These exercises are in qa format. You should answer each question before looking at the solution. You are welcome to submit your results along with self-critiques.
This is called an 'open' qa since the entire document is open to you, rather than being sequenced as in a qa program.
Topics will be added regularly according to what is needed by students.
Topic 3 is directly relevant to the process of identifying the information given by a problem, in terms of the scheme involving v0, vf, a, `dt, `ds, vAve and `dv. It addresses the units of the various quantities, situations in which the quantities are identified, and the types of calculations that typically occur with these quantities.
There are four problems in this sequence.
`q001. Events are observed at clock times of 43.8, 44.9, 46.1 and 47.0 seconds. What are the corresponding time intervals?
Solution: The corresponding time intervals are the intervals between the given clock times.
Between clock times t = 43.8 sec and t = 44.9 sec the time interval is 44.9 sec – 43.8 sec = 1.1 sec.
The other intervals are 46.1 sec – 44.9 sec = 1.2 sec and 47.0 sec – 46.1 sec = .9 sec.
We therefore have three time intervals, 1.1 sec, 1.2 sec and 0.9 sec.
`q002 At the clock times observed in the preceding question, the positions of rolling ball are 32 cm, 40 cm, 44 cm and 47 cm. What therefore is the distance moved during each time interval?
Solution: As seen before there are three intervals.
On the first interval the ball moves from position x = 32 cm to position x = 40 cm, a distance of 40 cm – 32 cm = 8 cm.
On the second interval the ball moves a distance of 44 cm – 40 cm = 4 cm, and on the third interval it moves a distance of 47 cm – 44 cm = 3 cm.
`q003 How fast is the ball moving during each interval? According to the information given here, is the ball speeding up, slowing down or moving at an approximately constant rate?
Solution: On the first interval the ball moves 8 cm in 1.1 seconds, so that its average speed is
Ave speed = distance / time interval = 8 cm / (1.1 sec) = 7.3 cm/s, approximately.
On the second and third intervals the ball moves, respectively, 4 cm in 1.2 sec and 3 cm in 0.9 sec so that its approximate average speeds are
ave speed on first interval = distance/time interval = 4 cm / (1.2 sec) = 3.3 cm/sec, and
ave speed on second interval = distance/time interval = 3 cm / (0.9 sec) = 3.3 cm/sec.
The ball appears to slow down from the first interval to the second, that there is no significant difference between its average speeds on the second and third intervals.
`q004 If positions 32 cm, 40 cm, 44 cm and 47 cm are observed at clock times 73, 83, 87 and 89 seconds, what is the velocity for each interval? Is the ball speeding up, slowing down or moving at an approximately constant rate?
Solution: The ball moves distance is of 8.0, 4.0 and 3.0 cm on intervals lasting 10, 4.0 and 2.0 seconds. The average speeds are therefore
8 cm / (10 s) = .8 cm/s
4 cm / (4 s) = 1 cm/s and
3 cm / 2 s = 1.5 cm/s.
While at first it would appear that since the ball is moving less distance on each successive interval, it is slowing down. However the time intervals are also decreasing, and our calculations here show us that the speed of the ball is apparently increasing.
There are seven problems under Topic 2.
`q001. Sketch a set of x-y axes and mark the points (5, 7) and (11, 9). Sketch a straight line segment from the first point to the second. What is the ‘rise’ from the first point to the second? What is the ‘run’ from the first point to the second? What therefore is the slope of the segment?
Solution: From (5, 7) to (11, 9) the ‘rise’ is 9 – 7 = 2 and the ‘run’ is (11 – 5) = 6. The slope is therefore
Slop = rise / run = 2 / 6 = 1/3.
`q002 Extend the line segment in the preceding so that it intersects the y axis. Estimate as nearly as you can the coordinates of the point where the line intersects this axis. What is your estimate?
Solution: Your sketch should show the line extending to the left, intercepting the y axis around y coordinate 5. Even if you used a rough sketch you should be within 1 unit or so of y = 5.
If you used a very accurate sketch you will see that the y coordinate at the point where the line intersects the y-axis is about 5.3; the actual intercept is 5 1/3, or 5.333… . Your sketch won't be accurate enough to show the exact intercept.
`q003 The equation of the straight line between the points (5, 7) and (11, 9) is of the form y = m x + b, where m is the slope of the line. Substitute the x and y coordinates of the first point into this form, replacing the x in the equation with the x coordinates of the first point and the y in the equation with the y coordinate of the first point. Also substitute your value of the slope for m. Solve the resulting equation for b.
Solution: Substituting the coordinates of the first point, we see that we must substitute x = 5 and y = 7. Recall that the slope is m = 1/3, so that the equation y = m x + b becomes
7 = 1/3 * 5 + b.
There are a number of ways to go about solving for b. All involve adding or subtracting the same quantity from both sides, or multiplying by dividing by the same quantity on both sides. The order in which the steps are performed may differ from one individual to another.
In this solution we will proceed as follows:
Multiply both sides by 3 (this will have the effect of ‘clearing’ the denominator, i.e., giving us an equation without denominators):
3 * 7 = 3 * (1/3 * 5 + b). We apply the distributive law to the right-hand side to get
3 * 7 = 3 * (1/3 * 5) + 3 * b. We perform the multiplications and end up with (note that 3 * 1/3 = 3 / 3 = 1)
21 = 5 + 3 * b. Subtracting 5 from both sides we obtain
21 – 5 = 5 + 3 * b – 5, and doing the arithmetic in both sides we obtain
16 = 3 * b. Dividing both sides by 3 we have
16/3 = 3 * b / 3, which simplifies to give us
16/3 = b. Since 16/3 = 5 1/3, we have
B = 16/3 = 5 1/3.
`q004 Repeat the preceding exercise, but this time substitute the coordinates of the second point.
Solution: The second point is (9, 11), giving us the equation
9 = 1/3 * 11 + b.
This equation is easily solved using steps very similar to those in the preceding. We obtain
b = 16/3 = 5 1/3,
just as before.
`q005 The values of b obtained in the preceding should be identical. Your estimate of the coordinates of the point where the line intersects the y axis should be approximately equal to the value of b, though there was some estimation involved when you found these coordinates. How close what is your estimate to the actual value of b?
Solution: The actual value of b is 5 1/3, a point 1/3 of the way between y = 5 and y = 6 on the y axis.
Anything from about y = 4 to y = 6 would be reasonable for a rough sketch.
`q006 The form of the equation of the straight line is, as before, y = m x + b. Leave the y and the x in the equation, but replace m and b with the values you have obtained for the slope and the y-intercept.
Solution: We have obtained m = 1/3 and b = 16/3. Our equation y = m x + b therefore becomes
y = 1/3 x + 16/3.
The improper fraction 16/3 is preferable to the mixed number 5 1/3, since before you can do arithmetic with the mixed number you have to convert it to fractional form.
Decimal approximations are not appropriate here, since any decimal approximation of these fractions is a nonterminating decimal, which cannot be written accurately in decimal form.
`q007 Sketch another graph, this time plotting the points (5, 7), (7, 8), (9, 10) and (11, 9).
Solution: One reasonable line passes through the points (4, 5) and (12, 10). This is not the best possible line, and your line might well be closer on the average and the points of this line. The best line is unlikely to pass through points having two whole-number coordinates (i.e., it’s unlikely that both the x and y coordinates of any point on the line are both whole numbers), so if your points have fractional or decimal coordinates it's probably because your estimates are more precise than the one given here.
For the line passing through the points (4, 7) and (12, 10), the rise is 10 – 7 = 3 and the run is 12 – 4 = 8, so the slope is 3/8. Since we are dealing with a relatively imprecise estimate, it does no harm to use the decimal equivalent of 3/8, which is .375. Since the estimate isn't that great, it does little further harm to simply round this off to .4.
Extending this line to the y axis we will intercept the axis around the point where y = 5.5.
Plugging this slope and y intercept into the for y = m x + b we obtain the equation
y = .4 x + 5.5.
This equation is a reasonable, but not completely accurate best fit to the four original data points. It does not actually pass through any of these data points, but it comes reasonably close to all of them.
Compare your equation with this one. Do you think it does a better job than this one?
Question: This is a 3-part question.
Part I: Which of the following quantities could possibly be a displacement and why?
14 cm
81 seconds
22 cm/s
7 cm/s^2
5 cm^2 / s^2
13 cm^2 / s
9 seconds
44 s / cm
14 s^2 / cm
Part II: Which of the following quantities could possibly be an average velocity and why?
14 cm
81 seconds
22 cm/s
7 cm/s^2
5 cm^2 / s^2
13 cm^2 / s
9 seconds
44 s / cm
14 s^2 / cm
Part III: Which of the following quantities could possibly be an average acceleration and why?
14 cm
81 seconds
22 cm/s
7 cm/s^2
5 cm^2 / s^2
13 cm^2 / s
9 seconds
44 s / cm
14 s^2 / cm
solution:
14 cm has units of distance and could represent a position or a change in position.
81 seconds has units of time and could represent a clock time or a change in clock time.
22 cm/s has units of distance / time and could represent an initial, a final or an average velocity, or a change in velocity.
7 cm/s^2 has units of distance / time^2 and could represent an acceleration.
5 cm^2 / s^2 has units of distance^2 / time^2, and could not represent a position, a change in position, a clock time, a change in clock time, a velocity of any type, or an acceleration.
13 cm^2 / s has units of distance^2 / time and could not represent a position, a change in position, a clock time, a change in clock time, a velocity of any type, or an acceleration.
9 seconds has units of time and could represent a clock time or a change in clock time.
44 s / cm has units of time / distance and could not represent a position, a change in position, a clock time, a change in clock time, a velocity of any type, or an acceleration.
14 s^2 / cm has units of time^2 / distance and could not represent a position, a change in position, a clock time, a change in clock time, a velocity of any type, or an acceleration.
... situations ...
Question: The following consists of a series of questions asking you to identify the possible meanings of a series of calculations:
Part 1: Which of the following calculations could correspond to the calculation of a displacement and why?
40 cm/s * 5 s
20 cm/s / (4 s)
30 cm / (6 seconds)
( 10 cm + 4 cm ) / 2
( 4 cm/s + 8 cm/s) / 2
8 cm / (5 cm/s^2)
14 cm / (3 cm/s)
(12 cm - 4 cm) / (6 s)
12 cm/s^2 * 5 s
(16 cm/s - 12 cm/s) / (8 s)
8 cm/s + 2 cm/s^2 * 10 s
The slope of a v vs. t graph between two points.
The area of a v vs. t graph beneath a line segment connecting two points.
The average altitude of a v vs. t graph.
The rise of a position vs. clock time graph.
The run of a position vs. clock time graph.
The slope of a position vs. clock time graph.
The area beneath a position vs. clock time graph.
The average 'altitude' of a position vs. clock time graph.
The rise of an acceleration vs. clock time graph.
The run of an acceleration vs. clock time graph.
The slope of an acceleration vs. clock time graph.
The area beneath an acceleration vs. clock time graph.
The average 'altitude' of an acceleration vs. clock time graph.
Part II: Which of the following calculations could correspond to the calculation of an average velocity and why?
40 cm/s * 5 s
20 cm/s / (4 s)
30 cm / (6 seconds)
( 10 cm + 4 cm ) / 2
( 4 cm/s + 8 cm/s) / 2
8 cm / (5 cm/s^2)
14 cm / (3 cm/s)
(12 cm - 4 cm) / (6 s)
12 cm/s^2 * 5 s
(16 cm/s - 12 cm/s) / (8 s)
8 cm/s + 2 cm/s^2 * 10 s
The slope of a v vs. t graph between two points.
The area of a v vs. t graph beneath a line segment connecting two points.
The average altitude of a v vs. t graph.
The rise of a position vs. clock time graph.
The run of a position vs. clock time graph.
The slope of a position vs. clock time graph.
The area beneath a position vs. clock time graph.
The average 'altitude' of a position vs. clock time graph.
The rise of an acceleration vs. clock time graph.
The run of an acceleration vs. clock time graph.
The slope of an acceleration vs. clock time graph.
The area beneath an acceleration vs. clock time graph.
The average 'altitude' of an acceleration vs. clock time graph.
Part III: Which of the following calculations
could correspond to the calculation of an average acceleration and why?
40 cm/s * 5 s
20 cm/s / (4 s)
30 cm / (6 seconds)
( 10 cm + 4 cm ) / 2
( 4 cm/s + 8 cm/s) / 2
8 cm / (5 cm/s^2)
14 cm / (3 cm/s)
(12 cm - 4 cm) / (6 s)
12 cm/s^2 * 5 s
(16 cm/s - 12 cm/s) / (8 s)
8 cm/s + 2 cm/s^2 * 10 s
The slope of a v vs. t graph between two points.
The area of a v vs. t graph beneath a line segment connecting two points.
The average altitude of a v vs. t graph.
The rise of a position vs. clock time graph.
The run of a position vs. clock time graph.
The slope of a position vs. clock time graph.
The area beneath a position vs. clock time graph.
The average 'altitude' of a position vs. clock time graph.
The rise of an acceleration vs. clock time graph.
The run of an acceleration vs. clock time graph.
The slope of an acceleration vs. clock time graph.
The area beneath an acceleration vs. clock time graph.
The average 'altitude' of an acceleration vs. clock time graph.
*******************
Solution:
40 cm/s * 5 s and could represent the product of an average velocity and the time interval, and could hence represent a displacement (ave vel = displacement / time interval so displacement = ave vel * time interval).
20 cm/s / (4 s) could represent a change in velocity divided by a change in clock time and could therefore represent an average acceleration (aAve = change in velocity / change in clock time)30 cm / (6 seconds) could represent a displacement divided by the time interval, and hence could represent an average rate of change in position with respect to clock time, i.e., an average velocity.
( 10 cm + 4 cm ) / 2 represents the average value of two positions. The result has units of position and cannot be any kind of a velocity or acceleration, and can certainly not be a time interval or a clock time.
8 cm / (5 cm/s^2) has units of s^2 (cm / (cm/s^2) = cm * (s^2 / cm) = (cm / cm) * s^2 = s^2). This is not a unit of clock time, position, displacement, any kind of velocity, or acceleration.
14 cm / (3 cm/s) could represent a 14 cm displacement divided by a 3 cm/s average velocity, and could therefore represent a time interval. The units of this calculation work out to cm / (cm/s) = cm * (s / cm) = (cm / cm) * s = s, which agree with the units of the time interval.
( 4 cm/s + 8 cm/s) / 2 could represent the average value of two velocities, and provided acceleration is uniform could then represent an average velocity.
(12 cm - 4 cm) / (6 s) could represent the difference of two positions, i.e., a change in position, divided by a time interval. This would correspond to an average velocity (vAve = `dx / `dt).
12 cm/s^2 * 5 s could represent the product of an acceleration and a time interval. This calculation could therefore represent a change in velocity (since aAve = `dv / `dt, it follows that `dv = aAve * `dt).
(16 cm/s - 12 cm/s) / (8 s) could represent the difference of two velocities, i.e., a change in velocity, divided by a time interval. This would correspond to an average acceleration (aAve = `dv / `dt). Also the units of the calculation are (cm/s) / s = (cm/s) * (1/s) = cm/s^2, in agreement with the units of acceleration.
8 cm/s + 2 cm/s^2 * 10 s has units of cm/s + cm/s, which gives us consistent units of cm/s. This can therefore be a valid calculation. The 8 cm/s could represent an initial velocity, while the 2 cm/s^2 could represent an average acceleration and 10 s a time interval. The product could therefore represent a change in velocity, which would be added to the initial quantity (the initial velocity), resulting in a final velocity.
The slope of a v vs. t graph between two points represents rise / run; rise represent change in v and run represents change in t so the slope represents change in v / change in t, which is the average rate of change of velocity with respect to clock time, i.e., average acceleration.
The area of a v vs. t graph beneath a line segment connecting two points represents the product of the midpoint velocity and the time interval. If acceleration is uniform, then the straight line segment accurately depicts the velocity function so the midpoint velocity is the same as the average velocity, and the area therefore represent the product of average velocity and time interval, i.e., the displacement.
The average altitude of a v vs. t graph represents the average value of v, which is the average velocity.
The rise of a position vs. clock time graph represents the change in the vertical coordinate. The vertical coordinate is position, so the rise represents change in position.
The run of a position vs. clock time graph represents the change in the horizontal coordinate. The horizontal coordinate is clock time, so the run represents the change in clock time.
The slope of a position vs. clock time graph represents the rise of the graph between two points divided by the run. The slope therefore represents change in position/change in clock time, which is the average rate of change of position with respect to clock time. This is the definition of average velocity.
The area beneath a position vs. clock time graph.
The average altitude of a position vs. clock time graph represents the average value of the vertical coordinate, which in this case is position. The average position is not generally a quantity of interest in the analysis of motion.
The rise of an acceleration vs. clock time graph represents the change in the vertical coordinate of the graph. The vertical coordinate represents the acceleration, so the rise represents the change in the acceleration. In a uniform acceleration situation, the acceleration doesn't change, so the graph will have no rise. The graph will therefore be a horizontal line segment.
The run of an acceleration vs. clock time graph represents the change in the horizontal coordinate. The horizontal coordinate is to clock time, so the run represents the change in clock time.
The slope of an acceleration vs. clock time graph represents the rise divided by the run between two points. Since the rise and run are respectively the changes in acceleration and clock time, the slope therefore represents change in acceleration / change in clock time, which by definition is the average rate of change of acceleration with respect to clock time. If acceleration is uniform, there will be no change in acceleration in this quantity will therefore be zero. If the acceleration is not uniform, this quantity represents the average 'jerk' over the corresponding time interval (when acceleration changes, for example someone suddenly presses down harder on the accelerator or the brakes, you feel a 'jerk').
The area beneath an acceleration vs. clock time graph represents the product of the average acceleration and the change in clock time. The average acceleration is the average rate of change of velocity with respect to clock time, i.e., change in velocity / change in clock time, so that the product gives you change in velocity / change in clock time * change in clock time = change in velocity.
The average altitude of an acceleration vs. clock time graph represents the average value of the vertical coordinate, which in this case is the acceleration. The average altitude therefore represents the average acceleration. In the case of uniform acceleration the graph consists of a horizontal line segment, so that all altitude their equal, and in particular the average altitude is equal to the initial and final altitude.
`q001: Graph the points (4, 9) and (7, 19) on a set of coordinate axes and sketch the straight line segment which runs from the first point to the second. Then sketch a straight vertical line segment from the first point down to the horizontal axis, and another from the second point down to the horizontal axis. Describe your sketch.
Answer: Your sketch should consist of a point with horizontal coordinate 4 and
vertical coordinate 9, and another with horizontal coordinate 7 and vertical
coordinate 19. The line segment connecting these points slopes upward into the
right. The first vertical line segment runs from the point (4, 9) to the point
(4, 0) on the horizontal axis, and the second vertical segment runs from the
point (7, 19) the point (7, 0) on the horizontal axis. Along with the segment
which runs along the horizontal axis from (4, 0) to (7, 0), and these line
segments form a trapezoid with vertical 'altitudes' of 9 and 21, and a 'width'
of (7 - 4) = 3. (What we call the altitudes here are actually the bases of the
trapezoid, and what we call the width is actually the altitude of the trapezoid,
when using standard geometric terminology. However when referring to trapezoids
of this nature on graphs, it makes more sense to referred to the vertical line
segments as altitudes.)
`q002: What is the rise between the points (4, 9) and (7, 21)?
Answer: The second coordinate of an ordered pair is
traditionally graphed on the ' vertical' axis, and the difference from the first
vertical coordinate to the second is called the 'rise'. The 'rise' between these
points is (21 - 9) = 12.
`q003: What is the run between the points (4, 9) and (7, 21)?
Answer:
The first coordinate of an ordered pair is traditionally
graphed on the 'horizontal' axis, and the difference from the first horizontal
coordinate to the second is called the 'run'. The 'run' between these points is
(7 - 4) = 3.
`q004: What therefore is the slope between these points?
Answer:
As seen above the rise is 12 and the run is 3, so the slope
is slope = rise / run = 12 / 3 = 3.
`q005: What is the area of the trapezoid defined by these points?
Answer:
The area of a 'graph trapezoid' is the same as that of a
rectangle having the same width, whose altitude is equal to the midpoint
altitude of the trapezoid. The midpoint altitude of the trapezoid is simply the
average of its two altitudes. For this reason we say that the midpoint altitude
of the trapezoid is the average altitude of the trapezoid. The two altitudes of
this trapezoid are 9 and 21, so the average altitude is their average (9 + 21) /
2 = 30 / 2 = 15. The width of the trapezoid can be measured from the point (4,
0) to the point (7, 0) and is 7 - 4 = 3 (you may also note that this is the same
as the 'run' between the two points). The area of a rectangle is the product of
its length and width, so the area of this trapezoid (being equal to the area of
the rectangle) is therefore area = ave altitude * width = 15 * 3 = 45.
`q006: What is the rise between the points (4 sec, 9 cm) and (7 sec, 21 cm)?
Answer:
The second coordinate of an ordered pair is traditionally
graphed on the ' vertical' axis, and the difference from the first vertical
coordinate to the second is called the 'rise'. The 'rise' between these points
is (21 cm - 9 cm) = 12 cm.
`q007: What is the run between the points (4 sec, 9 cm) and (7 sec, 21 cm)?
Answer:
The first coordinate of an ordered pair is traditionally
graphed on the 'horizontal' axis, and the difference from the first horizontal
coordinate to the second is called the 'run'. The 'run' between these points is
(7 sec - 4 sec) = 3 sec.
`q008: What therefore is the slope between these points?
Answer:
As seen above the rise is 12 cm and the run is 3 sec, so
the slope is slope = rise / run = 12 cm / (3 sec) = 3 cm/sec.
`q009: What is the area of the trapezoid defined by this graph?
Answer:
The area is equal to the average altitude of the trapezoid
multiplied by its width. The altitudes of this trapezoid are 9 cm and 21 cm, so
the average altitude is (9 cm + 21 cm) / 2 = 15 cm. The width is (7 sec - 3 sec)
= 4 sec. So its area is 15 cm * 4 sec = 60 cm * s. Note that the unit cm * s is
an essential aspect of the area and cannot be omitted. Note also that the unit
cm * s has no general interpretation in terms of motion. It is not be confused
with the unit cm / s, which is the unit of the slope (obtained by dividing a
quantity with unit cm by a quantity with unit s).
`q010: What is the rise between the points (4 sec, 9 cm / sec) and (7 sec, 21
cm / sec)?
Answer:
The second coordinate of an ordered pair is traditionally
graphed on the ' vertical' axis, and the difference from the first vertical
coordinate to the second is called the 'rise'. The 'rise' between these points
is (21 cm / sec - 9 cm / sec) = 12 cm / sec.
`q011: What is the run between the points (4 sec, 9 cm / sec) and (7 sec, 21
cm / sec)?
Answer:
The first coordinate of an ordered pair is traditionally
graphed on the 'horizontal' axis, and the difference from the first horizontal
coordinate to the second is called the 'run'. The 'run' between these points is
(7 sec - 4 sec) = 3 sec.
`q012: What therefore is the slope between these points?
Answer:
As seen above the rise is 12 cm / sec and the run is 3 sec,
so the slope is slope = rise / run = (12 cm /s) / (3 sec) = 3 (cm/sec) / sec = 3
(cm/s) * (1/s) = 3 (cm * 1) / (s * s) = 3 cm / s^2.
`q013: What is the area of the trapezoid defined by this graph?
Answer:
The area is equal to the average altitude of the trapezoid
multiplied by its width. The altitudes of this trapezoid are 9 cm/s and 21 cm/s,
so the average altitude is (9 cm/s + 21 cm/s) / 2 = 15 cm/s. The width is (7 sec
- 3 sec) = 4 sec. So its area is 15 cm/s * 4 sec = 60 cm/s * s = 60 cm ( s / s)
= 60 cm. Note that the product of 15 cm/s and 4 s can be interpreted as the
product of an average velocity of 15 cm/s and the time interval of 4 s, so that
with this interpretation the area represents the corresponding 60 cm
displacement. In general, if our graph represents velocity versus clock time for
an interval, the area under the graph represents the corresponding displacement.
`q014: What is the rise between the points (4 ft, 9 lb) and (7 ft, 21 lb)?
Answer:
The second coordinate of an ordered pair is traditionally
graphed on the ' vertical' axis, and the difference from the first vertical
coordinate to the second is called the 'rise'. The 'rise' between these points
is (21 lb - 9 lb) = 12 lb.
`q015: What is the run between the points (4 ft, 9 lb) and (7 ft, 21 lb)?
Answer:
The first coordinate of an ordered pair is traditionally
graphed on the 'horizontal' axis, and the difference from the first horizontal
coordinate to the second is called the 'run'. The 'run' between these points is
(7 ft - 4 ft) = 3 ft.
`q016: What therefore is the slope between these points?
Answer:
As seen above the rise is 12 lb and the run is 3 ft, so the
slope is slope = rise / run = 12 lb / (3 ft) = 3 lb/ft.
`q017: What is the area of the trapezoid defined by this graph?
Answer:
The area is equal to the average altitude of the trapezoid
multiplied by its width. The altitudes of this trapezoid are 9 lb and 21 lb, so
the average altitude is (9 lb + 21 lb) / 2 = 15 lb. The width is (7 ft - 3 ft) =
4 ft. So its area is 15 lb * 4 ft = 60 lb * ft. Note that the product of 15 lb
and 4 ft can be interpreted as the product of an average force of 15 lb and a
displacement of 4 ft. As you will see later in the course this is a very
important quantity, related to work and energy. You don't have to remember this
interpretation, but you should understand that the units of the quantities
represented on a graph determine the units of its slope and area, and that the
meanings of the quantities represented on a graph determine the meanings of the
slope and area. In some situations that can be difficult to understand meanings
and interpretations, but the first step in the process is to include the units
in every calculation and to do the algebra of the units.
`q018: In general what does the rise between two points of a graph represent?
Answer:
The rise represents the change in the quantity represented
by the vertical coordinate.
`q019: In general what does the run between two points of a graph represent?
Answer:
The run represents the change in the quantity represented
by the horizontal coordinate.
`q020: In general what does the slope between two points of a graph
represent?
Answer:
The slope represents the change in the quantity represented
by the vertical coordinate, divided by the change in the quantity represented by
the horizontal coordinate. If the graph represents quantity A vs. quantity B,
then the slope between two points represents change in A / change in B, which is
by definition the average rate at which quantity A changes with respect to
quantity B.
`q021: In general what does the average of the vertical quantities
represented by two points represent?
Answer:
If the graph is a straight line, this average represents
the average value of the quantity represented by the vertical coordinate. If the
graph for the interval between the points is nearly a straight line, then this
average is near the average value of the quantity, and hence represents the
approximate average value of the quantity represented by the vertical
coordinate.
`q022: In general what does the width of the trapezoid defined by two points
represent?
Answer:
The width is the same as the 'run' between the two points
and represents the change in the quantity represented by the horizontal
coordinate.
`q023: In general what does the area of the trapezoid defined by two points
represent?
Answer:
The area represents the product of the average 'graph
altitude' and the width of the trapezoid. If it makes sense to multiply the
average value of the 'vertical' quantity by the change in the 'horizontal'
quantity, then the area of the trapezoid is probably an important quantity.