but are not assured) will be between the initial and final density and will lie somewhere around halfway between the two.
When you multiply the base of the rectangle by the altitude to get the area, your result will represent the product of the average altitude and the base of the original trapezoid.
In this case we obtain
area = 11 * 15 = 165, representing 11 grams / cm * 15 cm = 165 grams / cm * cm.
The units of the product are grams / cm * cm = grams
Thus the area represents the quantity
area = 165 grams / cm * cm = 165 grams.
This makes perfect sense.
Recall that 11 grams / cm is the approximate average density and 15 cm the time interval.
Moving at an approximate average density of 11 grams / cm for 15 cm, an object would move approximately 165 grams.
We can generalize this:
Any 'graph trapezoid' which represents density vs. position will have an area equal to that of an equal-area rectangle which represents the product of the approximate average density and the time interval.
This product is the displacement of an object traveling at that average density, during the time interval defined by the trapezoid.
The slope of this trapezoid will represent the change in density / change in position, which is the average acceleration on the interval defined by the trapezoid.
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