Michael Jordan’s book.. Considering how the ending of the book was written, is it an effective ending? Did the ending surprise you? Why? Why not? Did the ending leave you with unanswered questions? Explain.

Michael Jordan’s book.. Considering how the ending of the book was written, is it an effective ending? Did the ending surprise you? Why? Why not? Did the ending leave you with unanswered questions? Explain.

The book is divided into the six Reading Sections namely … Read More...
Hobbes argues that “Nature hath made men. . . .equal.” What sort of equality is he talking about? How are people equal?

Hobbes argues that “Nature hath made men. . . .equal.” What sort of equality is he talking about? How are people equal?

People are equivalent both physically and psychologically. Hobbes’s stats that … Read More...
AERN 45350 Avionics Name: _______________________________ 1 | P a g e Homework Set One (40 Points) Due: 25 September 2015 General Instructions: Answer the following questions, submitting your answers on Blackboard. SHOW YOUR WORK on any math problems. Consider the following voltage signal: V t 12sin377t 1. What is the peak voltage of the signal [Volts]? 2. What is the angular frequency [rad/sec]? 3. What is the frequency of the signal [Hz]? 4. What is the period of the signal [sec/cycle]? In a heterodyne receiver, the intermediate frequency of the receiver is 21.4 MHz. 5. What is the local oscillator frequency (f1) if the tuned frequency (f2) is 120.9 MHz? 6. If the local oscillator frequency (f1) is 145.7 MHz, what is the tuned frequency (f2)? The gain of a power amplifier is 5. 7. If 30W are coming in, what is the power going out? 8. What is the gain in decibels (dB)? The attenuation of a voltage attenuator is 10. 9. If 120V are coming in, what is the voltage going out? 10. What is the loss in decibels (dB)? 11. What is the component of the ILS that provides the extended centerline of the runway? 12. What is the component of the ILS that provides vertical guidance to the runway? 13. If the aircraft is on the correct trajectory, the airplane will arrive at the outer marker on the ILS corresponding to intercepting what? 14. If the aircraft is on the correct trajectory, the airplane will arrive at the middle marker on the ILS corresponding to reaching what? 15. All marker beacons transmit at what frequency? 16. Why doesn’t this cause problems (all marker beacons transmitting on the same frequency)? 17. What are the four components to an ILS? 18. What is the most common ILS category? 19. Which ILS category requires aircraft with the “auto-land” feature? 20. An attenuator leads to a power ratio of 0.5. What is that in decibels (dB)?

AERN 45350 Avionics Name: _______________________________ 1 | P a g e Homework Set One (40 Points) Due: 25 September 2015 General Instructions: Answer the following questions, submitting your answers on Blackboard. SHOW YOUR WORK on any math problems. Consider the following voltage signal: V t 12sin377t 1. What is the peak voltage of the signal [Volts]? 2. What is the angular frequency [rad/sec]? 3. What is the frequency of the signal [Hz]? 4. What is the period of the signal [sec/cycle]? In a heterodyne receiver, the intermediate frequency of the receiver is 21.4 MHz. 5. What is the local oscillator frequency (f1) if the tuned frequency (f2) is 120.9 MHz? 6. If the local oscillator frequency (f1) is 145.7 MHz, what is the tuned frequency (f2)? The gain of a power amplifier is 5. 7. If 30W are coming in, what is the power going out? 8. What is the gain in decibels (dB)? The attenuation of a voltage attenuator is 10. 9. If 120V are coming in, what is the voltage going out? 10. What is the loss in decibels (dB)? 11. What is the component of the ILS that provides the extended centerline of the runway? 12. What is the component of the ILS that provides vertical guidance to the runway? 13. If the aircraft is on the correct trajectory, the airplane will arrive at the outer marker on the ILS corresponding to intercepting what? 14. If the aircraft is on the correct trajectory, the airplane will arrive at the middle marker on the ILS corresponding to reaching what? 15. All marker beacons transmit at what frequency? 16. Why doesn’t this cause problems (all marker beacons transmitting on the same frequency)? 17. What are the four components to an ILS? 18. What is the most common ILS category? 19. Which ILS category requires aircraft with the “auto-land” feature? 20. An attenuator leads to a power ratio of 0.5. What is that in decibels (dB)?

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SUPPLY CHAIN MANAGEMENT AT BOSE CORPORATION Bose Corporation, headquartered in Framingham, Massachusetts, offers an excellent example of integrated supply chain management. Bose, a producer of audio premium speakers used in automobiles, high-fidelity systems, and consumer and commercial broadcasting systems, was founded in 1964 by Dr. Bose of MIT. Bose currently maintains plants in Massachusetts and Michigan as well as Canada, Mexico, and Ireland. Its purchasing organization, while decentralized, has some overlap that requires coordination between sites. It manages this coordination by using conference calls between managers, electronic communication, and joint problem solving. The company is moving toward single sourcing many of its 800 to 1,000 parts, which include corrugated paper, particle board and wood, plastic injected molded parts, fasteners, glues, woofers, and fabric. Some product components, such as woofers, are sourced overseas. For example, at the Hillsdale, Michigan, plant, foreign sourcing accounts for 20% of purchases, with the remainder of suppliers located immediately within the state of Michigan. About 35% of the parts purchased at this site are single sourced, with approximately half of the components arriving with no incoming inspection performed. In turn, Bose ships finished products directly to Delco, Honda, and Nissan and has a record of no missed deliveries. Normal lead time to customers is 60 working days, but Bose can expedite shipments in one week and airfreight them if necessary. The company has developed a detailed supplier performance system that measures on-time delivery, quality performance, technical improvements, and supplier suggestions. A report is generated twice a month from this system and sent to the supplier providing feedback about supplier performance. If there is a three-week trend of poor performance, Bose will usually establish a specific goal for improvement that the supplier must attain. Examples include 10% delivery improvement every month until 100% conformance is achieved, or 5% quality improvement until a 1% defect level is reached over a four-month period. In one case, a supplier sent a rejected shipment back to Bose without explanation and with no corrective action taken. When no significant improvement occurred, another supplier replaced the delinquent supplier. Bose has few written contracts with suppliers. After six months of deliveries without rejects, Bose encourages suppliers to apply for a certificate of achievement form, signifying that they are qualified suppliers. One of the primary criteria for gaining certification involves how well the supplier responds to corrective action requests. One of the biggest problems observed is that suppliers often correct problems on individual parts covered by a corrective action form without extending these corrective actions to other part families and applicable parts. Bose has adopted a unique system of marrying just-in-time (JIT) purchasing with global sourcing. Approximately half of the dollar value of Bose’s total purchases are made overseas, with the majority of the sourcing done in Asia. Because foreign sourcing does not support just-in-time deliveries, Bose “had to find a way to blend low inventory with buying from distant sources,” says the director of purchasing and logistics for Bose. Visualizing itself as a customer-driven organization, Bose now uses a sophisticated transportation system—what Bose’s manager of logistics calls “the best EDI system in the country.” Working closely with a national less-than-truckload carrier for the bulk of its domestic freight movements, including shipments arriving at a U.S. port from oversees, Bose implemented an electronic data interchange (EDI) system that does much more than simple tracking. The system operates close to real time and allows two-way communication between every one of the freight handler’s 230 terminals and Bose. Information is updated several times daily and is downloaded automatically, enabling Bose to perform shipping analysis and distribution channel modeling to achieve reliable lowest total cost scenarios. The company can also request removal from a terminal of any shipment that it must expedite with an air shipment. This state-of-the-art system provides a snapshot of what is happening on a daily basis and keeps Bose’s managers on top of everyday occurrences and decisions. Management proactively manages logistics time elements in pursuit of better customer service. The next step is to implement this system with all major suppliers rather than just with transportation suppliers. In the future, Bose plans to automate its entire materials system. Perhaps one of the most unique features of Bose’s procurement and logistics system is the development of JIT II. The basic premise of JIT II is simple: The person who can do the best job of ordering and managing inventory of a particular item is the supplier himself. Bose negotiated with each supplier to provide a full-time employee at the Bose plant who was responsible for ordering, shipping, and receiving materials from that plant, as well as managing on-site inventories of the items. This was done through an EDI connection between Bose’s plant and the supplier’s facility. Collocating suppliers and buyers was so successful that Bose is now implementing it at all plant locations. In fact, many other companies have also begun to implement collocation of suppliers. Assignment Questions The following assignment questions relate to ideas and concepts presented throughout this text. Answer some or all of the questions as directed by your instructor. 1. Discuss how the strategy development process might work at a company like Bose. 2. What should be the relationship between Bose’s supply management strategy and the development of its performance measurement system? 3. Why is purchased quality so important to Bose? 4. Can a just-in-time purchase system operate without total quality from suppliers? 5. Why can some components arrive at the Hillsdale, Michigan, plant with no incoming inspection required? 6. Discuss the reasons why Bose has a certificate of achievement program for identifying qualified suppliers. 7. Bose is moving toward single sourcing many of its purchased part requirements. Discuss why the company might want to do this. Are there any risks to that approach? 8. Discuss some of the difficulties a company like Bose might experience when trying to implement just-in-time purchasing with international suppliers. 9. Why does Bose have to source so much of its purchase requirements from offshore suppliers? 10. What makes the JIT II system at Bose unique? Why would a company pursue this type of system? 11. Why is it necessary to enter into a longer-term contractual arrangement when pursuing arrangements like the one Bose has with its domestic transportation carrier? 12. Why is it important to manage logistics time elements proactively when pursuing higher levels of customer service? 13. What role does information technology play at Bose? 14. What advantages do information technology systems provide to Bose that might not be available to a company that does not have these systems? 15. Why has Bose developed its supplier performance measurement system? 16. Do you think the performance measurement systems at Bose are computerized or manual? Why?

SUPPLY CHAIN MANAGEMENT AT BOSE CORPORATION Bose Corporation, headquartered in Framingham, Massachusetts, offers an excellent example of integrated supply chain management. Bose, a producer of audio premium speakers used in automobiles, high-fidelity systems, and consumer and commercial broadcasting systems, was founded in 1964 by Dr. Bose of MIT. Bose currently maintains plants in Massachusetts and Michigan as well as Canada, Mexico, and Ireland. Its purchasing organization, while decentralized, has some overlap that requires coordination between sites. It manages this coordination by using conference calls between managers, electronic communication, and joint problem solving. The company is moving toward single sourcing many of its 800 to 1,000 parts, which include corrugated paper, particle board and wood, plastic injected molded parts, fasteners, glues, woofers, and fabric. Some product components, such as woofers, are sourced overseas. For example, at the Hillsdale, Michigan, plant, foreign sourcing accounts for 20% of purchases, with the remainder of suppliers located immediately within the state of Michigan. About 35% of the parts purchased at this site are single sourced, with approximately half of the components arriving with no incoming inspection performed. In turn, Bose ships finished products directly to Delco, Honda, and Nissan and has a record of no missed deliveries. Normal lead time to customers is 60 working days, but Bose can expedite shipments in one week and airfreight them if necessary. The company has developed a detailed supplier performance system that measures on-time delivery, quality performance, technical improvements, and supplier suggestions. A report is generated twice a month from this system and sent to the supplier providing feedback about supplier performance. If there is a three-week trend of poor performance, Bose will usually establish a specific goal for improvement that the supplier must attain. Examples include 10% delivery improvement every month until 100% conformance is achieved, or 5% quality improvement until a 1% defect level is reached over a four-month period. In one case, a supplier sent a rejected shipment back to Bose without explanation and with no corrective action taken. When no significant improvement occurred, another supplier replaced the delinquent supplier. Bose has few written contracts with suppliers. After six months of deliveries without rejects, Bose encourages suppliers to apply for a certificate of achievement form, signifying that they are qualified suppliers. One of the primary criteria for gaining certification involves how well the supplier responds to corrective action requests. One of the biggest problems observed is that suppliers often correct problems on individual parts covered by a corrective action form without extending these corrective actions to other part families and applicable parts. Bose has adopted a unique system of marrying just-in-time (JIT) purchasing with global sourcing. Approximately half of the dollar value of Bose’s total purchases are made overseas, with the majority of the sourcing done in Asia. Because foreign sourcing does not support just-in-time deliveries, Bose “had to find a way to blend low inventory with buying from distant sources,” says the director of purchasing and logistics for Bose. Visualizing itself as a customer-driven organization, Bose now uses a sophisticated transportation system—what Bose’s manager of logistics calls “the best EDI system in the country.” Working closely with a national less-than-truckload carrier for the bulk of its domestic freight movements, including shipments arriving at a U.S. port from oversees, Bose implemented an electronic data interchange (EDI) system that does much more than simple tracking. The system operates close to real time and allows two-way communication between every one of the freight handler’s 230 terminals and Bose. Information is updated several times daily and is downloaded automatically, enabling Bose to perform shipping analysis and distribution channel modeling to achieve reliable lowest total cost scenarios. The company can also request removal from a terminal of any shipment that it must expedite with an air shipment. This state-of-the-art system provides a snapshot of what is happening on a daily basis and keeps Bose’s managers on top of everyday occurrences and decisions. Management proactively manages logistics time elements in pursuit of better customer service. The next step is to implement this system with all major suppliers rather than just with transportation suppliers. In the future, Bose plans to automate its entire materials system. Perhaps one of the most unique features of Bose’s procurement and logistics system is the development of JIT II. The basic premise of JIT II is simple: The person who can do the best job of ordering and managing inventory of a particular item is the supplier himself. Bose negotiated with each supplier to provide a full-time employee at the Bose plant who was responsible for ordering, shipping, and receiving materials from that plant, as well as managing on-site inventories of the items. This was done through an EDI connection between Bose’s plant and the supplier’s facility. Collocating suppliers and buyers was so successful that Bose is now implementing it at all plant locations. In fact, many other companies have also begun to implement collocation of suppliers. Assignment Questions The following assignment questions relate to ideas and concepts presented throughout this text. Answer some or all of the questions as directed by your instructor. 1. Discuss how the strategy development process might work at a company like Bose. 2. What should be the relationship between Bose’s supply management strategy and the development of its performance measurement system? 3. Why is purchased quality so important to Bose? 4. Can a just-in-time purchase system operate without total quality from suppliers? 5. Why can some components arrive at the Hillsdale, Michigan, plant with no incoming inspection required? 6. Discuss the reasons why Bose has a certificate of achievement program for identifying qualified suppliers. 7. Bose is moving toward single sourcing many of its purchased part requirements. Discuss why the company might want to do this. Are there any risks to that approach? 8. Discuss some of the difficulties a company like Bose might experience when trying to implement just-in-time purchasing with international suppliers. 9. Why does Bose have to source so much of its purchase requirements from offshore suppliers? 10. What makes the JIT II system at Bose unique? Why would a company pursue this type of system? 11. Why is it necessary to enter into a longer-term contractual arrangement when pursuing arrangements like the one Bose has with its domestic transportation carrier? 12. Why is it important to manage logistics time elements proactively when pursuing higher levels of customer service? 13. What role does information technology play at Bose? 14. What advantages do information technology systems provide to Bose that might not be available to a company that does not have these systems? 15. Why has Bose developed its supplier performance measurement system? 16. Do you think the performance measurement systems at Bose are computerized or manual? Why?

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1 ACTIVITY PURPOSE The purpose of this activity is to give you practice preparing a four-week work schedule. PROCESS Follow the steps listed below to prepare a schedule. 1. Read the Information Sheet: Scheduling Employees. 2. The pay week for this medical record service runs Sunday – Saturday. The pay period is two pay weeks. Each full-time employee cannot work more than 40 hours per pay week, or 80 hours per pay period. Each part-time employee works 20 hours per pay week – 40 hours per pay period. 3. The first Friday of the four – week period is a holiday. 4. The medical record service has 24 hour coverage, seven days a week. All full-time employees work a five day pay week, eight hours per day, with rotating weekend coverage. Part-time employees work four hours Monday – Friday, except for their rotation weekend. On those days they work an eight hour shift. Remember to adjust their time accordingly. 5. The Assistant Director and all supervisors, except the Tumor Registry Supervisor, should be scheduled for rotating weekend coverage. 2 6. All employees, except the Tumor Registry employees, should be scheduled on a rotating basis for weekend coverage. 7. For weekend and holiday coverage, there needs to be at least two clerks and one transcriptionist on days and evenings, one clerk and one transcriptionist at night. 8. The Department Director has scheduled a two – week vacation for the first two full weeks of the four – week schedule. 9. Employees who work holidays must take the holiday time within the pay period in which the holiday occurs. 10.Use the following marks on the schedule: X – work eight hours V – vacation H – holiday D – day off 4 – hours for part-time employees 3 PERSONNEL OF HUFFMAN MEMORIAL MEDICAL RECORD DEPARTMENT DAYS (7:00 A.M. – 3:30 P.M.) Director Diane Lucas Assistant Director JoAnn DeWitt Coding 1 Supervisor – Nina Long 3 Coding/PAS Clerks – Cheryl Newman Pam Rogers Janet Bennett Transcription 1 Supervisor – 6 Transcribers – Jessica DuBois Eileen Andrews Iris Williams Diane Henderson Vivian Thomas Lois Fisher Emma Daily Filing/Retrieval 1 Supervisor – 4 Clerks – 1 Part-time Clerk – Bill James Darlene Cook Janice Stivers Larry Patterson Don Williamson Susan Evanston Tumor Registry 1 Supervisor – 1 Clerk – 1 Part-time Clerk – Mabel Smith Pauline Erskine Suzanne Chapman EVENING (3:00 P.M. – 11:00 P.M.) Transcription 1 Part-time – Beth Richman Filing/Retrieval 1 Supervisor – 2 Clerks – 1 Part-time Clerk – Daniel Johnson Harry Skinner Matthew Scott Anne Madison NIGHTS (11:00 P.M. – 7:00 A.M.) Transcription 3 Transcribers – Louise Wilson Jane Matters Nancy Lipman Filing/Retrieval 2 Clerks – Lily Jamison Helen Benson 4 INFORMATION SHEET SCHEDULING EMPLOYEES In addition to the planning, organizing and controlling of a medical record service, managers must accurately plan the work pattern for employees. This plan must insure that all duties are adequately covered, all shifts have sufficient numbers of people to perform duties, and employees are given appropriate days off. Scheduling encompasses both short term and long term plans. Short term scheduling involves planning work on a daily and/or weekly basis. Long term scheduling generally covers a four – to six – week time period, as well as yearly planning for holidays. In larger health care facilities with the medical record service providing 24 hour service, seven days a week, advanced planning is a requisite to a smooth operation. In smaller facilities with shorter hours of service, the schedule is less complex. The number of employees needed for weekend work for those facilities open on weekends is totally dependent upon the weekend workload. A volume of seventy (70) to ninety (90) discharges per day generally requires two (2) medical record clerks to process those discharges, as well as to perform the other daily responsibilities of the medical record service. It is also advisable to schedule a supervisor during the weekend in the event that any problems arise which a clerk might not be able to handle (i.e. medico-legal questions, irate patients or physicians). If you work in a department that has an active work 5 measurement program, valuable scheduling information can be obtained from the data reported. In planning for holidays, it is important to remember to: 1. obtain employee preferences for which holidays they might choose to work; 2. keep track of who has worked which holidays; 3. if a holiday occurs on a Friday or a Monday and the employee must work on the holiday, try to give them a Friday or Monday off to compensate. It is important for you to be fair in terms of assigning employees weekend work and scheduling Holidays. Everyone should share the responsibility equally. If you have all supervisors work one weekend per month, then that schedule should be followed. If you have clerks working every other weekend, then that pattern should be followed consistently. When preparing a schedule it is best to put in all the “givens” first. For example, if you have vacations scheduled for the four weeks you’re preparing, then those should be marked in first. Also included in this category would be employees who do not work weekends (i.e. personnel in the Tumor Registry). Once all work times have been scheduled, you must be certain that an employee receives two (2) days off for every seven (7) days. If an employee works more than forty (40) hours in one (1) week, the facility must pat time-an-a-half for all hours over forty. Some facilities are experimenting with a variety of scheduling techniques: flex time and the four-day work week. Both techniques have been 6 heavily debated. The final questions regarding these nontraditional alternatives end up being: 1. Are your employees willing to try it? 2. Are you ready to handle the extra planning these alternatives may warrant? 3. Do you have the necessary resources, including equipment, to accommodate a nontraditional scheduling alternative? 4. Will administrator of the facility support your proposal? Once you have established answers to those questions you are ready to embark on a new technique of scheduling. Scheduling employees can be one of the most challenging tasks that a manager faces. Whether you elect to try one of the nontraditional alternatives or use the five-day work week, the manager must: 1. be fair; 2. apply all guidelines to every employee consistently 3. utilize all available data to arrive at appropriate numbers for weekend and holiday staffing requirements; and 4. maximize the utilization of equipment and resources.

1 ACTIVITY PURPOSE The purpose of this activity is to give you practice preparing a four-week work schedule. PROCESS Follow the steps listed below to prepare a schedule. 1. Read the Information Sheet: Scheduling Employees. 2. The pay week for this medical record service runs Sunday – Saturday. The pay period is two pay weeks. Each full-time employee cannot work more than 40 hours per pay week, or 80 hours per pay period. Each part-time employee works 20 hours per pay week – 40 hours per pay period. 3. The first Friday of the four – week period is a holiday. 4. The medical record service has 24 hour coverage, seven days a week. All full-time employees work a five day pay week, eight hours per day, with rotating weekend coverage. Part-time employees work four hours Monday – Friday, except for their rotation weekend. On those days they work an eight hour shift. Remember to adjust their time accordingly. 5. The Assistant Director and all supervisors, except the Tumor Registry Supervisor, should be scheduled for rotating weekend coverage. 2 6. All employees, except the Tumor Registry employees, should be scheduled on a rotating basis for weekend coverage. 7. For weekend and holiday coverage, there needs to be at least two clerks and one transcriptionist on days and evenings, one clerk and one transcriptionist at night. 8. The Department Director has scheduled a two – week vacation for the first two full weeks of the four – week schedule. 9. Employees who work holidays must take the holiday time within the pay period in which the holiday occurs. 10.Use the following marks on the schedule: X – work eight hours V – vacation H – holiday D – day off 4 – hours for part-time employees 3 PERSONNEL OF HUFFMAN MEMORIAL MEDICAL RECORD DEPARTMENT DAYS (7:00 A.M. – 3:30 P.M.) Director Diane Lucas Assistant Director JoAnn DeWitt Coding 1 Supervisor – Nina Long 3 Coding/PAS Clerks – Cheryl Newman Pam Rogers Janet Bennett Transcription 1 Supervisor – 6 Transcribers – Jessica DuBois Eileen Andrews Iris Williams Diane Henderson Vivian Thomas Lois Fisher Emma Daily Filing/Retrieval 1 Supervisor – 4 Clerks – 1 Part-time Clerk – Bill James Darlene Cook Janice Stivers Larry Patterson Don Williamson Susan Evanston Tumor Registry 1 Supervisor – 1 Clerk – 1 Part-time Clerk – Mabel Smith Pauline Erskine Suzanne Chapman EVENING (3:00 P.M. – 11:00 P.M.) Transcription 1 Part-time – Beth Richman Filing/Retrieval 1 Supervisor – 2 Clerks – 1 Part-time Clerk – Daniel Johnson Harry Skinner Matthew Scott Anne Madison NIGHTS (11:00 P.M. – 7:00 A.M.) Transcription 3 Transcribers – Louise Wilson Jane Matters Nancy Lipman Filing/Retrieval 2 Clerks – Lily Jamison Helen Benson 4 INFORMATION SHEET SCHEDULING EMPLOYEES In addition to the planning, organizing and controlling of a medical record service, managers must accurately plan the work pattern for employees. This plan must insure that all duties are adequately covered, all shifts have sufficient numbers of people to perform duties, and employees are given appropriate days off. Scheduling encompasses both short term and long term plans. Short term scheduling involves planning work on a daily and/or weekly basis. Long term scheduling generally covers a four – to six – week time period, as well as yearly planning for holidays. In larger health care facilities with the medical record service providing 24 hour service, seven days a week, advanced planning is a requisite to a smooth operation. In smaller facilities with shorter hours of service, the schedule is less complex. The number of employees needed for weekend work for those facilities open on weekends is totally dependent upon the weekend workload. A volume of seventy (70) to ninety (90) discharges per day generally requires two (2) medical record clerks to process those discharges, as well as to perform the other daily responsibilities of the medical record service. It is also advisable to schedule a supervisor during the weekend in the event that any problems arise which a clerk might not be able to handle (i.e. medico-legal questions, irate patients or physicians). If you work in a department that has an active work 5 measurement program, valuable scheduling information can be obtained from the data reported. In planning for holidays, it is important to remember to: 1. obtain employee preferences for which holidays they might choose to work; 2. keep track of who has worked which holidays; 3. if a holiday occurs on a Friday or a Monday and the employee must work on the holiday, try to give them a Friday or Monday off to compensate. It is important for you to be fair in terms of assigning employees weekend work and scheduling Holidays. Everyone should share the responsibility equally. If you have all supervisors work one weekend per month, then that schedule should be followed. If you have clerks working every other weekend, then that pattern should be followed consistently. When preparing a schedule it is best to put in all the “givens” first. For example, if you have vacations scheduled for the four weeks you’re preparing, then those should be marked in first. Also included in this category would be employees who do not work weekends (i.e. personnel in the Tumor Registry). Once all work times have been scheduled, you must be certain that an employee receives two (2) days off for every seven (7) days. If an employee works more than forty (40) hours in one (1) week, the facility must pat time-an-a-half for all hours over forty. Some facilities are experimenting with a variety of scheduling techniques: flex time and the four-day work week. Both techniques have been 6 heavily debated. The final questions regarding these nontraditional alternatives end up being: 1. Are your employees willing to try it? 2. Are you ready to handle the extra planning these alternatives may warrant? 3. Do you have the necessary resources, including equipment, to accommodate a nontraditional scheduling alternative? 4. Will administrator of the facility support your proposal? Once you have established answers to those questions you are ready to embark on a new technique of scheduling. Scheduling employees can be one of the most challenging tasks that a manager faces. Whether you elect to try one of the nontraditional alternatives or use the five-day work week, the manager must: 1. be fair; 2. apply all guidelines to every employee consistently 3. utilize all available data to arrive at appropriate numbers for weekend and holiday staffing requirements; and 4. maximize the utilization of equipment and resources.

Chapter 03 Reading Questions Due: 11:59pm on Friday, May 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Chapter 3 Reading Quiz Question 1 Part A Isotopes of an element differ from each other by the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 2 Part A Which one of the following statements about pH is correct? ANSWER: Correct Lemon juice is an acid. Chapter 3 Reading Quiz Question 17 Part A In which form are water molecules most closely bonded to each other? ANSWER: number of electrons number of neutrons types of electrons number of protons Stomach acid has more OH- ions than H+ ions. Baking soda has more H+ ions than OH- ions. Lemon juice has more H+ ions than OH- ions. Seawater is slightly acidic. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 1 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 16 Part A Which one of the following is a molecule but NOT a compound? ANSWER: Correct Oxygen is a molecule made up of just one element. Therefore, it is not a compound. Chapter 3 Reading Quiz Question 3 Part A Which one of the following is a carbohydrate and one of Earth’s most abundant organic molecule? ANSWER: Correct equally closely bonded in water vapor and ice solid ice forming part of an Antarctic sheet liquid water a few degrees above the freezing point water vapor above a boiling pot of water CH4 O2 CO2 H2O oil protein cellulose DNA Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 2 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 4 Part A Which one of the following is a protein that functions as a catalyst? ANSWER: Correct Chapter 3 Reading Quiz Question 18 Part A The process of translation involves the use of _____. ANSWER: Chapter 3 Reading Quiz Question 5 Part A The cooling effect of sweating best represents _____. ANSWER: glucose cellulose enzyme RNA proteins to make lipids lipids to make carbohydrates carbohydrates to make proteins nucleic acids to make proteins latent heat transfer conduction radiation convection Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 3 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 6 Part A When plants use sunlight in photosynthesis, the plants are using a form of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 8 Part A Which of the following converts mass to energy? ANSWER: Correct Chapter 3 Reading Quiz Question 19 Part A When a windmill turns to generate electricity, the amount of kinetic energy input _____. ANSWER: chemical energy in sunlight nuclear fission electromagnetic radiation conduction conduction the breaking of chemical bonds nuclear fission photosynthesis Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 4 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 20 Part A Which of the following best represents kinetic energy? ANSWER: Correct Chapter 3 Reading Quiz Question 21 Part A Which of the following processes reduces entropy? ANSWER: Correct Chapter 3 Reading Quiz Question 9 is unrelated to the amount of electrical energy produced is more than the amount of electrical energy produced equals the amount of electrical energy produced is less than the amount of electrical energy produced a charged battery gunpowder in a bullet the energy in the wax molecules of a candle a hot burner on a stove burning gasoline in an automobile engine photosynthesis in a leaf a person walking up a flight of stairs cell respiration in a leaf Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 5 of 9 5/21/2014 7:58 PM Part A Which one of the following planets is a gas giant? ANSWER: Correct Chapter 3 Reading Quiz Question 10 Part A What is the main driving force that causes Earth’s tectonic plates to drift? ANSWER: Correct Chapter 3 Reading Quiz Question 23 Part A In which of the following locations would you expect to find large quantities of young rocks? ANSWER: Venus Jupiter Mars Mercury Heat from Earth’s core causes the mantle rock to circulate. The weight of the tectonic plates causes them to sink and melt. Currents of magma from the core of Earth circulate just beneath the tectonic plates. Electromagnetic radiation from the sun heats the tectonic plates, causing them to expand. the Appalachian Mountains the Himalayas deep in the central parts of India the Mid-Atlantic Ridge Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 6 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 12 Part A The oxygen-rich atmosphere of Earth is mainly the result of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 13 Part A A scientist working on the chemical reactions in the ozone layer is studying the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 24 Part A The total amount of moisture in the air is highest when relative humidity is _____. ANSWER: volcanic activity chemical reactions between the early Earth atmosphere and iron photosynthetic organisms erosion of rocks into soil troposphere thermosphere stratosphere mesosphere Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 7 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 15 Part A You are enjoying a spring day but expect a storm to arrive soon . As the storm arrives and the rain begins to fall, you notice that the temperature drops dramatically. Most likely, you have just experienced the arrival of a _____. ANSWER: Correct Chapter 3 Reading Quiz Question 25 Part A Every day tremendous amounts of the sun’s energy strikes Earth. Why doesn’t Earth overheat? ANSWER: Correct Earth’s energy budget is balanced. Over the course of a year, the energy input is equal to the energy output. Chapter 3 Reading Quiz Question 7 low and temperatures are low high and temperatures are high high and temperatures are low low and temperatures are high cold front Hadley cell intertropical convergence stratospheric event The energy is ultimately radiated back to space. Much of the heat melts rocks, forming lava deep inside of Earth. Most of the energy is used in photosynthesis to help plants grow and survive. The energy mostly is absorbed in various weather systems. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 8 of 9 5/21/2014 7:58 PM Part A How many calories are required to heat up 1,000 grams of liquid water (about 1 liter) from 20 °C to 70 °C? ANSWER: Correct Chapter 3 Reading Quiz Question 14 Part A Hadley cells near the Equator consist of _____. ANSWER: Correct Score Summary: Your score on this assignment is 85.5%. You received 19.67 out of a possible total of 23 points. 100 1,000 5,000 50,000 rising dry air associated with deserts and falling moist air that produces precipitation and rainforests rising moist air that produces precipitation and rainforests, and falling dry air associated with deserts warm, moist air rising up the sides of mountains and cool, dry air descending on the leeward sides cool, dry air rising up the sides of mountains and warm, moist air descending on the leeward sides Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 9 of 9 5/21/2014 7:58 PM

Chapter 03 Reading Questions Due: 11:59pm on Friday, May 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Chapter 3 Reading Quiz Question 1 Part A Isotopes of an element differ from each other by the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 2 Part A Which one of the following statements about pH is correct? ANSWER: Correct Lemon juice is an acid. Chapter 3 Reading Quiz Question 17 Part A In which form are water molecules most closely bonded to each other? ANSWER: number of electrons number of neutrons types of electrons number of protons Stomach acid has more OH- ions than H+ ions. Baking soda has more H+ ions than OH- ions. Lemon juice has more H+ ions than OH- ions. Seawater is slightly acidic. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 1 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 16 Part A Which one of the following is a molecule but NOT a compound? ANSWER: Correct Oxygen is a molecule made up of just one element. Therefore, it is not a compound. Chapter 3 Reading Quiz Question 3 Part A Which one of the following is a carbohydrate and one of Earth’s most abundant organic molecule? ANSWER: Correct equally closely bonded in water vapor and ice solid ice forming part of an Antarctic sheet liquid water a few degrees above the freezing point water vapor above a boiling pot of water CH4 O2 CO2 H2O oil protein cellulose DNA Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 2 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 4 Part A Which one of the following is a protein that functions as a catalyst? ANSWER: Correct Chapter 3 Reading Quiz Question 18 Part A The process of translation involves the use of _____. ANSWER: Chapter 3 Reading Quiz Question 5 Part A The cooling effect of sweating best represents _____. ANSWER: glucose cellulose enzyme RNA proteins to make lipids lipids to make carbohydrates carbohydrates to make proteins nucleic acids to make proteins latent heat transfer conduction radiation convection Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 3 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 6 Part A When plants use sunlight in photosynthesis, the plants are using a form of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 8 Part A Which of the following converts mass to energy? ANSWER: Correct Chapter 3 Reading Quiz Question 19 Part A When a windmill turns to generate electricity, the amount of kinetic energy input _____. ANSWER: chemical energy in sunlight nuclear fission electromagnetic radiation conduction conduction the breaking of chemical bonds nuclear fission photosynthesis Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 4 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 20 Part A Which of the following best represents kinetic energy? ANSWER: Correct Chapter 3 Reading Quiz Question 21 Part A Which of the following processes reduces entropy? ANSWER: Correct Chapter 3 Reading Quiz Question 9 is unrelated to the amount of electrical energy produced is more than the amount of electrical energy produced equals the amount of electrical energy produced is less than the amount of electrical energy produced a charged battery gunpowder in a bullet the energy in the wax molecules of a candle a hot burner on a stove burning gasoline in an automobile engine photosynthesis in a leaf a person walking up a flight of stairs cell respiration in a leaf Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 5 of 9 5/21/2014 7:58 PM Part A Which one of the following planets is a gas giant? ANSWER: Correct Chapter 3 Reading Quiz Question 10 Part A What is the main driving force that causes Earth’s tectonic plates to drift? ANSWER: Correct Chapter 3 Reading Quiz Question 23 Part A In which of the following locations would you expect to find large quantities of young rocks? ANSWER: Venus Jupiter Mars Mercury Heat from Earth’s core causes the mantle rock to circulate. The weight of the tectonic plates causes them to sink and melt. Currents of magma from the core of Earth circulate just beneath the tectonic plates. Electromagnetic radiation from the sun heats the tectonic plates, causing them to expand. the Appalachian Mountains the Himalayas deep in the central parts of India the Mid-Atlantic Ridge Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 6 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 12 Part A The oxygen-rich atmosphere of Earth is mainly the result of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 13 Part A A scientist working on the chemical reactions in the ozone layer is studying the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 24 Part A The total amount of moisture in the air is highest when relative humidity is _____. ANSWER: volcanic activity chemical reactions between the early Earth atmosphere and iron photosynthetic organisms erosion of rocks into soil troposphere thermosphere stratosphere mesosphere Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 7 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 15 Part A You are enjoying a spring day but expect a storm to arrive soon . As the storm arrives and the rain begins to fall, you notice that the temperature drops dramatically. Most likely, you have just experienced the arrival of a _____. ANSWER: Correct Chapter 3 Reading Quiz Question 25 Part A Every day tremendous amounts of the sun’s energy strikes Earth. Why doesn’t Earth overheat? ANSWER: Correct Earth’s energy budget is balanced. Over the course of a year, the energy input is equal to the energy output. Chapter 3 Reading Quiz Question 7 low and temperatures are low high and temperatures are high high and temperatures are low low and temperatures are high cold front Hadley cell intertropical convergence stratospheric event The energy is ultimately radiated back to space. Much of the heat melts rocks, forming lava deep inside of Earth. Most of the energy is used in photosynthesis to help plants grow and survive. The energy mostly is absorbed in various weather systems. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 8 of 9 5/21/2014 7:58 PM Part A How many calories are required to heat up 1,000 grams of liquid water (about 1 liter) from 20 °C to 70 °C? ANSWER: Correct Chapter 3 Reading Quiz Question 14 Part A Hadley cells near the Equator consist of _____. ANSWER: Correct Score Summary: Your score on this assignment is 85.5%. You received 19.67 out of a possible total of 23 points. 100 1,000 5,000 50,000 rising dry air associated with deserts and falling moist air that produces precipitation and rainforests rising moist air that produces precipitation and rainforests, and falling dry air associated with deserts warm, moist air rising up the sides of mountains and cool, dry air descending on the leeward sides cool, dry air rising up the sides of mountains and warm, moist air descending on the leeward sides Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 9 of 9 5/21/2014 7:58 PM

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During deployment processing, soldiers undergo medical screening.After a wait, each soldier see same dicaltechnician who reviews his or her records.If the records review shows no problems—the soldier is physically qualified to deploy—then the soldier departs to the next step in deployment processing.It he records do reveal a potential medical issue,the soldier instead sees a doctor,who assesses the soldier’s condition and determines both deployability and for those who are medically disqualified, treatment needs.In rare cases ,the doctor in itiate a medical board to evaluate the soldier for retention in the military. Currently, 80 soldiers per hour arrive for deployment screening,and 80% of them pass there cords review.On average,20 people are waiting for the medical records review,which takes 6 minutes. When the records review indicates a soldier must see a doctor,the soldier reports to a waiting room,where an average of 8 soldiers are waiting. After a wait,the soldier sees a doctor, who reviews the soldier’s condition and either approves the soldier for deployment(75%of the time) or disapproves deployment and conducts a morein-deptxeam to determine treatment(20%of the time)or the need for a medical board(5%ofthetime).Each doctor’s exam takes,on average,6minutes if the solider is medically able to deploy(the doctor pretty much replicates the records review),15 minutes if the soldier requires some kind of treatment,and 30 minutes in those rare cases that require the doctor to initiate a medical review board. Assume the process is stable;that is,average inflow rate equals average outflow rate.[Finally,thisisNOTaqueuingproblem.] a. On average,how long does a soldier spend in the deployment process?

During deployment processing, soldiers undergo medical screening.After a wait, each soldier see same dicaltechnician who reviews his or her records.If the records review shows no problems—the soldier is physically qualified to deploy—then the soldier departs to the next step in deployment processing.It he records do reveal a potential medical issue,the soldier instead sees a doctor,who assesses the soldier’s condition and determines both deployability and for those who are medically disqualified, treatment needs.In rare cases ,the doctor in itiate a medical board to evaluate the soldier for retention in the military. Currently, 80 soldiers per hour arrive for deployment screening,and 80% of them pass there cords review.On average,20 people are waiting for the medical records review,which takes 6 minutes. When the records review indicates a soldier must see a doctor,the soldier reports to a waiting room,where an average of 8 soldiers are waiting. After a wait,the soldier sees a doctor, who reviews the soldier’s condition and either approves the soldier for deployment(75%of the time) or disapproves deployment and conducts a morein-deptxeam to determine treatment(20%of the time)or the need for a medical board(5%ofthetime).Each doctor’s exam takes,on average,6minutes if the solider is medically able to deploy(the doctor pretty much replicates the records review),15 minutes if the soldier requires some kind of treatment,and 30 minutes in those rare cases that require the doctor to initiate a medical review board. Assume the process is stable;that is,average inflow rate equals average outflow rate.[Finally,thisisNOTaqueuingproblem.] a. On average,how long does a soldier spend in the deployment process?

6+0.75*6+.2*15+.05*30 = 15 miutes     Timeindeploymentsystem:15 (minutes)
Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

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Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 57 5/9/2014 8:02 PM Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 57 5/9/2014 8:02 PM Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? = = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 57 5/9/2014 8:02 PM Hint 1. What is the general quadratic equation? The general quadratic equation is , where , , and are constants. Depending on the value of the discriminant, , the equation may have two real valued 1. solutions if , 2. one real valued solution if , or 3. two complex valued solutions if . In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. No; there is no chance he is going to get aboard. Yes; he will get a second chance Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 57 5/9/2014 8:02 PM ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: C and E E and F A and F C and D B and D C and D A and F E and F A and B E and D Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 57 5/9/2014 8:02 PM Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 2. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. ANSWER: A and F A and E D and B C and D E and F Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 57 5/9/2014 8:02 PM Correct Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. = 5 positive negative Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 57 5/9/2014 8:02 PM ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth’s gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, , is constant. 1. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of its velocity, , changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time corresponds to the moment just after the ball is launched from position and . Its launch velocity, also called the initial velocity, is . Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time with velocity . Its position at this moment is denoted by or since it is at its maximum height. The other point, at time with velocity , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is , also known as ( since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence . Part A , = -2,-5 , Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 57 5/9/2014 8:02 PM How do the speeds , , and (at times ,

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 57 5/9/2014 8:02 PM Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 57 5/9/2014 8:02 PM Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? = = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 57 5/9/2014 8:02 PM Hint 1. What is the general quadratic equation? The general quadratic equation is , where , , and are constants. Depending on the value of the discriminant, , the equation may have two real valued 1. solutions if , 2. one real valued solution if , or 3. two complex valued solutions if . In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. No; there is no chance he is going to get aboard. Yes; he will get a second chance Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 57 5/9/2014 8:02 PM ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: C and E E and F A and F C and D B and D C and D A and F E and F A and B E and D Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 57 5/9/2014 8:02 PM Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 2. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. ANSWER: A and F A and E D and B C and D E and F Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 57 5/9/2014 8:02 PM Correct Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. = 5 positive negative Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 57 5/9/2014 8:02 PM ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth’s gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, , is constant. 1. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of its velocity, , changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time corresponds to the moment just after the ball is launched from position and . Its launch velocity, also called the initial velocity, is . Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time with velocity . Its position at this moment is denoted by or since it is at its maximum height. The other point, at time with velocity , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is , also known as ( since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence . Part A , = -2,-5 , Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 57 5/9/2014 8:02 PM How do the speeds , , and (at times ,

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