3.7: Free Fall
- Use the kinematic equations with the variables y and g to analyze free-fall motion.
- Describe how the values of the position, velocity, and acceleration change during a free fall.
- Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.
An interesting application of Equation 3.3.2 through Equation 3.5.22 is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.
Gravity
The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure \(\PageIndex{1}\).
In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.
For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.
Acceleration due to gravity is so important that its magnitude is given its own symbol, g. It is constant at any given location on Earth and has the average value
\[g = 9.81\; m/s^{2}\; (or\; 32.2\; ft/s^{2}) \ldotp\]
Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value +g or −g depends on how we define our coordinate system. If we define the upward direction as positive, then a = −g = −9.8 m/s 2 , and if we define the downward direction as positive, then a = g = 9.8 m/s 2 .
One-Dimensional Motion Involving Gravity
The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g. We represent vertical displacement with the symbol y.
We assume here that acceleration equals −g (with the positive direction upward).
\[v =v _{0} - gt \label{3.15}\]
\[y = y_{0} + v_{0} t - \frac{1}{2} gt^{2} \label{3.16}\]
\[v^{2} = v_{0}^{2} - 2 g(y - y_{0}) \label{3.17}\]
- Decide on the sign of the acceleration of gravity. In Equation \ref{3.15} through Equation \ref{3.17}, acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
- Draw a sketch of the problem. This helps visualize the physics involved.
- Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
- Decide which of Equation \ref{3.15} through Equation \ref{3.17} are to be used to solve for the unknowns.
Figure \(\PageIndex{2}\) shows the positions of a ball, at 1-s intervals, with an initial velocity of 4.9 m/s downward, that is thrown from the top of a 98-m-high building. (a) How much time elapses before the ball reaches the ground? (b) What is the velocity when it arrives at the ground?
Strategy
Choose the origin at the top of the building with the positive direction upward and the negative direction downward. To find the time when the position is −98 m, we use Equation \ref{3.16}, with y 0 = 0, v 0 = −4.9 m/s, and g = 9.8 m/s 2 .
Solution
- Substitute the given values into the equation: $$y = y_{0} + v_{0} t - \frac{1}{2} gt^{2}$$ $$-98.0\; m = 0 - (4.9\; m/s)t - \frac{1}{2} (9.8\; m/s^{2}) t^{2} \ldotp$$This simplifies to $$t^{2} + t - 20 = 0 \ldotp$$This is a quadratic equation with roots t = −5.0 s and t = 4.0 s. The positive root is the one we are interested in, since time t = 0 is the time when the ball is released at the top of the building. (The time t = −5.0 s represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
- Using Equation \ref{3.15}, we have $$v =v _{0} - gt = -4.9\; m/s - (9.8\; m/s^{2})(4.0\; s) = -44.1\; m/s \ldotp$$
Significance
For situations when two roots are obtained from a quadratic equation in the time variable, we must look at the physical significance of both roots to determine which is correct. Since t = 0 corresponds to the time when the ball was released, the negative root would correspond to a time before the ball was released, which is not physically meaningful. When the ball hits the ground, its velocity is not immediately zero, but as soon as the ball interacts with the ground, its acceleration is not g and it accelerates with a different value over a short time to zero velocity. This problem shows how important it is to establish the correct coordinate system and to keep the signs of g in the kinematic equations consistent.
A batter hits a baseball straight upward at home plate and the ball is caught 5.0 s after it is struck Figure \(\PageIndex{3}\). (a) What is the initial velocity of the ball? (b) What is the maximum height the ball reaches? (c) How long does it take to reach the maximum height? (d) What is the acceleration at the top of its path? (e) What is the velocity of the ball when it is caught? Assume the ball is hit and caught at the same location.
Strategy
Choose a coordinate system with a positive y-axis that is straight up and with an origin that is at the spot where the ball is hit and caught.
Solution
- Equation \ref{3.16} gives $$y = y_{0} + v_{0} t - \frac{1}{2} gt^{2}$$ $$0 = 0 + v_{0} (5.0\; s)- \frac{1}{2} (9.8\; m/s^{2}) (5.0\; s)^{2} \ldotp$$which gives v 0 = 24.5 m/sec.
- At the maximum height, v = 0. With v 0 = 24.5 m/s, Equation \ref{3.17} gives $$v^{2} = v_{0}^{2} - 2 g(y - y_{0})$$ $$0 = (24.5\; m/s^{2}) - 2 (9.8\; m/s^{2})(y - 0)$$or $$y = 30.6\; m \ldotp$$
- To find the time when v = 0 , we use Equation \ref{3.15}: $$v = v_{0} - gt$$ $$0 = 24..5\; m/s - (9.8\; m/s^{2})t \ldotp$$This gives t = 2.5 s. Since the ball rises for 2.5 s, the time to fall is 2.5 s.
- The acceleration is 9.8 m/s 2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s 2 downward.
- The velocity at t = 5.0 s can be determined with Equation \ref{3.15}: $$\begin{split} v & = v_{0} - gt \\ & = 24.5\; m/s - 9.8\; m/s^{2} (5.0\; s) \\ & = -24.5\; m/s \ldotp \end{split}$$
Significance
The ball returns with the speed it had when it left. This is a general property of free fall for any initial velocity. We used a single equation to go from throw to catch, and did not have to break the motion into two segments, upward and downward. We are used to thinking of the effect of gravity is to create free fall downward toward Earth. It is important to understand, as illustrated in this example, that objects moving upward away from Earth are also in a state of free fall.
A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?
A small rocket with a booster blasts off and heads straight upward. When at a height of 5.0 km and velocity of 200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity of the booster at a height of 6.0 km? Neglect air resistance.
Strategy
We need to select the coordinate system for the acceleration of gravity, which we take as negative downward. We are given the initial velocity of the booster and its height. We consider the point of release as the origin. We know the velocity is zero at the maximum position within the acceleration interval; thus, the velocity of the booster is zero at its maximum height, so we can use this information as well. From these observations, we use Equation \ref{3.17}, which gives us the maximum height of the booster. We also use Equation \ref{3.17} to give the velocity at 6.0 km. The initial velocity of the booster is 200.0 m/s.
Solution
- From Equation \ref{3.17}, \(v^{2} = v_{0}^{2} - 2 g(y - y_{0})\). With v = 0 and y 0 = 0, we can solve for y: $$y = \frac{v_{0}^{2}}{-2g} = \frac{(2.0 \times 10^{2}\; m/s)^{2}}{-2(9.8\; m/s^{2})} = 2040.8\; m \ldotp$$This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
- An altitude of 6.0 km corresponds to y = 1.0 x 10 3 m in the coordinate system we are using. The other initial conditions are y 0 = 0, and v 0 = 200.0 m/s. We have, from Equation \ref{3.17}, $$v^{2} = (200.0\; m/s)^{2} - 2(9.8\; m/s^{2})(1.0 \times 10^{3}\; m) \Rightarrow v = \pm 142.8\; m/s \ldotp$$
Significance
We have both a positive and negative solution in (b). Since our coordinate system has the positive direction upward, the +142.8 m/s corresponds to a positive upward velocity at 6000 m during the upward leg of the trajectory of the booster. The value v = −142.8 m/s corresponds to the velocity at 6000 m on the downward leg. This example is also important in that an object is given an initial velocity at the origin of our coordinate system, but the origin is at an altitude above the surface of Earth, which must be taken into account when forming the solution.
Visit this site to learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y = bx) to see how they add to generate the polynomial curve.


2.7: Plausibility checks
A plausibility check is a calculation or validation procedure that is performed to verify if the obtained solution to a certain problem is correct.
The main takeaways from this section are the following:
- If your answer passes all plausibility checks, then the answer is probably correct.
- If your answer does not pass a plausibility check, then the answer is probably wrong and the failed check will likely provide you information where the mistake is, such that you can fix it.
- Therefore, if you learn properly to perform plausibility checks, you will not need a solution anymore to check if your answer is correct.
- Note that you can also perform a plausibility check at every intermediate step of the solution. This helps to find the step with the mistake.
- For professional engineers there is no solution manual. To avoid making mistakes, which can have a huge impact, they always need to check the plausibility of their calculations, to be close to \(100 \%\) sure of their answer.
- Therefore checking the correctness of answers (without solution manual) is one of the most important skills an engineer should acquire.
- For this reason we do not give students answers to all problems, since if one uses a solution manual to verify the correctness of answers, one never learns how to properly perform plausibility checks.
- If you, despite the plausibility checks, do not manage to solve a problem, always ask explanation from your professor, teacher or colleagues.
Here are some examples of questions you can ask yourself to check the plausibility of an answer:
- Does the quantity in the answer have the right units?
- Are units on both sides of the equation equal? (Sec. 2.2)
- Is the sign of the answer correct when comparing it to the question? E.g. in a problem where a ball is falling downward due to gravity, and the \(y\) axis points upward, is \(v_{y}\) negative?
- Does the value of the answer make sense? (not more than 10 times smaller or larger than expected?). E.g. in a problem where a mass is accelerating due to gravity over a surface, its acceleration can never exceed the free fall value \(g\).
- When you substitute the answer in the equations used, do you get back the given values? E.g. if you have found the solution \(x(t)=c_{1} t^{2}\), from the initially given acceleration, check by twice differentiating that \(a(t)=2 c_{1}\).
- When projecting forces on axes, does the length of the vector correspond to that of the components? E.g. \(v_{x}^{2}+v_{y}^{2}+v_{z}^{2}=|\overrightarrow{\boldsymbol{v}}|^{2}\).
- Does the net force vector \(\sum \overrightarrow{\boldsymbol{F}}\) point in the same direction as the acceleration \(\overrightarrow{\boldsymbol{a}}\) ?
- Do the chosen axes form a right-handed coordinate system?
- Does \(\sum \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}}\) and the other laws of Newton and Euler hold if you substitute the final answer back into them?
- Can methods of work and energy be used to double check the solution?
- Are all given constraint equations satisfied if you substitute the final answer into them?
- Is there any other information given that can be verified with the answer?
Of course there are many more plausibility checks that can be performed, which really depend on the type of problem under consideration.


6.2: Western Europe and Byzantium circa 500-1000 CE
December 25, 800 CE, the cavernous interior of St. Peter’s Church smelled faintly of incense and the open space of the nave was packed with the people of Rome. At the eastern end of the church, King Charles of the Franks knelt before the pope. The Frankish king wore the dress of a Roman patrician: a tunic of multi-colored silk, embroidered trousers, and a richly embroidered cloak clasped with a golden brooch at his shoulder. As King Charles knelt, the pope placed a golden crown, set with pearls and precious stones of blue, green, and red, on the king’s head. He stood to his full height of six feet and the people gathered in the church cried out, “Hail Charles, Emperor of Rome!” The inside of the church was filled with cheers. For the first time in three centuries, the city of Rome had an emperor.
Outside of the church, the city of Rome itself told a different story. The great circuit of walls built in the third century by the emperor Aurelian still stood as a mighty bulwark against attackers. Much of the land within those walls lay empty. Although churches of all sorts could be found, pigs, goats, and other livestock roamed through the open fields and streets. Where once the Roman forum had been a bustling market, filled with merchants from as far away as India, now the crumbling columns of long-abandoned temples looked out over a broad, grassy field where shepherds grazed their flocks.
The fountains that had once given drinking water to millions of inhabitants now went unused and choked with weeds. The once-great baths that had echoed with the lively conversation of thousands of bathers stood only as tumbled down piles of stone that served as quarries for the men and women who looked to repair their modest homes. The Coliseum was now honeycombed with houses built into the tunnels that had once admitted crowds to the games in the arena.
And yet within this city of ruins, a new Rome sprouted from the ruins of the old. Just outside the city walls and across the Tiber River, St. Peter’s Basilica rose as the symbol of Peter, prince of the Apostles. The golden-domed Pantheon still stood, now a church of the Triune God rather than a temple of the gods of the old world. And, indeed, all across Western Europe, a new order had arisen on the wreck of the Roman state. Although this new order shared the universal ideals of Rome, its claims rested upon the foundations of the Christian faith. How this post-Roman world had come about is the subject to which we shall turn.


8: Water and Minerals
Drinking water, also known as potable water or improved drinking water, is water safe enough for drinking and food preparation. Globally, in 2012, 89% of people had access to water suitable for drinking. Nearly 4 billion had access to tap water while another 2.3 billion had access to wells or public taps. 1.8 billion people still use an unsafe drinking water source which may be contaminated by feces. This can result in infectious diarrhea such as cholera and typhoid among others. We will also discuss the category of essential nutrients called Minerals, also known as Dietary elements or mineral nutrients. Minerals are naturally occurring chemical elements required by living organisms for survival. Many of the minerals we need are plentiful in a typical scoop of dirt!
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- 8.1: Water
- Drinking water, also known as potable water or improved drinking water, is water safe enough for drinking and food preparation. Globally, in 2012, 89% of people had access to water suitable for drinking. Nearly 4 billion had access to tap water while another 2.3 billion had access to wells or public taps. 1.8 billion people still use an unsafe drinking water source which may be contaminated by feces.This can result in infectious diarrhea such as cholera and typhoid among others.
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- 8.2: Minerals - Basic Concepts
- The latest National Health and Nutrition Examination Survey, covering the period from 2005 to 2008, reports that about 50 percent of Americans consume sugary drinks daily. Excess consumption of sugary soft drinks have been scientifically proven to increase the risk for dental caries, obesity, Type 2 diabetes, and cardiovascular disease. In addition to sugary soft drinks, beverages containing added sugars include fruit drinks, sports drinks, energy drinks and sweetened bottled waters.
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- 8.3: Overview of Fluid and Electrolyte Balance
- A human body is made up of mostly water. An adult consists of about 37 to 42 liters of water, or about eighty pounds. Fortunately, humans have compartmentalized tissues; otherwise we might just look like a water balloon! Newborns are approximately 70 percent water. Adult males typically are composed of about 60 percent water and females are about 55 percent water. (This gender difference reflects the differences in body-fat content, since body fat is practically water-free.
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- 8.4: Water’s Importance to Vitality
- Add all the ways you use water every day and you still will not come close to the countless uses water has in the human body. Of all the nutrients, water is the most critical as its absence proves lethal within a few days. Organisms have adapted numerous mechanisms for water conservation. Water uses in the human body can be loosely categorized into four basic functions: transportation vehicle, medium for chemical reactions, lubricant/shock absorber, and temperature regulator.
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- 8.5: Regulation of Water Balance
- As the food enters your stomach, gastric juice is secreted. As it enters the small intestine, pancreatic juice is secreted. Each of these fluids contains a great deal of water. How is that water replaced in these organs? What happens to the water now in the intestines? In a day, there is an exchange of about 10 liters of water among the body’s organs. The osmoregulation of this exchange involves complex communication between the brain, kidneys, and endocrine system.
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- 8.6: Major Minerals
- Similarly to vitamins, minerals are essential to human health and can be obtained in our diet from different types of food. Minerals are abundant in our everyday lives. From the soil in your front yard to the jewelry you wear on your body, we interact with minerals constantly. There are 20 essential minerals that must be consumed in our diets to remain healthy. The amount of each mineral found in our bodies vary greatly and therefore, so does consumption of those minerals.


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2: The Handbooks from Philosophers
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2.1: What is a Classical Education?
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In this essay, Montaigne begins by saying that he does not know how to educate children but then goes on to give a series of practical recommendations based on his philosophical positions on the nature of knowledge and teaching. His philosophical system is not entirely clear throughout the essay, but the reflection questions will ask you to investigate what might have been his underlying philosophical positions in traditional realms such as epistemology, ontology, and ethics.
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2.2: What Kind of Subjects are Worth Investigating?
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In this text, Descartes outlines several of his rules for rigorous thinking. The two principal questions that he seeks to answer are 1) what kind of things deserve the attention of rigorous and extended thinking and 2) on what topics can we obtain certainty? Descartes’ claims about what objects are worthy of attention and why they are worthy of attention stand in fairly stark contrast to the underlying mindset behind exploration-based curricula of the 21st century.
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2.3: How to Educate a Gentleman?
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In this section Locke lays out several of his thoughts on the best way to edcuate a young gentleman. He speaks primarily of how a tutor should act toward a young puil. Much of his attention is focused on how avoiding physical punishment and helping students develope their own motivation.
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2.4: How is the Desire to Learn Cultivated?
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In this text, Rousseau vigorously attempts to make the case for allowing nature to educate a child by working to help the child keep their curiosity. He also disagrees with Locke directly and indirectly. If you have not read the essay by Locke, it may be best to read it first.
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2.5: How Can Girls Receive an Equal Education?
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A Vindication of the Rights of Woman (1792), written Mary Wollstonecraft, is one of the earliest works of feminist philosophy. In it, Wollstonecraft responds to those educational and political theorists of the 18th century who believed that women should not receive a rational education.


7.12: Chronic Lower Respiratory Diseases
Also referred to as chronic obstructive pulmonary diseases (COPD), this category includes emphysema, chronic bronchitis, and other chronic lower respiratory diseases (except asthma). Although this is currently the 6th leading cause of death for Americans behind heart disease, stroke, cancer, COVID-19 and accidents, it still is a significant group of diseases that contributes to both disability and mortality. These diseases affect the lungs and bronchi; causing inflammation, reducing elasticity and oxygen exchange within the air sacs (alveoli), and creating more mucus and phlegm. This causes coughing, wheezing, and trouble breathing - limiting the amount of activity that the person can do. COPD is progressive, meaning it gets worse over time (National Heart, Lung, and Blood Institute, 2023).


10.10: Introduction to Factoring Polynomials
- Find the greatest common factor of two or more expressions
- Factor the greatest common factor from a polynomial
Before you get started, take this readiness quiz.
- Factor 56 into primes. If you missed this problem, review Example 2.9.1 .
- Multiply: −3(6a + 11). If you missed this problem, review Example 7.4.9 .
- Multiply: 4x 2 (x 2 + 3x − 1). If you missed this problem, review Example 10.4.5 .
Find the Greatest Common Factor of Two or More Expressions
Earlier we multiplied factors together to get a product. Now, we will be reversing this process; we will start with a product and then break it down into its factors. Splitting a product into factors is called factoring.
In The Language of Algebra we factored numbers to find the least common multiple (LCM) of two or more numbers. Now we will factor expressions and find the greatest common factor of two or more expressions. The method we use is similar to what we used to find the LCM.
The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions.
First we will find the greatest common factor of two numbers
Find the greatest common factor of 24 and 36.
Solution
| Step 1 : Factor each coefficient into primes. Write all variables with exponents in expanded form. | Factor 24 and 36. | |
| Step 2 : List all factors—matching common factors in a column. | ||
| In each column, circle the common factors. | Circle the 2, 2, and 3 that are shared by both numbers. | |
| Step 3 : Bring down the common factors that all expressions share. | Bring down the 2, 2, 3 and then multiply. | |
| Step 4 : Multiply the factors. | The GCF of 24 and 36 is 12. |
Notice that since the GCF is a factor of both numbers, 24 and 36 can be written as multiples of 12.
\[\begin{split} 24 &= 12 \cdot 2 \\ 36 &= 12 \cdot 3 \end{split}\]
Find the greatest common factor: 54, 36.
- Answer
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18
Find the greatest common factor: 48, 80.
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16
In the previous example, we found the greatest common factor of constants. The greatest common factor of an algebraic expression can contain variables raised to powers along with coefficients. We summarize the steps we use to find the greatest common factor.
Step 1. Factor each coefficient into primes. Write all variables with exponents in expanded form.
Step 2. List all factors—matching common factors in a column. In each column, circle the common factors.
Step 3. Bring down the common factors that all expressions share.
Step 4. Multiply the factors.
Find the greatest common factor of 5x and 15.
Solution
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Factor each number into primes. Circle the common factors in each column. Bring down the common factors. |
The GCF of 5x and 15 is 5.
Find the greatest common factor: 7y, 14.
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7
Find the greatest common factor: 22, 11m.
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11
In the examples so far, the greatest common factor was a constant. In the next two examples we will get variables in the greatest common factor.
Find the greatest common factor of 12x 2 and 18x 3 .
Solution
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Factor each coefficient into primes and write the variables with exponents in expanded form. Circle the common factors in each column. Bring down the common factors. Multiply the factors. |
The GCF of 12x 2 and 18x 3 is 6x 2 .
Find the greatest common factor: 16x 2 , 24x 3 .
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\(8x^2\)
Find the greatest common factor: 27y 3 , 18y 4 .
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\(9y^3\)
Find the greatest common factor of 14x 3 , 8x 2 , 10x.
Solution
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Factor each coefficient into primes and write the variables with exponents in expanded form. Circle the common factors in each column. Bring down the common factors. Multiply the factors. |
The GCF of 14x 3 and 8x 2 , and 10x is 2x.
Find the greatest common factor: 21x 3 , 9x 2 , 15x.
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3x
Find the greatest common factor: 25m 4 , 35m 3 , 20m 2 .
- Answer
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\(5m^2\)
Factor the Greatest Common Factor from a Polynomial
Just like in arithmetic, where it is sometimes useful to represent a number in factored form (for example, 12 as 2 • 6 or 3 • 4), in algebra it can be useful to represent a polynomial in factored form. One way to do this is by finding the greatest common factor of all the terms. Remember that you can multiply a polynomial by a monomial as follows:
\[\begin{split} 2(x &+ 7) \quad factors \\ 2 \cdot x &+ 2 \cdot 7 \\ 2x &+ 14 \quad product \end{split}\]
Here, we will start with a product, like 2x + 14, and end with its factors, 2(x + 7). To do this we apply the Distributive Property “in reverse”.
If a, b, c are real numbers, then a(b + c) = ab + ac and ab + ac = a(b + c).
The form on the left is used to multiply. The form on the right is used to factor.
So how do we use the Distributive Property to factor a polynomial? We find the GCF of all the terms and write the polynomial as a product!
Factor: 2x + 14.
Solution
| Step 1 : Find the GCF of all the terms of the polynomial. | Find the GCF of 2x and 14. | |
| Step 2 : Rewrite each term as a product using the GCF. | Rewrite 2x and 14 as products of their GCF, 2.$$\begin{split} 2x &= 2 \cdot x \\ 14 &= 2 \cdot 7 \end{split}$$ | $$\begin{split} 2x &+ 14 \\ \textcolor{red}{2} \cdot x &+ \textcolor{red}{2} \cdot 7 \end{split}$$ |
| Step 3 : Use the Distributive Property 'in reverse' to factor the expression. | 2(x + 7) | |
| Step 4 : Check by multiplying the factors. | Check. | $$\begin{split} 2(x &+ 7) \\ 2 \cdot x &+ 2 \cdot 7 \\ 2x &+ 14\; \checkmark \end{split}$$ |
Factor: 4x + 12.
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4(x + 3)
Factor: 6a + 24.
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6(a + 4)
Notice that in Example 10.84, we used the word factor as both a noun and a verb:
| Noun | 7 is a factor of 14 |
| Verb | Factor 2 from 2x + 14 |
Step 1. Find the GCF of all the terms of the polynomial.
Step 2. Rewrite each term as a product using the GCF.
Step 3. Use the Distributive Property ‘in reverse’ to factor the expression.
Step 4. Check by multiplying the factors.
Factor: 3a + 3.
Solution
| Rewrite each term as a product using the GCF. | $$\textcolor{red}{3} \cdot a + \textcolor{red}{3} \cdot 1$$ |
| Use the Distributive Property 'in reverse' to factor the GCF. | $$3(a+1)$$ |
| Check by multiplying the factors to get the original polynomial. | $$\begin{split} 3(a &+ 1) \\ 3 \cdot a &= 3 \cdot 1 \\ 3a &+ 3\; \checkmark \end{split}$$ |
Factor: 9a + 9.
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9(a + 1)
Factor: 11x + 11.
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11(x + 1)
The expressions in the next example have several factors in common. Remember to write the GCF as the product of all the common factors.
Factor: 12x − 60.
| Rewrite each term as a product using the GCF. | $$\textcolor{red}{12} \cdot x - \textcolor{red}{12} \cdot 5$$ |
| Factor the GCF. | $$12(x-5)$$ |
| Check by multiplying the factors. | $$\begin{split} 12(x &- 5) \\ 12 \cdot x &- 12 \cdot 5 \\ 12x &- 60\; \checkmark \end{split}$$ |
Factor: 11x − 44.
- Answer
-
11(x - 4)
Factor: 13y − 52.
- Answer
-
13(y - 4)
Now we’ll factor the greatest common factor from a trinomial. We start by finding the GCF of all three terms.
Factor: 3y 2 + 6y + 9.
Solution
| Rewrite each term as a product using the GCF. | $$\textcolor{red}{3} \cdot y^{2} + \textcolor{red}{3} \cdot 2y + \textcolor{red}{3} \cdot 3$$ |
| Factor the GCF. | $$3(y^{2} + 2y + 3)$$ |
| Check by multiplying. | $$\begin{split} 3(y^{2} &+ 2y + 3) \\ 3 \cdot y^{2} &+ 3 \cdot 2y + 3 \cdot 3 \\ 3y^{2} &+ 6y + 9\; \checkmark \end{split}$$ |
Factor: 4y 2 + 8y + 12.
- Answer
-
\(4\left(y^{2}+2 y+3\right) \)
Factor: 6x 2 + 42x − 12.
- Answer
-
\( 6\left(x^{2}+7 x-2\right) \)
In the next example, we factor a variable from a binomial.
Factor: 6x 2 + 5x.
Solution
| Find the GCF of 6x 2 and 5x and the math that goes with it. | |
| Rewrite each term as a product. | $$\textcolor{red}{x} \cdot 6x + \textcolor{red}{x} \cdot 5$$ |
| Factor the GCF. | $$x(6x + 5)$$ |
| Check by multiplying. | $$\begin{split} x(6x &+ 5) \\ x \cdot 6x &+ x \cdot 5 \\ 6x^{2} &+ 5x\; \checkmark \end{split}$$ |
Factor: 9x 2 + 7x.
- Answer
-
\( x(9x+7) \)
Factor: 5a 2 − 12a.
- Answer
-
a(5a - 12)
When there are several common factors, as we’ll see in the next two examples, good organization and neat work helps!
Factor: 4x 3 − 20x 2 .
Solution
| Rewrite each term. | $$\textcolor{red}{4x^{2}} \cdot x - \textcolor{red}{4x^{2}} \cdot 5$$ |
| Factor the GCF. | $$4x^{2} (x-5)$$ |
| Check. | $$\begin{split} 4x^{2} (x &- 5) \\ 4x^{2} \cdot x &- 4x^{2} \cdot 5 \\ 4x^{3} &- 20x^{2}\; \checkmark \end{split}$$ |
Factor: 2x 3 + 12x 2 .
- Answer
-
\( 2 x^{2}(x+6) \)
Factor: 6y 3 − 15y 2 .
- Answer
-
\( 3 y^{2}(2 y-5) \)
Factor: 21y 2 + 35y.
Solution
| Find the GCF of 21y 2 and 35y. | |
| Rewrite each term. | $$\textcolor{red}{7y} \cdot 3y + \textcolor{red}{7y} \cdot 5$$ |
| Factor the GCF. | $$7y(3y + 5)$$ |
Factor: 18y 2 + 63y.
- Answer
-
9y(2y + 7)
Factor: 32k 2 + 56k.
- Answer
-
8k(4k + 7)
Factor: 14x 3 + 8x 2 − 10x.
Solution
Previously, we found the GCF of 14x 3 , 8x 2 , and 10x to be 2x.
| Rewrite each term using the GCF, 2x. | $$\textcolor{red}{2x} \cdot 7x^{2} + \textcolor{red}{2x} \cdot 4x - \textcolor{red}{2x} \cdot 5$$ |
| Factor the GCF. | $$2x(7x^{2} + 4x - 5)$$ |
| Check. | $$\begin{split} 2x(7x^{2} &+ 4x - 5) \\ 2x \cdot 7x^{2} &+ 2x \cdot 4x - 2x \cdot 5 \\ 14x^{3} &+ 8x^{2} - 10x\; \checkmark \end{split}$$ |
Factor: 18y 3 − 6y 2 − 24y.
- Answer
-
\(6 y\left(3 y^{2}-y-4\right)\)
Factor: 16x 3 + 8x 2 − 12x.
- Answer
-
\(4 x\left(4 x^{2}+2 x-3\right)\)
When the leading coefficient, the coefficient of the first term, is negative, we factor the negative out as part of the GCF.
Factor: −9y − 27.
Solution
| When the leading coefficient is negative, the GCF will be negative. Ignoring the signs of the terms, we first find the GCF of 9y and 27 is 9. | |
| Since the expression −9y − 27 has a negative leading coefficient, we use −9 as the GCF. | |
| Rewrite each term using the GCF. | $$\textcolor{red}{-9} \cdot y + (\textcolor{red}{-9}) \cdot 3$$ |
| Factor the GCF. | $$-9(y+3)$$ |
| Check. | $$\begin{split} -9(y &+ 3) \\ -9 \cdot y &+ (-9) \cdot 3 \\ -9y &- 27\; \checkmark \end{split}$$ |
Factor: −5y − 35.
- Answer
-
-5(y + 7)
Factor: −16z − 56.
- Answer
-
-8(2z + 7)
Pay close attention to the signs of the terms in the next example.
Factor: −4a 2 + 16a.
Solution
| The leading coefficient is negative, so the GCF will be negative. | |
| Since the leading coefficient is negative, the GCF is negative, −4a. | |
| Rewrite each term. | $$\textcolor{red}{-4a} \cdot a - (\textcolor{red}{-4a}) \cdot 4$$ |
| Factor the GCF. | $$-4a(a-4)$$ |
| Check on your own by multiplying. |
Factor: −7a 2 + 21a.
- Answer
-
-7a(a - 3)
Factor: −6x 2 + x.
- Answer
-
-x(6x - 1)
Factor GCF
Factor a Binomial
Identify GCF
Practice Makes Perfect
Find the Greatest Common Factor of Two or More Expressions
In the following exercises, find the greatest common factor.
- 40, 56
- 45, 75
- 72, 162
- 150, 275
- 3x, 12
- 4y, 28
- 10a, 50
- 5b, 30
- 16y, 24y 2
- 9x, 15x 2
- 18m 3 , 36m 2
- 12p 4 , 48p 3
- 10x, 25x 2 , 15x 3
- 18a, 6a 2 , 22a 3
- 24u, 6u 2 , 30u 3
- 40y, 10y 2 , 90y 3
- 15a 4 , 9a 5 , 21a 6
- 35x 3 , 10x 4 , 5x 5
- 27y 2 , 45y 3 , 9y 4
- 14b 2 , 35b 3 , 63b 4
Factor the Greatest Common Factor from a Polynomial
In the following exercises, factor the greatest common factor from each polynomial.
- 2x + 8
- 5y + 15
- 3a − 24
- 4b − 20
- 9y − 9
- 7x − 7
- 5m 2 + 20m + 35
- 3n 2 + 21n + 12
- 8p 2 + 32p + 48
- 6q 2 + 30q + 42
- 8q 2 + 15q
- 9c 2 + 22c
- 13k 2 + 5k
- 17x 2 + 7x
- 5c 2 + 9c
- 4q 2 + 7q
- 5p 2 + 25p
- 3r 2 + 27r
- 24q 2 − 12q
- 30u 2 − 10u
- yz + 4z
- ab + 8b
- 60x − 6x 3
- 55y − 11y 4
- 48r 4 − 12r 3
- 45c 3 − 15c 2
- 4a 3 − 4ab 2
- 6c 3 − 6cd 2
- 30u 3 + 80u 2
- 48x 3 + 72x 2
- 120y 6 + 48y 4
- 144a 6 + 90a 3
- 4q 2 + 24q + 28
- 10y 2 + 50y + 40
- 15z 2 − 30z − 90
- 12u 2 − 36u − 108
- 3a 4 − 24a 3 + 18a 2
- 5p 4 − 20p 3 − 15p 2
- 11x 6 + 44x 5 − 121x 4
- 8c 5 + 40c 4 − 56c 3
- −3n − 24
- −7p − 84
- −15a 2 − 40a
- −18b 2 − 66b
- −10y 3 + 60y 2
- −8a 3 + 32a 2
- −4u 5 + 56u 3
- −9b 5 + 63b 3
Everyday Math
- Revenue A manufacturer of microwave ovens has found that the revenue received from selling microwaves a cost of p dollars each is given by the polynomial −5p 2 + 150p. Factor the greatest common factor from this polynomial.
- Height of a baseball The height of a baseball hit with velocity 80 feet/second at 4 feet above ground level is −16t 2 + 80t + 4, with t = the number of seconds since it was hit. Factor the greatest common factor from this polynomial.
Writing Exercises
- The greatest common factor of 36 and 60 is 12. Explain what this means.
- What is the GCF of y 4 , y 5 , and y 10 ? Write a general rule that tells how to find the GCF of y a , y b , and y c .
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?
Contributors and Attributions
-
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


22: Carbonyl Alpha-Substitution Reactions
When you have completed Chapter 22, you should be able to
- fulfill all of the detailed objectives listed under each individual section.
- design multi‑step syntheses in which the reactions introduced in this unit are used in conjunction with any of the reactions described in previous units.
- solve road‑map problems that require a knowledge of carbonyl alpha‑substitution reactions.
- define, and use in context, the key terms introduced.
Alpha‑substitution reactions are the third major type of reaction that you will study in your investigation of the chemistry of carbonyl compounds. As you will see, these reactions proceed through the formation of the enol form of the carbonyl compound.
After a brief review of keto‑enol tautomerism, we begin our discussion of alpha‑substitution reactions by looking at the methods used to introduce a halogen atom into a carbonyl‑containing compound—reactions in which the enol form of the carbonyl compound is directly involved. After discussing the factors that influence the formation and stability of enolate anions, we examine some halogenation reactions in which an enolate ion is formed as an intermediate.
The chapter concludes with a study of the alkylation of enolate anions. These reactions are of tremendous use in organic syntheses, as they provide a method of forming new carbon‑carbon bonds, and hence facilitate the laboratory preparation of increasingly complex compounds.


14.1: ADAPT Account Types
Adapt currently had three account types: (1) instructors, (2) graders, and (3) students. Below outlines how these accounts are created.
Instructor Accounts
Main Features:
- Make courses and assignments
- Add assessments to assignments
- Set and edit course preferences
- Grade assignments
- Add Graders and give access to assignments
How to access: Instructor account access code. Contact the LibreTexts development team for access<EMAIL_ADDRESS>).
- Go to https://adapt.libretexts.org/
- Click "Register" in the top right corner
- Select "Instructor" from the drop-down menu
-
Fill out your name, email, your unique access code, create password and choose your time zone from the drop-down menu
Note: Please contact Delmar Larsen to obtain your instructor account access code. - Click "Submit"
Grader Accounts
Grader accounts are designed to facilitate streamlined open-ended grading. These accounts are designed to handle the grading of questions that autograding technologies are poorly crafted to address and require human review. Main features of this account is the abiility to
- grade assignments they are given access to
- provide students with feedback
- Answer students questions when contacted via ADAPT
Graders cannot make or edit assignments. Only instructor accounts have that ability.
How to access: Direct invitation from the instructor via ADAPT
Instructors Instructions
- Open Course Preferences
- Select "Graders" in the menu on the left
- Click "Invite Grader"
- Enter Grader’s email address
- Choose sections to give Grader access to or click "select all" to give access to all sections
- Click "Submit". Invited Grader will receive an email with the access code to join.
Graders Instructions
- Open your invitation sent via email and copy access code
- Go to https://adapt.libretexts.org
- Click "Register" in the top right corner
-
Select "Grader" from the drop-down menu
Note: if you already have a student or instructor account you will need to make a separate account to be a grader using another email - Fill out your information to make your account
- After your account is created you will be brought to My Courses page
- Click "Add Course"
- Paste the Access Code you received via email
- Click Submit
Grader Permissions
After the grader has made their account and registered, you can set permissions for which assignments they can access.
Under Course Properties select “Grader Permissions”. You can give grading access to individual assignments and graders, or you can select "give all graders access" to give all graders access to all assignments
Student Account
Main Features:
- Join Courses
- View Assignments
- Submit Answers
- View grader's feedback
- Contact Grader
How to access: Access code distrbiuted the instructor.
Instructors Instructions
- Open Course Preferences
- Select "Sections" in the menu on the left
-
Copy access code.
Note: each section will have a different access code - Give this access code to the students
Students Instructions
- Make sure you have an access code from your instructor. Each course and each section will have a different access code.
- Go to https://adapt.libretexts.org
- Click Register in the top right corner
- Select "Student" from the drop-down menu
- Fill out your information to make your account
- After your account is created you will be brought to My Courses page.
-
Click "Enroll in Course"
- Paste the Access Code your instructor has provided
- Click "Submit"


13.9.9: Use Direct and Inverse Variation
-
- Last updated
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Learning Objectives
By the end of this section, you will be able to:
- Solve direct variation problems
- Solve inverse variation problems
Note
Before you get started, take this readiness quiz.
If you miss a problem, go back to the section listed and review the material.
-
Find the multiplicative inverse of −8
If you missed this problem, review Exercise 1.10.13 . -
Solve for n: 45=20n
If you missed this problem, review Exercise 2.2.1 . -
Evaluate \(5x^2\) when x=10
If you missed this problem, review Exercise 1.3.25 .
When two quantities are related by a proportion, we say they are proportional to each other. Another way to express this relation is to talk about the variation of the two quantities. We will discuss direct variation and inverse variation in this section.
Solve Direct Variation Problems
Lindsay gets paid $15 per hour at her job. If we let s be her salary and h be the number of hours she has worked, we could model this situation with the equation
s=15h
Lindsay’s salary is the product of a constant, 15, and the number of hours she works. We say that Lindsay’s salary varies directly with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.
Definition: DIRECT VARIATION
For any two variables x and y , y varies directly with x if
y=kx, where \(n \ne 0\)
In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates x and y . Then we can use that equation to find values of y for other values of x .
How to Solve Direct Variation Problems
Example \(\PageIndex{1}\)
If y varies directly with x and y=20 when x=8, find the equation that relates x and y .
- Answer
-
Example \(\PageIndex{2}\)
If y varies directly as x and y=3, when x=10, find the equation that relates x and y .
- Answer
-
\(y=\frac{3}{10}x\)
Example \(\PageIndex{3}\)
If y varies directly as x and y=12 when x=4, find the equation that relates x and y .
- Answer
-
y=3x
We’ll list the steps below.
Definition: SOLVE DIRECT VARIATION PROBLEMS
- Write the formula for direct variation.
- Substitute the given values for the variables.
- Solve for the constant of variation.
- Write the equation that relates x and y.
Now we’ll solve a few applications of direct variation.
Example \(\PageIndex{4}\)
When Raoul runs on the treadmill at the gym, the number of calories, c , he burns varies directly with the number of minutes, m , he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.
- Write the equation that relates c and m .
- How many calories would he burn if he ran on the treadmill for 25 minutes?
- Answer
-
1.
The number of calories, c varies directly with the number of minutes, m, on the treadmill, and c=315 when m=18 Write the formula for direct variation. We will use c in place of y and mm in place of x. Substitute the given values for the variables. Solve for the constant of variation. Write the equation that relates c and m. Substitute in the constant of variation. 2.
Find c when m=25.
Write the equation that relates c and m. Substitute the given value for m. Simplify. Raoul would burn 437.5 calories if he used the treadmill for 25 minutes.
Example \(\PageIndex{5}\)
The number of calories, c , burned varies directly with the amount of time, t, spent exercising. Arnold burned 312 calories in 65 minutes exercising.
- Write the equation that relates c and t .
- How many calories would he burn if he exercises for 90 minutes?
- Answer
-
- c=4.8t
- 432 calories
Example \(\PageIndex{6}\)
The distance a moving body travels, d , varies directly with time, t , it moves. A train travels 100 miles in 2 hours
- Write the equation that relates d and t .
- How many miles would it travel in 5 hours?
- Answer
-
- d=50t
- 250 miles
In the previous example, the variables c and m were named in the problem. Usually that is not the case. We will have to name the variables in the next example as part of the solution, just like we do in most applied problems.
Example \(\PageIndex{7}\)
The number of gallons of gas Eunice’s car uses varies directly with the number of miles she drives. Last week she drove 469.8 miles and used 14.5 gallons of gas.
- Write the equation that relates the number of gallons of gas used to the number of miles driven.
- How many gallons of gas would Eunice’s car use if she drove 1000 miles?
- Answer
-
1.
The number of gallons of gas varies directly with the number of miles driven. First we will name the variables. Let g= number of gallons of gas.
m= number of miles driven.Write the formula for direct variation. We will use g in place of y and m in place of x. Substitute the given values for the variables Solve for the constant of variation. We will round to the nearest thousandth. Write the equation that relates g and m. Substitute in the constant of variation. 2.
Find g when m=1000. Write the equation that relates g and m. g=0.031m Substitute the given value for m. g=0.031(1000) Simplify. g=31 Eunice’s car would use 31 gallons of gas if she drove it 1,000 miles. Notice that in this example, the units on the constant of variation are gallons/mile. In everyday life, we usually talk about miles/gallon.
Example \(\PageIndex{8}\)
The distance that Brad travels varies directly with the time spent traveling. Brad travelled 660 miles in 12 hours,
- Write the equation that relates the number of miles travelled to the time.
- How many miles could Brad travel in 4 hours?
- Answer
-
- m=55h
- 220 miles
Example \(\PageIndex{9}\)
The weight of a liquid varies directly as its volume. A liquid that weighs 24 pounds has a volume of 4 gallons.
- Write the equation that relates the weight to the volume.
- If a liquid has volume 13 gallons, what is its weight?
- Answer
-
- w=6v
- 78 pounds
In some situations, one variable varies directly with the square of the other variable. When that happens, the equation of direct variation is \(y=kx^2\).
Example \(\PageIndex{10}\)
The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 4” will support a maximum load of 75 pounds.
- Write the equation that relates the maximum load to the cross-section.
- What is the maximum load that can be supported by a beam with diagonal 8”?
- Answer
-
1.
The maximum load varies directly with the square of the diagonal of the cross-section. Name the variables. Let L= maximum load.
c= the diagonal of the cross-sectionWrite the formula for direct variation, where y varies directly with the square of x. We will use L in place of y and c in place of x. Substitute the given values for the variables. Solve for the constant of variation. Write the equation that relates L and c. Substitute in the constant of variation. 2.
Find L when c=8. Write the equation that relates L and c. \(L=4.6875c^2\) Substitute the given value for c. \(L=4.6875(8)^2\) Simplify. L=300 A beam with diagonal 8” could supporta maximum load of 300 pounds.
Example \(\PageIndex{11}\)
The distance an object falls is directly proportional to the square of the time it falls. A ball falls 144 feet in 3 seconds.
- Write the equation that relates the distance to the time.
- How far will an object fall in 4 seconds?
- Answer
-
- \(d=16t^2\)
- 256 feet
Example \(\PageIndex{12}\)
The area of a circle varies directly as the square of the radius. A circular pizza with a radius of 6 inches has an area of 113.04 square inches.
- Write the equation that relates the area to the radius.
- What is the area of a pizza with a radius of 9 inches?
- Answer
-
- \(A=3.14r^2\)
- 254.34 square inches
Solve Inverse Variation Problems
Many applications involve two variable that vary inversely . As one variable increases, the other decreases. The equation that relates them is \(y=\frac{k}{x}\).
Definition: INVERSE VARIATION
For any two variables x and y , y varies inversely with x if
y=\(\frac{k}{x}\),where \(k \ne 0\)
The word ‘inverse’ in inverse variation refers to the multiplicative inverse. The multiplicative inverse of x is \(\frac{1}{x}\).
We solve inverse variation problems in the same way we solved direct variation problems. Only the general form of the equation has changed. We will copy the procedure box here and just change ‘direct’ to ‘inverse’.
Definition: SOLVE INVERSE VARIATION PROBLEMS
- Write the formula for inverse variation.
- Substitute the given values for the variables.
- Solve for the constant of variation.
- Write the equation that relates x and y.
Example \(\PageIndex{13}\)
If y varies inversely with x and y=20 when x=8 x and y .
- Answer
-
Write the formula for inverse variation. Substitute the given values for the variables. Solve for the constant of variation. Write the equation that relates x and y. Substitute in the constant of variation.
Example \(\PageIndex{14}\)
If p varies inversely with q and p=30 when q=12 find the equation that relates p and q.
- Answer
-
\(p=\frac{360}{q}\)
Example \(\PageIndex{15}\)
If y varies inversely with x and y=8 when x=2 find the equation that relates x and y.
- Answer
-
\(y=\frac{16}{x}\)
Example \(\PageIndex{16}\)
The fuel consumption (mpg) of a car varies inversely with its weight. A car that weighs 3100 pounds gets 26 mpg on the highway.
- Write the equation of variation.
- What would be the fuel consumption of a car that weighs 4030 pounds?
- Answer
-
1.
The fuel consumption varies inversely with the weight. First we will name the variables. Let f= fuel consumption.
w= weight.Write the formula for inverse variation. We will use f in place of y and w in place of x. Substitute the given values for the variables. Solve for the constant of variation. Write the equation that relates f and w. Substitute in the constant of variation. 2.
Find f when w=4030. Write the equation that relates f and w. Substitute the given value for w. \(f=\frac{80,600}{4030}\) Simplify. f=20 A car that weighs 4030 pounds would have fuel consumption of 20 mpg.
Example \(\PageIndex{17}\)
A car’s value varies inversely with its age. Elena bought a two-year-old car for $20,000.
- Write the equation of variation.
- What will be the value of Elena’s car when it is 5 years old?
- Answer
-
- \(v=\frac{40,000}{a}\)
- $8,000
Example \(\PageIndex{18}\)
The time required to empty a pool varies inversely as the rate of pumping. It took Lucy 2.5 hours to empty her pool using a pump that was rated at 400 gpm (gallons per minute).
- Write the equation of variation.
- How long will it take her to empty the pool using a pump rated at 500 gpm?
- Answer
-
- \(t=\frac{1000}{r}\)
- 2 hours
Example \(\PageIndex{19}\)
The frequency of a guitar string varies inversely with its length. A 26” long string has a frequency of 440 vibrations per second.
- Write the equation of variation.
- How many vibrations per second will there be if the string’s length is reduced to 20” by putting a finger on a fret?
- Answer
-
1.
The frequency varies inversely with the length. Name the variables. Let f= frequency.
L=length.Write the formula for inverse variation. We will use f in place of y and L in place of x Substitute the given values for the variables. Solve for the constant of variation. Write the equation that relates f and L. Substitute in the constant of variation. 2.
Find f when L=20. Write the equation that relates f and L. \(f=\frac{11,440}{L}\) Substitute the given value for L. \(f=\frac{11,440}{20}\) Simplify. f=572 A 20” guitar string has frequency 572 vibrations per second.
Example \(\PageIndex{20}\)
The number of hours it takes for ice to melt varies inversely with the air temperature. Suppose a block of ice melts in 2 hours when the temperature is 65 degrees.
- Write the equation of variation.
- How many hours would it take for the same block of ice to melt if the temperature was 78 degrees?
- Answer
-
- \(h=\frac{130}{t}\)
- \(1\frac{2}{3}\) hours
Example \(\PageIndex{21}\)
The force needed to break a board varies inversely with its length. Richard uses 24 pounds of pressure to break a 2-foot long board.
- Write the equation of variation.
- How many pounds of pressure is needed to break a 5-foot long board?
- Answer
-
- \(F=\frac{48}{L}\)
- 9.6 pounds


7.5: Preparing for and Responding to Emergencies
Organizations should be prepared to respond to emergency situations. These situations may arise from a multitude of sources, including natural and man-made hazards. Large emergencies typically involve multiple infrastructure systems and large populations, so co-ordination with emergency services and managers is critical.
A few elements of preparation can be noted. First, a communications strategy should be in place. This strategy should include a means of communicating with infrastructure users and response personnel. For example, a broadcast strategy of emergency communications via email and text messaging should be in place. The broadcast mechanism should be regularly tested.
The second component of emergency preparation is a decision making strategy. Who is empowered to make decisions in the event of an emergency? Successors should also be identified in case the primary decision-maker is not available.
In an emergency, knowledge of available resources may be critical. Where and what resources and emergency supplies are positioned can be important information to aid effective decision making.
As noted earlier, pre-positioning personnel, equipment, and supplies can be a very effective preparation strategy. Temporary bridges, transformers, power lines and the like can be all readied for rapid deployment.
Temporary facilities may also be needed for emergency operations. Shelters can be assembled for individuals made homeless by fires or other destruction. Dirt runways can be used for the delivery of emergency supplies by air. Temporary cellular telephone and Wi-Fi stations can be moved in and installed rapidly.
Response to emergencies generally requires immediate attention from all the members of the infrastructure management team. Innovation, coordination and rapid decision making are necessary elements of a successful response!


3.5: Contingency Tables and Probability Trees
Contingency Tables
A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.
Example \(\PageIndex{1}\)
Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:
| Speeding violation in the last year | No speeding violation in the last year | Total | |
|---|---|---|---|
| Uses cell phone while driving | 25 | 280 | 305 |
| Does not use cell phone while driving | 45 | 405 | 450 |
| Total | 70 | 685 | 755 |
The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.
Calculate the following probabilities using the table.
a. Find P (Driver is a cell phone user).
- Answer
-
a. \(\frac{\text { number of cell phone users }}{\text { total number in study }}=\frac{305}{755}\)
b. Find P (Driver had no violation in the last year).
- Answer
-
b. \(\frac{\text { number that had no violation }}{\text { total number in study }}=\frac{685}{755}\)
c. Find P (Driver had no violation in the last year \(\cap\) was a cell phone user).
- Answer
-
c. \(\frac{280}{755}\)
d. Find P (Driver is a cell phone user \(\cup\) driver had no violation in the last year).
- Answer
-
d. \(\left(\frac{305}{755}+\frac{685}{755}\right)-\frac{280}{755}=\frac{710}{755}\)
e. Find P (Driver is a cell phone user \(|\) driver had a violation in the last year).
- Answer
-
e. \(\frac{25}{70}\) (The sample space is reduced to the number of drivers who had a violation.)
f. Find P (Driver had no violation last year \(|\) driver was not a cell phone user)
- Answer
-
f. \(\frac{405}{450}\) (The sample space is reduced to the number of drivers who were not cell phone users.)
Exercise \(\PageIndex{1}\)
Table \(\PageIndex{2}\) shows the number of athletes who stretch before exercising and how many had injuries within the past year.
| Injury in last year | No injury in last year | Total | |
|---|---|---|---|
| Stretches | 55 | 295 | 350 |
| Does not stretch | 231 | 219 | 450 |
| Total | 286 | 514 | 800 |
Table \(\PageIndex{2}\)
- What is P (athlete stretches before exercising)?
- What is P (athlete stretches before exercising|no injury in the last year)?
Example \(\PageIndex{2}\)
Table \(\PageIndex{3}\) shows a random sample of 100 hikers and the areas of hiking they prefer.
| Sex | The coastline | Near lakes and streams | On mountain peaks | Total |
|---|---|---|---|---|
| Female | 18 | 16 | ___ | 45 |
| Male | ___ | ___ | 14 | 55 |
| Total | ___ | 41 | ___ | ___ |
Table \(\PageIndex{3}\) Hiking Area Preference
a. Complete the table.
- Answer
-
a.
Sex The coastline Near lakes and streams On mountain peaks Total Female 18 16 11 45 Male 16 25 14 55 Total 34 41 25 100 Table \(\PageIndex{4}\) Hiking Area Preference
b. Are the events "being female" and "preferring the coastline" independent events?
Let F = being female and let C = preferring the coastline.
- Find \(P(F\cap C)\).
- Find P (F) P (C)
Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.
- Answer
-
b.
1. \(P(F\cap C)=\frac{18}{100}\) = 0.18 - 2. P (F) P (C) = \(\left(\frac{45}{100}\right)\left(\frac{34}{100}\right)\) = (0.45)(0.34) = 0.153
-
\(P(F\cap C)\) ≠ P (F) P (C), so the events F and C are not independent.
c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams.
- What word tells you this is a conditional?
- Fill in the blanks and calculate the probability: P (___|___) = ___.
- Is the sample space for this problem all 100 hikers? If not, what is it?
- Answer
-
c.
1.The word 'given' tells you that this is a conditional.
2. P (M|L) = \(\frac{25}{41}\)
3.No, the sample space for this problem is the 41 hikers who prefer lakes and streams.
d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks.
- Find P (F).
- Find P (P).
- Find \(P(F\cap P)\).
- Find \(P(F\cup P)\).
- Answer
-
d.
- P (F) = \(\frac{45}{100}\)
- P (P) = \(\frac{25}{100}\)
- \(P(F\cap P)\)= \(\frac{11}{100}\)
- \(P(F\cup P)\)= \(\frac{45}{100}+\frac{25}{100}-\frac{11}{100}=\frac{59}{100}\)
Exercise \(\PageIndex{2}\)
Table \(\PageIndex{5}\) shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path.
| Gender | Lake path | Hilly path | Wooded path | Total |
|---|---|---|---|---|
| Female | 45 | 38 | 27 | 110 |
| Male | 26 | 52 | 12 | 90 |
| Total | 71 | 90 | 39 | 200 |
- Out of the males, what is the probability that the cyclist prefers a hilly path?
- Are the events “being male” and “preferring the hilly path” independent events?
Example \(\PageIndex{3}\)
Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 1/15 and the probability he is not caught is 4/15. If he goes out the second door, the probability he gets caught by Alissa is 1/12 and the probability he is not caught is 3/12. The probability that Alissa catches Muddy coming out of the third door is 1/6 and the probability she does not catch Muddy is 1/6. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 1/3.
| Caught or not | Door one | Door two | Door three | Total |
|---|---|---|---|---|
| Caught | \(\frac{1}{15}\) | \(\frac{1}{12}\) | \(\frac{1}{6}\) | ____ |
| Not caught | \(\frac{4}{15}\) | \(\frac{3}{12}\) | \(\frac{1}{6}\) | ____ |
| Total | ____ | ____ | ____ | 1 |
- The first entry \(\frac{1}{15}=\left(\frac{1}{5}\right)\left(\frac{1}{3}\right)\) is \(P(Door One\cap Caught)\)
- The entry \(\frac{4}{15}=\left(\frac{4}{5}\right)\left(\frac{1}{3}\right)\) is \(P(Door One\cap Not Caught)\)
Verify the remaining entries.
a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.
- Answer
-
a.
Caught or not Door one Door two Door three Total Caught \(\frac{1}{15}\) \(\frac{1}{12}\) \(\frac{1}{6}\) \(\frac{19}{60}\) Not caught \(\frac{4}{15}\) \(\frac{3}{12}\) \(\frac{1}{6}\) \(\frac{41}{60}\)
Total \(\frac{5}{15}\) \(\frac{4}{12}\) \(\frac{2}{6}\) 1 Table \(\PageIndex{7}\) Door Choice
b. What is the probability that Alissa does not catch Muddy?
- Answer
-
b. \(\frac{41}{60}\)
c. What is the probability that Muddy chooses Door One \(\cap\) Door Two given that Muddy is caught by Alissa?
- Answer
-
c. \(\frac{9}{19}\)
Example \(\PageIndex{4}\)
Table \(\PageIndex{8}\) contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.
| Year | Robbery | Burglary | Rape | Vehicle | Total |
|---|---|---|---|---|---|
| 2008 | 145.7 | 732.1 | 29.7 | 314.7 | |
| 2009 | 133.1 | 717.7 | 29.1 | 259.2 | |
| 2010 | 119.3 | 701 | 27.7 | 239.1 | |
| 2011 | 113.7 | 702.2 | 26.8 | 229.6 | |
| Total |
TOTAL each column and each row. Total data = 4,520.7
- Find \(P(2009\cap Robbery)\).
- Find \(P(2010\cap Burglary)\).
- Find \(P(2010\cup Burglary)\).
- Find P (2011|Rape).
- Find P (Vehicle|2008).
- Answer
-
- 0.0294
- 0.1551
- 0.7165
- 0.2365
- 0.2575
Exercise \(\PageIndex{3}\)
Table \(\PageIndex{9}\) relates the weights and heights of a group of individuals participating in an observational study.
| Weight/height | Tall | Medium | Short | Totals |
|---|---|---|---|---|
| Obese | 18 | 28 | 14 | |
| Normal | 20 | 51 | 28 | |
| Underweight | 12 | 25 | 9 | |
| Totals |
- Find the total for each row and column
- Find the probability that a randomly chosen individual from this group is Tall.
- Find the probability that a randomly chosen individual from this group is Obese and Tall.
- Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.
- Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
- Find the probability a randomly chosen individual from this group is Tall and Underweight.
- Are the events Obese and Tall independent?
Tree Diagrams
Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams can be used to visualize and solve conditional probabilities.
Tree Diagrams
A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.
Example \(\PageIndex{5}\)
In an urn, there are 11 balls.  Draw two balls, one at a time, with replacement . "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.
The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:
R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3
The other outcomes are similar.
 Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space .
a. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...
- Answer
-
a. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3
b. Using the tree diagram, calculate P (RR).
- Answer
-
b. P (RR) = \(\left(\frac{3}{11}\right)\left(\frac{3}{11}\right) = \frac{9}{121}\)
c. Using the tree diagram, calculate P (RB \(\cup\) BR).
- Answer
-
c. \(P(RB \cup BR)\) = \(\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)+\left(\frac{8}{11}\right)\left(\frac{3}{11}\right)=\frac{48}{121}\)
d. Using the tree diagram, calculate P (R on 1st draw \(\cap\) B on 2nd draw).
- Answer
-
d. P (R on 1st draw \(\cap\) B on 2nd draw) = \(\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)=\frac{24}{121}\)
e. Using the tree diagram, calculate P (R on 2nd draw|B on 1st draw).
- Answer
-
e. P (R on 2nd draw|B on 1st draw) = P (R on 2nd|B on 1st) = \(\frac{24}{88} = \frac{3}{11}\)
This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. \(\frac{24}{88} = \frac{3}{11}\).
f. Using the tree diagram, calculate P (BB).
- Answer
-
f. P (BB) = \(\frac{64}{121}\)
g. Using the tree diagram, calculate P (B on the 2nd draw|R on the first draw).
- Answer
-
g. P (B on 2nd draw|R on 1st draw) = \(\frac{8}{11}\)
There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then \(\frac{24}{33}\).
Exercise \(\PageIndex{4}\)
In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF).
Example \(\PageIndex{6}\)
An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, \(\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}\).
NOTE
If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement , so that on the second draw there are ten marbles left in the urn.
Calculate the following probabilities using the tree diagram.
a. P (RR) = ________
- Answer
-
a. P (RR) = \(\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}\)
b. Fill in the blanks:
\(P(RB\cup BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)\) + (___)(___) = \(\frac{48}{110}\)
- Answer
-
b. \(P(RB\cup BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)+\left(\frac{8}{11}\right)\left(\frac{3}{10}\right)=\frac{48}{110}\)
c. P (R on 2nd|B on 1st) =
- Answer
-
c. P (R on 2nd|B on 1st) = \(\frac{3}{10}\)
d. Fill in the blanks.
P (R on 1st \(\cap\) B on 2nd) = (___)(___) = \(\frac{24}{100}\)
- Answer
-
d. \(P(R \text{ on 1st }\cap B \text{ on 2nd}) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)=\frac{24}{110}\)
e. Find P (BB).
- Answer
-
e. P (BB) = \(\left(\frac{8}{11}\right)\left(\frac{7}{10}\right)\)
f. Find P (B on 2nd|R on 1st).
- Answer
-
f. Using the tree diagram, P (B on 2nd|R on 1st) = P (R|B) = \(\frac{8}{10}\).
If we are using probabilities, we can label the tree in the following general way.
- P (R|R) here means P (R on 2nd|R on 1st)
- P (B|R) here means P (B on 2nd|R on 1st)
- P (R|B) here means P (R on 2nd|B on 1st)
- P (B|B) here means P (B on 2nd|B on 1st)
Exercise \(\PageIndex{5}\)
In a standard deck, there are 52 cards. Twelve cards are face cards ( F ) and 40 cards are not face cards ( N ). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.
- Find \(P(FN\cup NF)\).
- Find P (N|F).
-
Find
P
(at most one face card).
Hint: "At most one face card" means zero or one face card. -
Find
P
(at least on face card).
Hint: "At least one face card" means one or two face cards.
Example \(\PageIndex{7}\)
A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.
-
What is the probability that both kittens are tabby?
\(a \cdot\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) b \cdot\left(\frac{4}{9}\right)\left(\frac{4}{9}\right) c \cdot\left(\frac{4}{9}\right)\left(\frac{3}{8}\right) d \cdot\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)\) -
What is the probability that one kitten of each coloring is selected?
a.\(\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)\) b.\(\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)\) c.\(\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{9}\right)\) d.\(\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)\) - What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
- What is the probability of choosing two kittens of the same color?
  What is the probability that one ball of each coloring is selected?


2.2.13: H.G. Wells (1866-1946)
The Island of Dr. Moreau British Realism Herbert George Wells, generally referred to as H.G. Wells, was a prolific 19th-century British writer best known for his science fiction novels. He is often referred to, in fact, as the father of science fiction. Born into a lower middle class family, after his father's shop failed and the family went bankrupt, Wells held a variety of jobs as an adolescent, including working as a teaching assistant, apprentice draper, and pharmacy clerk. He would later use these experiences in his novels as the basis for social satire. After eventually winning a scholarship to Imperial College, Wells trained as a scientist; he was particularly interested in Darwinian theory. Wells was also an avid Socialist and an active member of the Fabian Society, an organization that advocated a long-term approach to the eventual Socialist revolution. Throughout his life, Wells suffered from various physical ailments, including diabetes, and his life was further complicated by a series of romantic affairs and two failed marriages. Wells, who published 51 novels as well as dozens of stories, story collections, works of non-fiction, and essays, is also best known for his novels The Time Machine , War of the Worlds , The Invisible Man , and When the Sleeper Awakes . An early science fiction novel, The Island of Dr. Moreau (1896) dramatizes the practice of vivisection, the practice of performing operations on live animals for the purpose of scientific research. Vivisection was a controversial topic in fin de siècle England, with a number of organizations formed to fight the practice as cruel and unethical. As the novel examines the ethics of vivisection, it also illustrates the possibility that civilization was spiraling downward into an increasingly degenerate state. Although Wells was an educated man and a scientist, he appears to be warning of the dangers that unregulated science can pose to the larger community. Consider while reading:
- Although this novel is considered an early science fiction text, it can also be described as a work of Gothic horror. What elements of the Gothic genre do you see here?
- What does this novel have in common with Mary Shelley's Frankenstein ?
- What do you think of the ending of the novel? What does the ending suggest about the nature of humankind?
Written by Anita Turlington


3.1.15: Bertolt Brecht (1898-1956)
Mother Courage and Her Children German Modernism German playwright and filmmaker Bertolt Brecht was born into a middle-class family in Augsburg, Bavaria, in 1898. His mother, a devout Protestant, influenced Brecht's work, which often employs themes and motifs related to Christian theology. A lifelong committed Marxist, Brecht worked with a series of collaborators to make anti-fascist films and adapt classic works like Christopher Marlowe's Edward II for the German stage. However, he is best known for his plays and the conceptual dramatic structures that he described initially as "epic" and later as "dialectical" theatre. According to Brecht, this type of theatre was designed to engage the audience directly in a dialogue about current issues by employing audience interaction, images, documentary and commentary effects, and choruses. He rejected the notion of theatrical productions as escapist entertainment. Unlike other playwrights who intended to shock audience members or engage them viscerally, Brecht sought primarily to promote theatre as contemplative experience in which audience members would come to understand themselves better in the context of contemporary events. Brecht's best known plays are The Threepenny Opera (1928), Mother Courage and Her Children (1938), The Good Woman of Szechuan (1939), and The Caucasian Chalk Circle (1943). One of the greatest anti-war plays of the twentieth century, Mother Courage was written by Brecht in response to the Nazi invasion of Poland. To demonstrate the devastating effects of war, Brecht sets the action of the play during the Thirty Years' War, 1618-1648, which involved all of Europe. Over the course of 12 scenes, he illustrates the personal toll of the war on Mother Courage, a woman who attempts to make a living from the war but pays a heavy price when she loses all three of her children to the war from which she attempts to profit. Consider while reading:
- Generally critics say that Brecht distances the audience from the characters because he has a different purpose in mind than having the audience empathize with the characters. What is that purpose? What are the themes that emerge from the play?
- How do we view Mother Courage by the end of the play? Is she a good mother? Is she a noble character? Why or why not?
- Is this play a tragedy? Do you see any techniques that are comparable to the traditions of classical (Greek) tragedy in its structure?
Written by Anita Turlington


8.4: IDS
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Another device that can be placed on the network for security purposes is an intrusion detection system, or IDS. An IDS does not add any additional security; instead, it provides the functionality to identify if the network is being attacked. An IDS can be configured to watch for specific types of activities and then alert security personnel if that activity occurs. An IDS also can log various types of traffic on the network for analysis later. An IDS is an essential part of any good security setup.
Using firewalls and other security technologies, organizations can effectively protect many of their information resources by making them invisible to the outside world. But what if an employee working from home requires access to some of these resources? What if a consultant is hired who needs to do work on the internal corporate network from a remote location? In these cases, a virtual private network (VPN) is called for.
A VPN allows a user who is outside of a corporate network to take a detour around the firewall and access the internal network from the outside. Through a combination of software and security measures, this lets an organization allow limited access to its networks while at the same time ensuring overall security.


29.3: Photon Energies and the Electromagnetic Spectrum
Learning Objectives
By the end of this section, you will be able to:
- Explain the relationship between the energy of a photon in joules or electron volts and its wavelength or frequency.
- Calculate the number of photons per second emitted by a monochromatic source of specific wavelength and power.
Ionizing Radiation
A photon is a quantum of EM radiation. Its energy is given by \(E = hf\) and is related to the frequency \(f\) and wavelength \(\lambda\) of the radiation by
\[\begin{align*} E &= hf \\[4pt] &= \dfrac{hc}{\lambda} \quad \text{(energy of a photon)} \end{align*}\]
where \(E\) is the energy of a single photon and \(c\) is the speed of light. When working with small systems, energy in eV is often useful. Note that Planck’s constant in these units is
\[h = 4.14 \times 10^{-15} \, eV \cdot s.\]
Since many wavelengths are stated in nanometers (nm), it is also useful to know that
\[hc = 1240 \, eV \cdot nm.\]
These will make many calculations a little easier.
All EM radiation is composed of photons. Figure \(\PageIndex{1}\) shows various divisions of the EM spectrum plotted against wavelength, frequency, and photon energy. Previously in this book, photon characteristics were alluded to in the discussion of some of the characteristics of UV, x rays, and \(\gamma\) rays, the first of which start with frequencies just above violet in the visible spectrum. It was noted that these types of EM radiation have characteristics much different than visible light. We can now see that such properties arise because photon energy is larger at high frequencies.
Photons act as individual quanta and interact with individual electrons, atoms, molecules, and so on. The energy a photon carries is, thus, crucial to the effects it has. Table \(\PageIndex{1}\) lists representative submicroscopic energies in eV. When we compare photon energies from the EM spectrum in Figure \(\PageIndex{1}\) with energies in the table, we can see how effects vary with the type of EM radiation.
| Rotational energies of molecules | \(10^{-5} \, eV\) |
| Vibrational energies of molecules | 0.1 eV |
| Energy between outer electron shells in atoms | 1 eV |
| Binding energy of a weakly bound molecule | 1 eV |
| Energy of red light | 2 eV |
| Binding energy of a tightly bound molecule | 10 eV |
| Energy to ionize atom or molecule | 10 to 1000 eV |
Gamma rays , a form of nuclear and cosmic EM radiation, can have the highest frequencies and, hence, the highest photon energies in the EM spectrum. For example, a \(\gamma\)-ray photon with \(f = 10^{21} \, Hz\) has an energy \(E = hf = 6.63 \times 10^{-13} \, J = 4.14 \, MeV\). This is sufficient energy to ionize thousands of atoms and molecules, since only 10 to 1000 eV are needed per ionization. In fact, \(\gamma\)-rays are one type of ionizing radiation , as are x rays and UV, because they produce ionization in materials that absorb them. Because so much ionization can be produced, a single \(\gamma\)-ray photon can cause significant damage to biological tissue, killing cells or damaging their ability to properly reproduce. When cell reproduction is disrupted, the result can be cancer, one of the known effects of exposure to ionizing radiation. Since cancer cells are rapidly reproducing, they are exceptionally sensitive to the disruption produced by ionizing radiation. This means that ionizing radiation has positive uses in cancer treatment as well as risks in producing cancer.
High photon energy also enables \(\gamma\) rays to penetrate materials, since a collision with a single atom or molecule is unlikely to absorb all the \(\gamma\) ray’s energy. This can make \(\gamma\) rays useful as a probe, and they are sometimes used in medical imaging. x rays , as you can see in Figure \(\PageIndex{1}\) , overlap with the low-frequency end of the \(\gamma\) ray range. Since x rays have energies of keV and up, individual x-ray photons also can produce large amounts of ionization. At lower photon energies, x rays are not as penetrating as \(\gamma\) rays and are slightly less hazardous. X rays are ideal for medical imaging, their most common use, and a fact that was recognized immediately upon their discovery in 1895 by the German physicist W. C. Roentgen (1845–1923). (See Figure \(\PageIndex{2}\) .) Within one year of their discovery, x rays (for a time called Roentgen rays) were used for medical diagnostics. Roentgen received the 1901 Nobel Prize for the discovery of x rays.
CONNECTIONS: CONSERVATION OF ENERGY
Once again, we find that conservation of energy allows us to consider the initial and final forms that energy takes, without having to make detailed calculations of the intermediate steps. Example \(\PageIndex{1}\) is solved by considering only the initial and final forms of energy.
While \(\gamma\) rays originate in nuclear decay, x rays are produced by the process shown in Figure \(\PageIndex{3}\). Electrons ejected by thermal agitation from a hot filament in a vacuum tube are accelerated through a high voltage, gaining kinetic energy from the electrical potential energy. When they strike the anode, the electrons convert their kinetic energy to a variety of forms, including thermal energy. But since an accelerated charge radiates EM waves, and since the electrons act individually, photons are also produced. Some of these x-ray photons obtain the kinetic energy of the electron. The accelerated electrons originate at the cathode, so such a tube is called a cathode ray tube (CRT), and various versions of them are found in older TV and computer screens as well as in x-ray machines.
Example \(\PageIndex{1}\): X-ray Photon Energy and X-ray Tube Voltage
Find the maximum energy in eV of an x-ray photon produced by electrons accelerated through a potential difference of 50.0 kV in a CRT like the one in Figure \(\PageIndex{3}\).
Strategy
Electrons can give all of their kinetic energy to a single photon when they strike the anode of a CRT. (This is something like the photoelectric effect in reverse.) The kinetic energy of the electron comes from electrical potential energy. Thus we can simply equate the maximum photon energy to the electrical potential energy—that is, \(hf = qV\). (We do not have to calculate each step from beginning to end if we know that all of the starting energy \(qV\) is converted to the final form \(hf.\))
Solution
The maximum photon energy is \(hf = qV\), where \(q\) is the charge of the electron and \(V\) is the accelerating voltage. Thus,
\[hf = (1.60 \times 10^{-19} \, C)(50.0 \times 10^3 \, V). \nonumber\]
From the definition of the electron volt, we know \(1 \, eV = 1.60 \times 10^{-19} \, J\), where \(1 \, J = 1 \, C \cdot V\). Gathering factors and converting energy to eV yields
\[hf = (50.0 \times 10^3)(1.60 \times 10^{-19} \, C \cdot V)\left(\dfrac{1 \, eV}{1.60 \times 10^{-19} \, C \cdot V} \right) \nonumber\]
Discussion
This example produces a result that can be applied to many similar situations. If you accelerate a single elementary charge, like that of an electron, through a potential given in volts, then its energy in eV has the same numerical value. Thus a 50.0-kV potential generates 50.0 keV electrons, which in turn can produce photons with a maximum energy of 50 keV. Similarly, a 100-kV potential in an x-ray tube can generate up to 100-keV x-ray photons. Many x-ray tubes have adjustable voltages so that various energy x rays with differing energies, and therefore differing abilities to penetrate, can be generated.
Figure \(\PageIndex{4}\) shows the spectrum of x rays obtained from an x-ray tube. There are two distinct features to the spectrum. First, the smooth distribution results from electrons being decelerated in the anode material. A curve like this is obtained by detecting many photons, and it is apparent that the maximum energy is unlikely. This decelerating process produces radiation that is called bremsstrahlung (German for braking radiation ). The second feature is the existence of sharp peaks in the spectrum; these are called characteristic x rays , since they are characteristic of the anode material. Characteristic x rays come from atomic excitations unique to a given type of anode material. They are akin to lines in atomic spectra, implying the energy levels of atoms are quantized. Phenomena such as discrete atomic spectra and characteristic x rays are explored further in Atomic Physics.
Ultraviolet radiation (approximately 4 eV to 300 eV) overlaps with the low end of the energy range of x rays, but UV is typically lower in energy. UV comes from the de-excitation of atoms that may be part of a hot solid or gas. These atoms can be given energy that they later release as UV by numerous processes, including electric discharge, nuclear explosion, thermal agitation, and exposure to x rays. A UV photon has sufficient energy to ionize atoms and molecules, which makes its effects different from those of visible light. UV thus has some of the same biological effects as rays and x rays. For example, it can cause skin cancer and is used as a sterilizer. The major difference is that several UV photons are required to disrupt cell reproduction or kill a bacterium, whereas single \(\gamma\)-ray and X-ray photons can do the same damage. But since UV does have the energy to alter molecules, it can do what visible light cannot. One of the beneficial aspects of UV is that it triggers the production of vitamin D in the skin, whereas visible light has insufficient energy per photon to alter the molecules that trigger this production. Infantile jaundice is treated by exposing the baby to UV (with eye protection), called phototherapy, the beneficial effects of which are thought to be related to its ability to help prevent the buildup of potentially toxic bilirubin in the blood.
Example \(\PageIndex{2}\): Photon Energy and Effects for UV
Short-wavelength UV is sometimes called vacuum UV, because it is strongly absorbed by air and must be studied in a vacuum. Calculate the photon energy in eV for 100-nm vacuum UV, and estimate the number of molecules it could ionize or break apart.
Strategy
Using the equation \(E = hf\)
and appropriate constants, we can find the photon energy and compare it with energy information in Table .
Solution
The energy of a photon is given by
\[E = hf = \dfrac{hc}{\lambda}. \nonumber\]
Using \(hc = 1240 \, eV \cdot nm\), we find that
\[E = \dfrac{hc}{\lambda} = \dfrac{1240 \, eV \cdot nm}{100 \, nm} = 12.4 \, eV. \nonumber\]
Discussion
According to Table \(\PageIndex{1}\), this photon energy might be able to ionize an atom or molecule, and it is about what is needed to break up a tightly bound molecule, since they are bound by approximately 10 eV. This photon energy could destroy about a dozen weakly bound molecules. Because of its high photon energy, UV disrupts atoms and molecules it interacts with. One good consequence is that all but the longest-wavelength UV is strongly absorbed and is easily blocked by sunglasses. In fact, most of the Sun’s UV is absorbed by a thin layer of ozone in the upper atmosphere, protecting sensitive organisms on Earth. Damage to our ozone layer by the addition of such chemicals as CFC’s has reduced this protection for us.
Visible Light
The range of photon energies for visible light from red to violet is 1.63 to 3.26 eV, respectively (left for this chapter’s Problems and Exercises to verify). These energies are on the order of those between outer electron shells in atoms and molecules. This means that these photons can be absorbed by atoms and molecules. A single photon can actually stimulate the retina, for example, by altering a receptor molecule that then triggers a nerve impulse. Photons can be absorbed or emitted only by atoms and molecules that have precisely the correct quantized energy step to do so. For example, if a red photon of frequency \(f\) encounters a molecule that has an energy step, \(\Delta E\), equal to \(hf\), then the photon can be absorbed. Violet flowers absorb red and reflect violet; this implies there is no energy step between levels in the receptor molecule equal to the violet photon’s energy, but there is an energy step for the red.
There are some noticeable differences in the characteristics of light between the two ends of the visible spectrum that are due to photon energies. Red light has insufficient photon energy to expose most black-and-white film, and it is thus used to illuminate darkrooms where such film is developed. Since violet light has a higher photon energy, dyes that absorb violet tend to fade more quickly than those that do not. (See Figure \(\PageIndex{5}\).) Take a look at some faded color posters in a storefront some time, and you will notice that the blues and violets are the last to fade. This is because other dyes, such as red and green dyes, absorb blue and violet photons, the higher energies of which break up their weakly bound molecules. (Complex molecules such as those in dyes and DNA tend to be weakly bound.) Blue and violet dyes reflect those colors and, therefore, do not absorb these more energetic photons, thus suffering less molecular damage.
Transparent materials, such as some glasses, do not absorb any visible light, because there is no energy step in the atoms or molecules that could absorb the light. Since individual photons interact with individual atoms, it is nearly impossible to have two photons absorbed simultaneously to reach a large energy step. Because of its lower photon energy, visible light can sometimes pass through many kilometers of a substance, while higher frequencies like UV, x ray, and \(\gamma\) rays are absorbed, because they have sufficient photon energy to ionize the material.
Example \(\PageIndex{3}\): How Many Photons per Second Does a Typical Light Bulb Produce?
Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second.
Strategy
Power is energy per unit time, and so if we can find the energy per photon, we can determine the number of photons per second. This will best be done in joules, since power is given in watts, which are joules per second.
Solution
The power in visible light production is 10.0% of 100 W, or 10.0 J/s. The energy of the average visible photon is found by substituting the given average wavelength into the formula
\[E = \dfrac{hc}{\lambda}. \nonumber\]
This produces
\[E = \dfrac{(6.63 \times 10^{-34} \, J \cdot s)(3.00 \times 10^8 \, m/s}{580 \times 10^{-9}} = 3.43 \times 10^{-19} \, J. \nonumber\]
The number of visible photons per second is thus
\[photon/s = \dfrac{10.0 \, J/s}{3.43 \times 10^{-19} \, J/photon} = 2.92 \times 10^{19} \, photon/s \nonumber\]
Discussion
This incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. It is also a verification of the correspondence principle—on the macroscopic scale, quantization becomes essentially continuous or classical. Finally, there are so many photons emitted by a 100-W lightbulb that it can be seen by the unaided eye many kilometers away.
Lower-Energy Photons
Infrared radiation (IR) has even lower photon energies than visible light and cannot significantly alter atoms and molecules. IR can be absorbed and emitted by atoms and molecules, particularly between closely spaced states. IR is extremely strongly absorbed by water, for example, because water molecules have many states separated by energies on the order of \(10^{-5} \, eV\) to \(10^{-2} \, eV\) well within the IR and microwave energy ranges. This is why in the IR range, skin is almost jet black, with an emissivity near 1—there are many states in water molecules in the skin that can absorb a large range of IR photon energies. Not all molecules have this property. Air, for example, is nearly transparent to many IR frequencies.
Microwaves are the highest frequencies that can be produced by electronic circuits, although they are also produced naturally. Thus microwaves are similar to IR but do not extend to as high frequencies. There are states in water and other molecules that have the same frequency and energy as microwaves, typically about \(10^{-5} \, eV\). This is one reason why food absorbs microwaves more strongly than many other materials, making microwave ovens an efficient way of putting energy directly into food.
Photon energies for both IR and microwaves are so low that huge numbers of photons are involved in any significant energy transfer by IR or microwaves (such as warming yourself with a heat lamp or cooking pizza in the microwave). Visible light, IR, microwaves, and all lower frequencies cannot produce ionization with single photons and do not ordinarily have the hazards of higher frequencies. When visible, IR, or microwave radiation is hazardous, such as the inducement of cataracts by microwaves, the hazard is due to huge numbers of photons acting together (not to an accumulation of photons, such as sterilization by weak UV). The negative effects of visible, IR, or microwave radiation can be thermal effects, which could be produced by any heat source. But one difference is that at very high intensity, strong electric and magnetic fields can be produced by photons acting together. Such electromagnetic fields (EMF) can actually ionize materials.
MISCONCEPTION ALERT: HIGH VOLTAGE POWER LINES
- Although some people think that living near high-voltage power lines is hazardous to one’s health, ongoing studies of the transient field effects produced by these lines show their strengths to be insufficient to cause damage. Demographic studies also fail to show significant correlation of ill effects with high-voltage power lines. The American Physical Society issued a report over 10 years ago on power-line fields, which concluded that the scientific literature and reviews of panels show no consistent, significant link between cancer and power-line fields. They also felt that the “diversion of resources to eliminate a threat which has no persuasive scientific basis is disturbing.”
It is virtually impossible to detect individual photons having frequencies below microwave frequencies, because of their low photon energy. But the photons are there. A continuous EM wave can be modeled as photons. At low frequencies, EM waves are generally treated as time- and position-varying electric and magnetic fields with no discernible quantization. This is another example of the correspondence principle in situations involving huge numbers of photons.
PHET EXPLORATIONS: COLOR VISION
Make a whole rainbow by mixing red, green, and blue light. Change the wavelength of a monochromatic beam or filter white light. View the light as a solid beam, or see the individual photons.
Summary
- Photon energy is responsible for many characteristics of EM radiation, being particularly noticeable at high frequencies.
- Photons have both wave and particle characteristics.
Glossary
- gamma ray
- also \(\gamma\)-ray; highest-energy photon in the EM spectrum
- ionizing radiation
- radiation that ionizes materials that absorb it
- x ray
- EM photon between \(\gamma\)-ray and UV in energy
- bremsstrahlung
- German for braking radiation ; produced when electrons are decelerated
- characteristic x rays
- x rays whose energy depends on the material they were produced in
- ultraviolet radiation
- UV; ionizing photons slightly more energetic than violet light
- visible light
- the range of photon energies the human eye can detect
- infrared radiation
- photons with energies slightly less than red light
- microwaves
- photons with wavelengths on the order of a micron \((\mu m)0\))