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Changes to the Earth's Surface
Quiz by Derrick Dormer
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Cohesion and Adhesion Water molecules stick to each other as a result of hydrogen bond- ing. An attractive force that holds molecules of a single substance together is known as cohesion. Cohesion due to hydrogen bonding between water molecules contributes to the upward movement of water from plant roots to their leaves. Related to cohesion is the surface tension of water. The cohe- sive forces in water resulting from hydrogen bonds cause the mol- ecules at the surface of water to be pulled downward into the liquid. As a result, water acts as if it has a thin “skin” on its sur- face. You can observe water’s surface tension by slightly overfill- ing a drinking glass with water. The water will appear to bulge above the rim of the glass. Surface tension also enables small crea- tures such as spiders and water-striders to run on water without breaking the surface. Adhesion is the attractive force between two particles of differ- ent substances, such as water molecules and glass molecules. A related property is capillarity (KAP-uh-LER-i-tee), which is the attrac- tion between molecules that results in the rise of the surface of a liquid when in contact with a solid. Together, the forces of adhe- sion, cohesion, and capillarity help water rise through narrow tubes against the force of gravity. Figure 2-11 shows cohesion and adhesion in the water-conducting tubes in the stem of a flower. Temperature Moderation Water has a high heat capacity, which means that water can absorb or release relatively large amounts of energy in the form of heat with only a slight change in temperature. This property of water is related to hydrogen bonding. Energy must be absorbed to break hydrogen bonds, and energy is released as heat when hydrogen bonds form. The energy that water initially absorbs breaks hydro- gen bonds between molecules. Only after these hydrogen bonds are broken does the energy begin to increase the motion of the water molecules, which raises the temperature of the water. When the temperature of water drops, hydrogen bonds reform, which releases a large amount of energy in the form of heat. Therefore, during a hot summer day, water can absorb a large quantity of energy from the sun and can cool the air without a large increase in the water’s temperature. At night, the gradually cooling water warms the air. In this way, the Earth’s oceans stabilize global temperatures enough to allow life to exist. Water’s high heat capac- ity also allows organisms to keep cells at an even temperature despite temperature changes in the environment. As a liquid evaporates, the surface of the liquid that remains behind cools down. A relatively large amount of energy is absorbed by water during evaporation, which significantly cools the surface of the remaining liquid. Evaporative cooling prevents organisms that live on land from overheating. For example, the evaporation of sweat from a person’s skin releases body heat and prevents over- heating on a hot day or during strenuous activity. Adhesion Cohesion Hydrogen bonds Cohesion, adhesion, and capillarity contribute to the upward movement of water from the roots of plants. FIGURE 2–11 www.scilinks.org Topic: Hydrogen Bonding Keyword: HM60777 mb06se_cols03.qxd 5/18/07 10:47 AM Page 41 42 CHAPTER 2 Density of Ice Unlike most solids, which are denser than their liquids, solid water is less dense than liquid water. This property is due to the shape of the water molecule and hydrogen bonding. The angle between the hydrogen atoms is quite wide. So, when water forms solid ice, the angles in the molecules cause ice crystals to have large amounts of open space, as shown in Figure 2-12. This open space lattice structure causes ice to have a low density. Because ice floats on water, bodies of water such as ponds and lakes freeze from the top down and not the bottom up. Ice insulates the water below from the cold air, which allows fish and other aquatic crea- tures to survive under the icy surface.
1. Settlements Importance of Rivers Fertile Land: The soil near rivers was great for farming, thanks to regular flooding that added nutrients. Trade and Travel: Rivers made moving things and people easy, which helped trade and communication. Protection: Rivers could act as natural barriers, making it harder for enemies to attack. Food: Rivers were full of fish and other food, adding to what people could eat. Energy: People used the river's flow to power machines, for example, grinding grain. Cleanliness: Rivers were used to wash away waste, keeping settlements cleaner. Culture: Rivers often had spiritual importance, and ceremonies and stories revolved around them. Common Geographic Features of Ancient Civilizations Mesopotamia: the Tigris and Euphrates Rivers in central Iraq Indus River Valley: the river runs in the northwestern part of India Nile River Valley: the major river of Egypt Yellow River Valley: a major river flowing through the southern part of China Rivers provided water, food, transportation, and shaped the way of life and development of these ancient civilizations. Impact of Mountains on Settlements Mountains served as barriers to early settlement due to the lack of technology to cross them. The Himalayan Mountains isolated much of India and China during their early development. Impact of Deserts on Migration Deserts posed significant challenges to people who wanted to migrate due to their harsh and unforgiving conditions. Notable deserts include the Empty Quarter in Saudi Arabia and the Sahara Desert in Africa. Changes in Migration and Cultural Blending Advancements in transportation technology post-Industrial Revolution increased cultural blending. Transportation advancements enabled global migration. Before, cultures were isolated, focusing on beliefs and local adaptations. The Industrial Revolution transformed migration and cultural blending. 2. How Humans Modify and Adapt to Their Environment Ways Humans Modify Their Environment Mining: Removing the earth's surface for precious metals. Irrigation: Diverting water for farming. Transportation: Moving goods with trains, cars, airplanes, and boats. Mining Strip mining removes large layers of the earth. Can impact the environment by removing plants and polluting water sources. Irrigation Diverting water for farming and urban development. Transportation Moving goods using trains, cars, airplanes, and boats. Human Adaptation to the Environment Adjusting to environmental conditions by changing behavior. Examples: Wearing specific clothing, using specific building materials. Human Modification of the Environment Changing the earth to meet human needs by physically altering the environment. Examples: Dams, canals, roads, bridges. Impact of Weather and Geological Events on Humans Events like earthquakes, hurricanes, and cold weather affect human settlements. Examples: Building earthquake-resistant buildings, creating levees, using ice for tourism. 3. Understanding Culture Introduction to Culture Culture refers to the way of life of a group of people who live in a particular place. It includes traditions, beliefs, values, and the way they do things. Cultural Characteristics Religious traditions Language Family values Laws Cultural characteristics make each culture unique. Cultural Representations Art Architecture Music Literature Cultural representations express a culture's creativity and show their beliefs and history to the world. Government and Culture Types of government reflect cultural beliefs and traditions. Examples: democratic republic, communist state. The way a country is governed tells a lot about its culture. Economic Systems and Cultures Economic systems reflect cultural values. Examples: bartering, modern economies (e.g., United States, China). How people earn and spend money also reflects their culture. Spread of Cultural Ideas Trade: Spreading ideas through interactions during trade. Travel: Visitors bringing new ideas. War: Conquering armies imposing beliefs. Cultural ideas spread through trade, travel, and war. Multicultural Societies Blending of multiple cultural and ethnic groups. Common in advanced societies with immigration. Multicultural societies create something new by bringing together different cultures. Cultural Adaptation Cultures can change and adapt by taking new ideas and blending them with their own traditions. Example: 'Tex-Mex' food, which blends Mexican and Texan traditions.
Based on the provided sources, here is a comprehensive extraction of the information regarding the water cycle, energy transfer, and Earth's wind systems, organized into key points: The Water Cycle and Its Reservoirs • Definition: The water cycle is the continuous movement of water among various reservoirs on Earth. • Water Reservoirs: These are storage locations for water and include: ◦ Oceans, seas, and lakes. ◦ Rivers, glaciers, soil, and rocks. ◦ The atmosphere and living organisms. • Total Volume: The total amount of water on Earth does not change, even when it changes state, because it is constantly being replaced or recycled through the cycle. Main Processes and Energy Transfer The movement of water through the cycle is driven by energy (thermal energy from the Sun) and force (gravity and wind). • Energy Gain (Absorption): ◦ Melting: Water changes from a solid state (ice) to a liquid state and gains energy. ◦ Evaporation: Liquid water changes into a gas state (water vapor) by gaining thermal energy. ◦ Transpiration: A specialized type of evaporation occurring in plants where water vapor is released through tiny holes in leaves called stomata. Approximately 10% of water vapor in the air comes from transpiration. • Energy Loss (Release): ◦ Condensation: Water vapor (gas) cools down and changes back into liquid water, releasing energy. ◦ Freezing: Liquid water changes into a solid state (ice) and loses energy. • Other Key Steps: ◦ Precipitation: Water falls back to Earth as rain, snow, sleet, or hail (snow pellets). ◦ Runoff: Water flows over Earth's surface into streams, rivers, and eventually larger bodies of water like oceans. ◦ Collection: Rainwater is collected in different water bodies to start the cycle again. Forces Driving Water Movement • Gravity: The main force that pulls water downward. It is responsible for: ◦ Bringing precipitation (rain and snow) from clouds to the surface. ◦ Moving ice in glaciers from higher to lower elevations. ◦ Causing liquid water to flow downhill into rivers and seas. ◦ Leakage: Pulling liquid water down into the ground to reach groundwater reservoirs. • Wind: Another force that affects water movement and transports water to different locations on Earth. Atmospheric Processes • Cloud Formation: Water vapor attaches to particles such as dust or smoke in the air and condenses into tiny droplets. When millions of these droplets join, they become heavy and fall as rain. • Convection: The transfer of heat in liquids and gases. ◦ Warm air/liquid: Becomes less dense, lighter, and rises upward. ◦ Cold air/liquid: Is more dense, heavier, and moves downward to replace the warm fluid. ◦ This process leads to convection currents, which help determine regional climates and drive wind and ocean currents. Solar Radiation and Climate The amount of solar energy reaching Earth differs from place to place, which affects the weather: • Hottest Regions (Equator): Sun rays fall perpendicular (vertical). Heat is concentrated on a small area, making the weather hot. • Moderate Regions: Sun rays fall semi-inclined. Heat is distributed over a larger area, making the weather warm. • Coolest Regions (Poles): Sun rays fall very slanted (inclined). Heat is spread over a very large area, making the weather very cold. Earth's Wind System • Wind Formation: Wind is generated when warm air (heated by the Sun) rises and is replaced by cooler air flowing from nearby areas. • Factors Affecting Wind: The amount of solar radiation and the rotation of Earth determine global wind directions. • Global Wind Cycle: Unequal heating between the equator and the poles generates a constant wind system. Warm air rises at the equator and moves toward the poles, while cold air from the poles moves toward the equator. • Importance: If there were no wind, the equator would become extremely hot, the poles would freeze solid, and many ecosystems would disappear. Practical Examples • Turkey’s Salt Lake: High evaporation in the summer can turn this large lake into a small puddle or dry it up completely. It is a critical site for flamingos, which migrate there to breed and feed on algae in the shallow, warm water.
5.7.B Recognize How Landforms Are The Result Of Changes To Earth's Surface
Weathering describes the breaking down or dissolving of rocks and minerals on the surface of the Earth. Water, ice, acids, salts, plants, animals, and changes in temperature are all agents of weathering. Once a rock has been broken down, a process called erosion transports the bits of rock and mineral away. No rock on Earth is hard enough to resist the forces of weathering and erosion. Together, these processes carved landmarks such as the Grand Canyon, in the U.S. state of Arizona. This massive canyon is 446 kilometers (277 miles) long, as much as 29 kilometers (18 miles) wide, and 1,600 meters (1 mile) deep. Weathering and erosion constantly change the rocky landscape of Earth. Weathering wears away exposed surfaces over time. The length of exposure often contributes to how vulnerable a rock is to weathering. Rocks, such as lavas, that are quickly buried beneath other rocks are less vulnerable to weathering and erosion than rocks that are exposed to agents such as wind and water, As it smoothes rough, sharp rock surfaces, weathering is often the first step in the production of soils. Tiny bits of weathered minerals mix with plants, animal remains, fungi, bacteria, and other organisms. A single type of weathered rock often produces infertile soil, while weathered materials from a collection of rocks is richer in mineral diversity and contributes to more fertile soil. Soils types associated with a mixture of weathered rock include glacial till, loess, and alluvial sediments. Weathering is often divided into the processes of mechanical weathering and chemical weathering. Biological weathering, in whichliving or once-living organisms contribute to weathering, can be a part of both processes. Mechanical Weathering Mechanical weathering, also called physical weathering and disaggregation, causes rocks to crumble. Water, in either liquid or solid form, is often a key agent of mechanical weathering. For instance, liquid water can seep into cracks and crevices in rock. If temperatures drop low enough, the water will freeze. When water freezes, it expands. The ice then works as a wedge. It slowly widens the cracks and splits the rock. When ice melts, liquid water performs the act of erosion by carrying away the tiny rock fragments lost in the split. This specific process (the freeze-thaw cycle) is called frost weathering or cryofracturing. Figure 4.3 Frost Wedging Temperature changes can also contribute to mechanical weathering in a process called thermal stress. Changes in temperature cause rock to expand (with heat) and contract (with cold). As this happens over and over again. the structure of the rock weakens. Over time, it crumbles. Rocky desert landscapes are particularly vulnerable to thermal stress. The outer layer of desert rocks undergo repeated stress as the temperature changes from day Eventually, Lo outer night. layersflake off in thin sheets, a process called exfoliation. Exfoliation contributes to the formation of bornhardts, one of the most dramatic features in landscapes formed by weathering and erosion. Bornhardts are tall, domed, isolated rocks often found areas. in tropical Sugarloaf Mountain, an iconic landmark in Rio de Janeiro, Brazil, is bornhardt. a Salt also works to weather rock in a process called haloclasty. Saltwater sometimes gets into the cracks and pores of rock. If the saltwater evaporates, salt crystals are left behind. As the crystals grow, they put pressure on the rock, slowly breaking it apart. Plants and animals can be agents of mechanical weathering. The seed of a tree may sprout in soil that has collected in a cracked rock. As the roots grow, they widen the cracks, eventually breaking the rock into pieces. Over time, trees can break apart even large rocks. Even small plants, such as mosses, can enlarge tiny cracks as they grow. Animals that tunnel underground, such as moles and prairie dogs, also work to break apart rock and soil. Other animals dig and trample rock aboveground, causing rock to slowly crumble. Chemical Weathering Chemical weathering changes the molecular structure of rocks and soil.For instance, carbon dioxide from the air or soil sometimes combines with water in a process called carbonation. This produces a weak acid, called carbonic acid, that can dissolve rock. Carbonic acid is especially effective at dissolving limestone. When carbonic acid seeps through limestone underground, it can open up huge cracks or hollow out vast networks of caves. Carlsbad Caverns National Park, in the U.S. state of New Mexico, includes more than 119 limestone caves created by weathering and erosion. The largest is called the Big Room.. With an area of about 33,210 square meters (357,469 square feet), the Big Room is the size of six football fields. Another type of chemical weathering works on rocks that contain iron. These rocks turn to rust in a process called oxidation. Rust is a compound created by the interaction of oxygen and iron in the presence of water. As rust expands, it weakens rock and helps break it apart. Another familiar form of chemical weathering is hydrolysis. In the process of hydrolysis, a new solution (a mixture of two or more substances) is formed as chemicals in rock interact with water. In many rocks, for example, sodium minerals interact with water to form a saltwater solution. Hydration and hydrolysis contribute to flared slopes, another dramatic example of a landscape formed by weathering and erosion. Flared slopes are sometimes nicknamed "wave rocks." Their c-shape is largely concave rock formations a result of subsurface weathering, in which hydration and hydrolysis wear away rocks beneath the landscape's surfaceWeathering and People Weathering is a natural process, but human activities can speed it up. For example, certain kinds of air pollution increase the rate of weathering Burning coal, natural and petroleum releases chemicals such as nitrogen oxide and gas, sulfur dioxide into the atmosphere. When these chemicals combine with sunlight and moisture, they change into acids. They then fall back to Earth as acid rain. Acid rain rapidly weathers limestone, marble, and other kinds of stone. The effects of acid rain can often be seen on gravestones, making names and other inscriptions impossible to read. Acid rain has also damaged many historic buildings and monuments. For example, at 71 meters (233 feet) tall, the Leshan Giant Buddha at Mount Emei, China is the world's largest statue of the Buddha. It was carved 1,300 years ago and sat unharmed for centuries. An innovative drainage system mitigates the natural process of erosion But in recent years, acid rain has turned the statue's nose black and made some of its hair crumble and fall.
Earth's Water Water Everywhere. Water fills oceans, lakes, and ponds. It flows in rivers, streams, and underground. It is even in the air. Some parts of Earth have snow and ice, which are frozen water. Water covers most of Earth's surface. Salt water in the oceans makes up much of Earth's water. Earth has much less fresh water. Many plants and animals need this fresh water to survive. Some of this fresh water is aboveground, while other fresh water is underneath Earth's surface. What are some ways you use Earth's water? Different Forms of Water. Liquid water is the most common state of Earth's water. It takes the shape of the container it is in. Liquid water is always moving even if you can't see it move. It flows in rivers and streams, and it crashes as ocean waves. Not all water is liquid. When liquid water gets very cold, it freezes to form ice. Ice is another state of water-solid water. Ice can float on liquid water. People form ice into different shapes. Artists even carve ice to make sculptures. Much of Earth's frozen water is at the North and South Poles, Earth's coldest areas. Some of Earth's water is in an invisible state as a gas called water vapor. While it's always invisible, water vapor is all around us. Changing Water. Earth's water is always changing from one state to another. When frozen water is heated, it melts and becomes liquid water. When liquid water is cooled, it freezes and becomes ice. Liquid water can become a gas, too. Have you ever seen a puddle of water dry up on a hot day? Energy from the Sun changed the liquid to a gas in a process called evaporation. Water evaporates from oceans, rivers, lakes, and puddles all over the world. When water vapor in the air cools down, it changes from a gas to a liquid. This process is called condensation. Clouds are made up of tiny drops of water formed by condensation. The tiny drops stick together, creating larger, heavier drops. Once they're large enough, they fall to the ground as rain or another type of precipitation. Water Is Important. Rain keeps plants alive and allows them to keep growing. People and other animals need water to survive. We also use it for other purposes, such as fighting fires. It is important to take care of Earth's water. Keeping waste and trash away from water keeps it from becoming dirty and unusable. Polluted water makes people, plants, and animals sick. Would you want to drink and play in polluted water?
Ect weather data. Trackingthe The forecasters have a lot of help! Meteorologists, or scientists who study weather, collect data from all over the world. They use automated systems at sea, on land, in the air, and in space to help track the weather. At sea, weather buoys collect data on coastal weather conditions. These data help keep people in coastal cities safe. On land, weather-monitoring stations track weather conditions in remote locations. They can send advance notice about incoming weather. Scientists also use Doppler radar towers to observe how storm clouds are moving. Weather balloons and weather satellites collect data from high above Earth’s surface. Weather satellites can track the weather over very large areas. They also help relay data from land-based tools to meteorologists around the world. Scientists launch weather balloons that carry tools, called radiosondes, into the atmosphere. A typical radiosonde measures air temperature, air pressure, and humidity. Weather satellites orbit Earth. They collect weather data, such as cloud cover, and track storms, such as hurricanes. These satellites use radio signals to transmit data back to Earth. Packets of air move across Earth’s surface. An air mass is a large body of air with the same temperature and humidity throughout. An air mass reflects the conditions of the place where it forms. Air masses that form over land are dry. Air masses that form over water are moist. Cold air masses form near the poles, and warm air masses form near the tropics. As air masses move across an area, they can collide. The boundary between two air masses is called a front. Weather changes take place at fronts. For example, as a warm front passes over an area, warm air replaces cooler air and the temperature rises. The movement of air masses and fronts explains why you might be chilly one day and warm the next. At a front, stormy conditions are common. Air masses may form over land or water. The masses that form over land near the poles will be cold and dry. The masses that form over water near the poles will be cold and moist. Which types of air masses will form near the tropics? A cold front forms where a cold air mass bumps into a warmer air mass. The denser cold air pushes the warm air up. Water vapor in the air cools, and large clouds form. Thunderstorms and heavy rain often take place. Cooler temperatures follow a cold front. A warm front forms where a warm air mass moves over a cold air mass. A warm front forms a wider area of clouds and rain than a cold front. Steady rain or snow may fall. Warmer temperatures follow a warm front.
Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.