Learning Healthcare Organizations Essay Example for Free

Learning Healthcare Organizations Essay There are two healthcare organizations that I will be discussing that have transformational change to promote/create learning organization. One is the Centers for Disease Control and Prevention (CDC), and the other one is International Agency for Research on Cancer (IARC). CDC is a federal agency under the Department of Health and Human Services that focuses national attention on developing and applying disease control and prevention. CDC collaborates to create the expertise, information, and tools that people and communities need to protect their health through health promotion, prevention of disease, injury and disability and preparedness for new health treats. Stakeholders at CDC are people invested in the program that are interested in the results of the evaluation, and/or with a stake in what will be done with the results of the evaluation. Representing their needs and interests throughout the process is fundamental to good program evaluation. Those involved in program operations are the management, program staff, partners, funding agencies and coalition members. Those served or affected by the program are patients or clients, advocacy group, community members, and elected official. And lastly, those who are intended users of the evaluation findings are persons in a position to make decisions about the program, such as partners, funding agencies, coalition members, and the general public or taxpayers. The Centers for Disease Control and Prevention (CDC) continues its long standing dedication to improving the health and wellness of all Americans with the Community Transformation Grant (CTG) program. The CTG program is funded by the Affordable Care Act’s Prevention and Public Health Fund and  awarded $103 million to 61 states and local government agencies, tribes, and territories, and nonprofit organizations in 36 states, along with nearly $4 million to 6 national networks of community-based organizations. Focusing on priorities for change for healthier living is improving health and wellness on tobacco-free living, active living and healthy eating, and high impact quality clinical and other preventive services to prevent and control high blood pressure and high cholesterol. Also, focusing on disease prevention and health promotion that includes social and emotional wellness and healthy and safe physical environments, which facilitate the early identification of mental health needs and access to quality services. Specific community interventions includes; promotes healthy eating by supporting local farmers and developing small grocery stores where people live, protecting people from secondhand smoke exposure, improving community environments to make it safe and easy for people to walk and ride bikes. The International Agency for Research on Cancer (IARC) is part of the World Health Organization. It coordinates and conducts both epidemiological and laboratory research into the causes of human cancer. IARC main objectives are; to monitor global cancer occurrence, identify the causes of cancer, elucidate the mechanism of carcinogenesis, and develop scientific strategies for cancer control. On February 3, 2014, the International Agency for Research on Cancer (IARC) released World Cancer Report 2014, a collaboration of over 250 leading scientist from more than 40 countries, describing multiple aspects of cancer research and control. The report says about half of all cancers could be avoided if current knowledge was adequately implemented. The stakeholders are the scientist’s that has been researching for the cure of different types of cancer; patient’s that are suffering and waiting for the cure, and the leadership of the World Health organization that implements the research. The IARC activities are mainly funded by the regular budget contributions paid by its participating states. The regular budget for the 2014-2015 biennium was approved in May 2013 at a level of 40 424 491 EUR. Recent changes in the epidemiology of head and neck cancer has new findings.  Overall, the incidence of head and neck cancer is increasing in women, whereas it is decreasing in men. Chewing tobacco is a newly recognized risk factor of great public health concern. The role of tobacco smoking and alcohol as the source of cancer has been reinforced. Head and neck cancer among women in developing countries should deserve more attention, as the mortality rates appears to be higher than those of women in developed countries. For never smokers and never drinkers, more research needs to be done to identify their risk factor patterns. While it is true that advances is medical science have led to continued improvements in medical care and health outcomes, the effectiveness of management options remains inadequate for informed medical care and health policy decision making. Frequently, the result is below an optimal level or standard and inefficient care as well as unsustainable cost. In order to maintain quality of care and cost containment, evidence of comparative clinical and cost effectiveness is necessary for healthcare organization. Examples of healthcare organization that I previously discussed have the institutional lessons learned from the process that is learn along the way. As Feinstein said â€Å"a strategic plan is not worth the paper it is printed on unless its underlying vision is embedded in the organization’s culture, (Feinstein W.L. The Institutional Change Process). The most essential element of organizational change is the alignment of all relevant stakeholders to the new directions. The following are critical to achieving momentum and the successful implementation of a vision for change such as: updating the executive’s leadership style, increasing staff involvement in achieving organizational plans, helping the board understand the scope of the change, and strengthening the agency-federation relationship. Enthusiasm, persistenc e, and commitment for change by the leadership are key. References Centers for Disease Control and Prevention. (2009). Prevention and control of seasonal influenza with vaccines. Recommendations of the Advisory Committee on Immunization Practices (ACIP), 2009. MMWR Early release, 58(Early release), 1-54. Chang, S., Collie, C. L. (2009). The future of cancer prevention: will our workforce be ready? Cancer Epidemiology Biomarkers Prevention, 18(9), 2348-2351. Feinstein, W. L. The Institutional Change Process: Lessons Learned Along the Way. Journal of Jewish Communal Service. Jewish Communal Service Association of North America (JCSA), 1999. James, J. (2009). Health Organizations Theory, Behavior, and Development: 273 Saudbery Jones and Bartlett Publishers. Oreg, Shaul; Berson, Yair. Personnel Psychology. Autumn2011, Vol. 64 Issue 3, p627-659. 33p. 1 Diagram, 2 Charts, 1 Graph. DOI: 10.1111/j.1744-6570.2011.01221.x. , Database: Business Source Elite Weiner, B. J. (2009). A theory of organizational readiness for change. Implement Sci, 4(1), 67.

Environmental Impact Assessment (EIA)

Environmental Impact Assessment (EIA) 1. Introduction Environmental Impact Assessment (EIA) is a procedure that requires developers to follow in order to be granted permission for a development and was first introduced in European Union (EU) in 1985 (Glasson, 1999). The guidelines and requirements of EIA come from a European Directive (85/33/EEC as amended by 97/11/EC). In this process, developer requires to compile an Environmental Statement (ES) where significant impacts and its effects on the environment as a result of a development are described including mitigation measures (Lee, 1995). However, there are weaknesses in EIA process. As a result of EIA weaknesses, Strategic Environmental Assessment (SEA) was introduced. SEA process was first introduced by EU Directive 2001/42/EC which environmental protection and sustainable development may be considered. It ensures that potential impact of proposed plans, policies and programmes on the environment are taking into consideration so that mitigation and communication between public and decision-makers are provided (Benson, 2003). Therefore, SEA is undertaken in the decision-making process of a development much earlier than EIA. In this section, the differences between EIA and SEA as well as the weaknesses of EIA that led the development of SEA in EU Directive in 2001 are being discussed. 2. Differences between EIA and SEA 2.1. Differences in provisions between SEA and EIA Directives Table 1: Summary of differences in action for EIA and SEA processes (Partidà ¡rio, 2000). SEA EIA Nature of action It is more strategic and contains visions and concepts in its action Actions towards the construction and operation level Assessment Involved evaluation Involved only assessment Focus More critical decision moments (decision windows) along with decision processes Only in project level Level of decision It involves policy and planning Only project level Relation to decision Facilitator Involved evaluator which often taking into consideration of administrative requirement Alternatives Broader and spatial balance of location, social and physical strategies, technologies and economics. More specific alternative in construction, operation, locations and design Scale of impacts Macroscopic involved local, regional, national and global Microscopic mainly involved local Scope of impacts Sustainability issues, economic and social issues may be more tangible than physical or ecological issues Environmental with a sustainability focus, physical or ecological issues, and also social and economic Time scale and review Long to medium term (after 5 years and then 7 years interval) Medium to short-term (after 5 years then silent continuing review). Key data sources State of the Environment Reports, Local Agenda 21, statistical data, policy and planning instruments Field work, sample analysis, statistical data Data Mainly descriptive and mixture with quantifiable More quantifiable Rigor of analysis (uncertainty) More uncertainty and less rigor Less uncertainty and more rigor Assessment benchmarks Sustainability benchmarks (criteria and objectives) Best practice and legal restrictions Public perception Vague/ distant More reactive Post-evaluation Other strategic actions and project planning Objective evidence in relation with construction and operation 2.2 Differences in procedural requirements of the EIA and SEA Directives. Table 2: Summary of the main differences between EIA and SEA Directives in procedural requirements ( Sources from : Sheate et al., 2005). STAGE SEA EIA Screening It requires consultation from the environmental authorities. Publicity: SEA does not need determination and reasons. It requires no consultation. Publicity: EIA requires determination and reasons. Environmental information/report Member States (MSs) have to ensure that sufficient quality and stronger emphasis on alternatives are provided in environmental reports (ERs). No quality control requires in EIA and only minimun information shoud be provided Consultation Involved relevant parties other than MSs such as public and autorities. Involved relevant parties other than MSs such as public and autorities and also consultation provision from Public Participation Directive. Decision-making All consultation comments and ER are to be taken into account. Consultation comments and environmental informations are included in decision-making. Info on decision More specific and detailed requirements. Information provision from Public Participation Directive are mademore specific in the requirement. Monitoring Long term monitoring required Not require monitoring 3. Weaknesses of Environmental Impact Assessment (EIA) 3.1. Lack of consideration of cumulative impacts EIA process is often facing difficulty in addressing cumulative impacts of a development. The significant impacts of a development especially issues on biodiversity, human health and cultural heritage are no included in their assessment (McDonald Brown 1995). For example in Scotland, several developments of wind farms proposed in close proximity have led to a very complicated EIA process (Glasson, 1999). Each developer required assessing the cumulative impact of landscape and visual application with those of neighboring projects (Benson, 2003). The planning process could face a delay due to this issue. 3.2. Insufficient public participation Public involvement has shown to be insufficient in EIA process (Gailus, 1995). In a recent research suggested that this is due to the attitude of the developer that discourages the participation of public in EIA process in the European Community. Due to lack of legislation and extensive for public involvement in Europe in the early 90s to influence the decision of a development, the general public is not aware of their rights and causes them not interested in the development involvement (Caddy, 1996). 3.3. Little monitoring and auditing process Previous study has showed that monitoring and auditing issues are still weak in EIA. Follow-up process is only performed by developers in a minority of cases (Arts and Nootebloom 1999). Monitoring process enables practioners to have better understanding for future extension, design and restoration projects (Frost, 1997). However, the river restoration process in United Kingdom (UK) was hampered by lack of monitoring process in EIA. 3.4. Inadequate consideration of alternatives The compilation of alternatives appears to be very limited in EIA report (Tesli, 2002). For example in Hungary for radioactive disposal, the report did not provide in details or rather limited in reducing the impacts of the project to the human health (Benson, 2003). It is important to include various alternatives to allow solutions being taken in a development. 3.5. The poor quality of environmental impact statements and reports The reports produced in EIA are often too complex in term of length and technical which is not easily understood by the public and decision makers (Lee, 1995). It is important for ES to be simple as it has to be made available to the public. 3.6. The timing of decisions The decision-making process in EIA project enters too late where the effects of policy and planning critical decisions are not being considered (Lee, 1995). This is due to the absence of systematic impact assessment process where the outcome of it could subsequently influence the project planning and design (Harrop Nixon 1999). 4. Conclusion There were various weaknesses have been identified in EIA process. Due to these weaknesses, SEA was developed in 2001 under EU Directive to strengthen the environmental assessment process. 5. References Arts, J. and Nootebloom, S. (1999) ‘Environmental Impact Assessment Monitoring and Auditing in: Petts, J. (ed.) Handbook of Environmental Impact Assessment Volume 1, Blackwell, Oxford: 229-251 Benson, J.F. (2003) ‘What is the alternative? Impact assessment tools and sustainable planning, Impact Assessment and Project Appraisal, 21 (4): 261-266 Caddy, J. (1996). Working Group on Environmental Studies, European University Institute, Florence. [Online] http://www.iue.it/WGES/Iss16/caddy.htm [Accessed: 29/01/2010]. Frost, R. (1997) Planning and Environmental Impact Assessment in Practice. Chapter 7 EIA monitoring and audit in Weston, J (ed). Longman, Harlow. pp 141 175. Gailus, J. (1995). Regional Environmental Centre: Hungary. [Online] http://www.rec.org/REC/Bulletin/Bull52/PublPart.html [Accessed: 29/01/2010] Glasson, J. (1999) ‘The First 10 Years of the UK EIA System: Strengths, Weaknesses, Opportunities and Threats, Planning Practice and Research, 14 (3): 363-375 Glasson, J. Therivel, R. and Chadwick, A. (1999) Introduction to Environmental Impact Assessment, Spon Press, London Harrop, O. and Nixon, A. (1999) Environmental Impact Assessment in Practice, Routledge, London Lee, N. (1995) ‘Environmental Assessment in the European Union: a tenth anniversary, Project Appraisal, 10 (2): 77-90 McDonald, G.T. and Brown, A.L. (1995) ‘Going Beyond Environmental Impact Assessment: environmental input to planning and design, Environmental Impact Assessment Review, 15: 483-495 Partidà ¡rio, M.R., 2000, Elements of an SEA framework improving the added-value of SEA, Environmental Impact Assessment Review, 20: 647-663. Sheate, W. Byron, H. Dagg, S. Cooper, L (2005), The Relationship between SEA and EIA Directives: Final Report to the European Commission. Imperial College London Tesli, A. (2002). The use of EIA and SEA relative to the objective of sustainable development, Norwegian Institute for Urban and Regional Research (NIBR). 1. Content of an SEA report as required by EU Directive The implementation of plans and programmes (PPs) in which likely significant effects produce by the project on the environment is the key requirement in preparation of an SEA report (European Parliament and Council of the European Union, 2001). Figure 1 shows the summary of the contents required by EU Directive (2001/42/EC) in producing SEA reports. Not all projects need to perform SEA process (Barth Fuder, 2002). The diagram in Figure 2 shows a set of set of criteria for application to PPs under the EU Directive (2001/42/EC). It specifies whether SEA is required or not according to the Directive. For simplicity, the developments of the PPs and reasonable alternative options of SEA are summarized in five key stages according to the government guidance in England (URL 1). 1.1. The key five stages Stage A: Context, Baseline and Scoping (SEA Directive Annex 1) Authority needs to include indicators, objectives and background information for SEA in the plan. The decisions of the scope can be decide by the authority including consultation on the statutory environmental bodies (URL 1). Stage B:Alternatives and Assessment (SEA Directive Article 5.1) Authority need to identify reasonable alternatives and assess the effects of the project on the environment. Ways of reducing, preventing and offsets the effects have to be provided as well (URL 1). Stage C: Preparing the Environmental Report Draft plan/programme which includes the information of the effects has to be presented as a key output of SEA process (URL 1). Stage D: Consultation (SEA Directive Article 6.2 and Annex 1) The draft plan and environmental report should be ready together for consultation where a statement are made from the consultation responses in order to produce an evolving plan (URL 1). Stage E: Monitoring (SEA Directive Article 10.1) The implementation of the plan where environment effects are produced needs monitoring process. It helps to provide more baseline information for future plans (URL 1). 2. Difficulties and limitations in fulfilling these requirements Table 1: Summary of the difficulties and limitations of SEA reports as required in EU Directive. Requirements issues Difficulties and limitations Availability and access to data Environmental data is often limited and not relevant because it is not collected and stored systematically. The process of data collection requires extensive resources and using these data are difficult because different departments tend to collect different set of data. The quality of good data is lacking and this issue has been reported by Member States such as Germany and Poland (European Commission, 2009). Best example of this issue is Poland. They are facing difficulties of generating and collecting data of affected area because of the implementation of a plan/programme. The current picture of the environment has to be identified especially in large areas but they indicate that it is very problematic (European Commission, 2009). Sometimes, the coverage areas of SEA are large (including few countries and produces large amount of alternatives (URL 1). This will increase the complexity of data collection and analysis (URL 1). Public Participation The availability of the data for the public is limited. In the UK, documents and information of the plan are not required to be published on their website until ER is finalized according to the draft Regulations (Partidà ¡rio, 1996). As a result, public participation is limited as not many public will travel to the plan-makers office to view the documents at a minimum time period for consultation. It is important to set up a website to facilitate the consultation process for the public to participate (Partidà ¡rio, 1996). Else, public is unaware that inspection of these documents are available to them and no feedback can be made (Kà ¸rnà ¸v Thissen, 2000). EA at higher levels of decision making As SEA involves higher levels of decision making, the implementation policy of PPs are subject to various departments decisions (Kà ¸rnà ¸v Thissen, 2000). For example, a Local Transport Plan requires policy from Regional Spatial Strategies, Aviation and Transport White Papers and Sustainable Communities Plan and Planning Policy Statements (Brown Thà ©rivel, 2000). Due to these requirements, a complex screening process has to be performed and decisions for PPS are even more very difficult in the assessment (Brown Thà ©rivel, 2000). Deciding on the level of detail of the environmental report The details of require information in SEA reports are vary due to lack of adaptation in assessment for the level of abstraction in PPs (European Commission, 2009). According to Member States (Latvia and Germany), the possible impacts of PPs are difficult to be included in SEA reports because of less information of the right scale and level required (European Commission, 2009). Therefore, important information for long term PPs is difficult as no appropriate spatial scale of information need to be included in the report (European Commission, 2009). Development of assessment methods As there are no specific guidelines, strong methodological background and lack of exchange for best practices, developing an effective assessment is very challenging (Brown Thà ©rivel, 2000). For example in Operational Programmes objectives, high level plans in a viable assessment do not necessary show the actual physical ground effects although strategic policies are implemented European Commission, 2009). Assessment of impacts Although SEA addressing the importance of cumulative impacts, there is no standard and effective assessment methodologies are being developed (Partidà ¡rio, 1996). The significant environmental impacts of PPs are difficult to assess and the identification of these aspects are limited (Partidà ¡rio, 1996). Monitoring and enforcement (Including issues of indicators) The assessment of plans is limited because there is no sustainability and environmental criteria developed in the monitoring programme (European Commission, 2009). Therefore, monitoring indicators (local agenda 21 for instance) is being used however it is difficult for monitoring process as mentioned by Member States like France. Environmental authorities have no proper enforcement tool to ensure that monitoring programme is being performed (European Commission, 2009). For example in the UK, no quality control body is being set up by the government to monitor the efficiency of monitoring process which is a limitation for SEA (Verheem, R. Tonk, J. 2000). Institutional and legal issues The supports for SEA process are still insufficient politically (European Commission, 2009). The bureaucratic prerogatives may hinder the effectiveness of SEA performance. As SEA process is relatively new, lack of human resources especially knowledgeable authorities is a major limitation for SEA (European Commission, 2009). 3. Comparison between requirement of SEA Directive (2001/42/EC) and Sustainability Appraisal (SA). In the United Kingdom (UK), SA and SEA are required in planning system and law for Government Plans and Programmes (Smith Sheate 2001). For most Development Plan Documents (DPD) and Supplementary Planning Documents (SPD), both SA and SEA process have to be carried out and include in the Local Development Framework (LDF) in the UK (Smith Sheate 2001). SA was developed to assess the likely economic, social and environmental impacts so proposed PPs can be implemented that leads to sustainable development unlike SEA which was previously described (Lee Kirkpatrick, 2000). The Planning and Compulsory Purchase Act (2004) and European Directive EC/2001/42 require both SA and SEA processes to be performed in any planning (Smith Sheate 2001). It is also requires by the Environmental Assessment Regulations for Plans and Programmes in UK (Smith Sheate 2001). In table 2, comparison of SA and SEA in terms of UK planning system for Government PPs are being summarized. Table 2: Comparison between SA and SEA requirements within the UK planning system Requirements Strategic Environmental Assessment Sustainability Appraisals Overall aims The aim of SEA is to raise the profile of environmental considerations as part of an advocative approach during decision-making process (Kà ¸rnà ¸v Thissen, 2000). In contrast to SEA, it is use as a support process in decision-making and representing an integrated approach that working towards in all aspects of sustainable development. Therefore, the interests at stake are remained neutral during this process (Minas, 2002). Focus Environmental effects (Lee Kirkpatrick, 2000) A full range of environmental, social and economic issues (Minas, 2002). Environmental/ Sustainability aspects It involves 15 components which is suggested in the 1993 guidance with additional social and economic factors to be considered in 1999 guidance (Thà ©rivel, Minas, 2002). Aspects of biodiversity, human health, cultural factors, water, landscape, population and material assets are considered primarily (Thà ©rivel, Minas, 2002). Report requirements There are no formal requirements for SA. In good practice guidance, identification of scoping and impacts stages is recommended (Thà ©rivel, Minas, 2002). The 1999 guidance also recommends that planners should provide and evaluates alternatives. Environmental baseline conditions should be considered as well. It involved extensive requirements of Annex 1 which have previously discussed (Barth Fuder, 2002). Methodology According to DETR Guide, the appraisal should emphasize on strategic options, alternatives, and policy impact matrices in achieving sustainable development. Therefore, the methodological statement is very brief if compare to SEA (Minas, 2002). In contrast to SA, heavy emphasis is in place on actually baseline data which set as a benchmark to assess the alternatives performance. It is also requires authorities to consult the final environmental report from the public on the scope of the assessment (Thà ©rivel, Minas, 2002). Timing The process is being carried out very early where every stage of the development plan process is considered as an important element (Thà ©rivel, Minas, 2002). In contrast to SA, it is usually being carried out during the preparation of a plan before the submission to the legislative procedure (Thà ©rivel, Minas, 2002). Involvement The appraisals are subjected to consultation with outside groups such as public consultation during the plan preparation. The appraisal usually made available on the internet where it is being carried out sporadically (Thà ©rivel, Minas, 2002). It is not required to make available on the internet. The consultation can be done during scoping stage by specified environmental authorities. Opportunity has to be given to the public to comment the draft plan (Partidà ¡rio, 1996). Documentation required No formal requirements according to the guidance A statement need to be produced by an authority to summarize all the considerations of the plan have been integrated. They need to provide a report of consultees opinion which taken during the consultation process and valid reasons on why the alternatives are being chosen (Partidà ¡rio, 1996). 4. Success of SEA report in delivering sustainable development objectives A wide range of processes has been integrated with SEA report. It has shown to be a systematic process where accountable decision making can be achieve due to the earlier evaluating process being taken with strong alternative visions (Wood Dejeddour, 1992). All these have incorporated in SEA policy, planning and program initiatives (PPPs) to ensure sustainable development with full integration of economic, social and political considerations (Partidà ¡rio Clarke, 2000). Although SEA shows to be a great tool in project level but the process of SEA is not easily accepted or consider as an effective solution due to its complexity (Thà ©rivel Partidà ¡rio, 2000). Recent research has shown that SEA produces both great opportunities as well as failures in sustainable development (Sadler, 1998). 4.1. Sustainability objectives are included in the integrated process of policy making and planning During the design stage of SEA, objectives of sustainability are being considered which provides a greater plan and policy in decision making (Thompson et al., 1995). It gradually delivers its objectives of sustainable development because the use of SEA articulates sustainability goals by enhancing the political action where substantive action can be taken (Partidà ¡rio, 1996). As the expectation of internal and external public of its delivery increasing, it offers the possibility of bringing better policy towards sustainable development into success. The transition of SEA involved two main steps: Information gathering and analysis work The core of strategic decision making Sustainable development consideration is being covered in a broader range where strategic level matters are separated from advisory conventionally-focused (Pezzoli, 1997). Sustainability criteria are used as the key bases to help in strategic decision making on the selection among best options available (Pezzoli, 1997). With these steps being taken, larger context of core policies and programmes in strategic assessment are identified in pursuing the objectives of national sustainability. Therefore, SEA report may be an important instrument in promoting sustainable development when it is fully integrated (Partidà ¡rio, 1996). 4.2. Operationalises sustainability principles In practical application, the concept of sustainability is very difficult because it faces high complexity and uncertain reality (Pezzoli, 1997). Interconnected generational boundaries and disciplinary of sustainability may further complicates the concept (Marsden, 1998). Therefore, the application of specific context of sustainability and commonly recognized principles are being clarified by SEA. SEA identified three key principle of sustainability: Integrated pursuit of ecological and socio-economic improvements Uncertainty imposes precautionary obligations Public choices involved SEA is a visible confirmation of commitment to sustainability as it offers broader exposure to notions such as natural capital and the precautionary principle (Pezzoli, 1997). Therefore, it can be translated into the language of politics of sustainability and functions as a heuristic device (Thà ©rivel et al., 1992). 4.3. Improves analysis of broad public purposes and alternatives With alternative technologies, lifestyle choices and better resources, SEA offers better possibility in achieving sustainable development because it has the capability to contemplating these factors that cannot be address at lower levels (Wood Dejeddour, 1992). Therefore, SEA report is the most effective and efficient point in finding alternatives and addressing the needs in pursuing the objectives of sustainability development (Thà ©rivel et al., 1992). 4.4. Facilitates proper attention to cumulative effects Strategic level proves to be the best way to deal with increasing number of cumulative impacts. The scope of SEA helps to identifying these impacts because of its space scales (Ortolano Shepherd, 1995; Scott, 1992). As SEA is performed in an early stage, this assessment allows assessors to provide more attention in a wider range of actions in a larger area. It allows them to provide a broader context of cumulative impacts in addressing each of the impacts (Thà ©rivel Partidà ¡rio 1996). According to Thà ©rivel Partidà ¡rio, undesirable activities as a results of a project can be removed before the project stage begins because these cumulative impacts have influence the project decision where SEA has incorporated environmental issues intrinsically during the planning stage. Earlier detection of these impacts helps to promote sustainable development. 4.5. Facilitates greater transparency and more effective public participation at the strategic level With the extensive involvement of public participation in SEA, it has improved the credibility and accountability of SEA in sustainability where they facilitating external scrutiny of decision (Wood Dejeddour, 1992). It creates increasing pressures in strategic decision making process to overcome bureaucratic fragmentation because in many jurisdictions, bureaucratic disorganisation and wastefulness in citizen has been declining. The expanding role of public provided in SEA allows an intrinsic connection between environmental sustainability and equity (George, 1999) to promote basic sustainability goals. 5. Conclusion There are many challenges ahead for SEA report. There are many difficulties in producing a good SEA report. Nevertheless, SEA report requirements did show success in achieving sustainable development goals. In order to ensure SEA report success, weaknesses and limitation needs to be considered and solution needed in order to overcome it. 6. References Barth, R. Fuder, A. (2002) Implementing Article 10 of the SEA Directive 2001/42/EC. Final Report : Freiburg, Darmstadt, Berlin. Brown, A L, and Thà ©rivel, R. (2000), â€Å"Principles to guide the development of strategic environmental assessment methodology†, Impact Assessment and Project Appraisal , 18(3), September, pages 183-189. Environment Agency (2004) SEA Good Practice Guidelines www.environment-agency.gov.uk/seaguidelines. Accessed on 27/01/10. European Parliament and Council of the European Union (2001) Directive 2001/42/EC on the assessment of the effects of certain plans and programmes on the environment Commission of the European Communities, Brussels. www.europa.eu.int/eur-lex/pri/en/oj/dat/2001/l_197/l_19720010721en00300037.pdf European Commission (2009) Study concerning the report on the application and

Definition and Features of Monopoly and Competition

Definition and Features of Monopoly and Competition Definition of Monopoly Monopoly is a well defined market structure where there is only one seller who controls the entire market supply, as there are no close substitutes for his product and there are no barriers to the entry of rival producers. This sole seller in the market is called â€Å"monopolist†. The term monopolist is derived from the Greek word â€Å"mono†, meaning â€Å"single†, and â€Å"polist† meaning seller. Thus the monopolist may be defined as the sole seller of a product which has no close substitutes. The monopolist is faced by a large number of competing buyers for his product. Evidently monopoly is the antithesis of competition on. In a monopoly market, the producer, being the sole seller, has no direct competitors in either the popular or technical sense. Thus, the monopoly market model is the opposite extreme of competition. Features of Monopoly The features of a monopoly are: The monopolist is the sole producer in the market. Thus, under monopoly, firm and industry are identical. There are no closely competitive substitutes for the product. So the buyers have no alternative or choice. They have either to buy the product or go without it. Monopoly is a complete negation of competition. A monopolist is a price maker and not a price taker. In fact his price fixing power is absolute. He is in a position to fix the price for the product, as he likes. He can vary the price from buyer to buyer. Thus in a competitive industry, there is a single ruling price, while in a monopoly, there may be differentials. A monopoly firm itself being the industry, it faces a downward-sloping demand curve for its product. That means it cannot sell more output unless the price is lowered. A pure monopolist has no immediate rivals due to certain barriers to entry in the field. There are legal, technological, economic or natural obstacles which may block the entry of new firms. Since a monopolist has a complete control over the market supply in the absence of a close or remote substitute for his product, he can fix the price as well as quantity of be sold in the market. Abuses of Monopoly Though a monopolist has complete freedom in determining his own price, there are some limits to his power. These are listed below: The demand curve of a monopolist slopes downwards. This is shown as demand curve DD of the monopolist in Figure. On such a curve, a monopolist cannot choose both Price and Output to be sold. He has to determine one of these quantities. If he chooses higher price P1 he has to be satisfied with smaller sales of quantity Q1. If he prefers larger output Q2 he will have to charge lower price P2. The second constraint on monopoly power arises out of the income and willingness of consumers. If the monopolist attempts to charge a price as high as Pn his sales fall to zero. So even though a monopolist has complete freedom to charge any high price this freedom is restricted by the consumer’s ability to purchase goods. Finally, monopoly power also depends upon elasticity of the demand curve. If the demand curve is rigid or less elastic the monopolist has a greater degree of control. As the demand curve becomes more flexible or flatter the monopolist’s control starts declining. This can be explained with the help of Figure. In the figure there are two demand curves. DD1 is rigid or less flexible showing greater monopoly control. DD2 is flatter or more flexible and depicts a lower degree of monopoly control. On rigid demand curve DD1 if the monopolist increases the price from P to P1 the fall in the quantity sold is as small as QQ1. On the flatter demand curve DD2 with the same rise in price, a fall in the quantity sold is as large as NN1. In case of a flexible demand curve there is a danger that even at a higher price, the total revenue of a monopolist may be smaller. This has been further explained in the table below: PRICE RIGID DEMAND D1 TOTAL REVENUE TR1 FLEXIBLE DEMAND D2 TOTAL REVENUE TR2 2 6 12 20 40 4 5 20 8 32 6 4 24 5 30 A monopolist attempts to raise his price from 2 to 4 to 6. As a result of this quantity demanded goes on falling. Yet in the case of Rigid Demand D1, with a fall in the demand from 6 to 5 to 4 Total Revenue TR1 increases from 12 to 20 to 24. With the Flexible Demand condition D2 the quantity demanded falls sharply from 20 to 8 to 5 causing Total Revenue TR2 to fall from 40 to 32 to 30. Hence the slope or the degree of flexibility of the demand curve governs the degree of monopoly power Monopoly market is restrictive and hence considered as an evil form of market. Monopoly is also a source of wastage. It underutilizes productive capacity and reduces Consumer’s Surplus. Underutilization of capacity may cause some workers to remain unemployed. These and other shortcomings can be analyzed and explained with the help of a comparative diagram. We find both competitive and monopoly equilibrium positions marketed by point e1 and e2 respectively. A competitive firm produces output Q1 and sells at price P1. A monopolist produces smaller output Q2 (Q2P1). Competition allows only normal profits to a firm as part of the average cost of production. A monopolist earns extra monopoly profits of the size CSRP2. Under competition output is produced at point e1 which is the lowest point on the average cost line. Therefore competition makes fuller utilization of the productive capacity. Under monopoly output is produced at point S which is on the falling phase of AC. This shows underutilization of the productive capacity. Finally, the size of the Consumer’s Surplus under competition is as large as De1P1 while that under monopoly is only DRP2. Hence under monopoly there is higher price, lower output, underutilization of productive capacity or wastage of resources and reduction in Consumer’s Surplus. Differences between Monopoly,  Equilibrium Competitive Equilibrium There are typical differences between the two types of market models their equilibrium positions. A comparative account of their differences is presented below: The demand curve of a competitive firm for its product is perfectly elastic. It is a horizontal straight line. It implies that the firm can sell any level of out put at the ruling market price. While the demand curve of the monopolistic for his product is relatively inelastic, it is a downward sloping curve. It suggests that the monopolist can sell more output only by lowering the price. To a competitive firm, price is given in the market. So at this price, average and marginal revenue will be the same. Hence, AR MR curves coincide and are represented through the demand curve which is a horizontal straight line. In the case of a monopoly, the downward sloping demand curve represents the AR curve. The MR curve also slopes downwards but it lies below the AR curve. If it is linear, then it lies half the distance between the price-axis and the demand curve. Under both perfect competition and monopoly, the equilibrium output is set at the point of equality between MC and MA. The competitive firm attains equilibrium only when the MC curve intersects the MR curve below. Thus, it is essential that MC must be rising at and near the equilibrium output. In fact, the falling cost curves caused by increasing returns to scale are incompatible with competitive equilibrium output, for the firm’s MR curve being horizontal, the falling MC curve can never lead to a competitive equilibrium position because as the firm will be inclined to expand its size until it becomes so large that its AR and MR curves ultimately begin to fall in order to cut the continuously falling MC curve. This means that the firm will become so large that competition will become imperfect and the individual firm would be in a position to influence the price of its product by altering its own output. In short, perfect competition will cease to exist when a firm increases i ts output to a very large extent in order to attain equilibrium under falling cost conditions. It may, therefore, be concluded that increasing returns to scale or a continuously downward sloping MC curve perfect competition are incompatible. It follows, thus, that a major difference between competitive equilibrium monopoly equilibrium is that while in the case of the former, the MC curve of the firm must be rising at or near the equilibrium level of output, in the case of the latter, this is not essential. A monopoly firm can attain equilibrium under any state of returns to scale or cost conditions, whether constant, rising or falling. The fundamental condition of monopoly equilibrium that must be satisfied is: MC=MR, and the MC curve must intersect the MR curve from below (yet it need not necessarily be rising). Again, when we compare the equilibrium conditions of the two models, we find that the fundamental rule of profit maximization is the same, i.e., equating MC with MR, the characteristic difference lies with respect to price as average revenue and MC. Under perfect competition, price=AR=MR; thus, at equilibrium output, MC=price. In monopoly, on the other hand MRMC. In a perfect normal equilibrium condition of a firm under competition in the long run only, normal profit is realized. In the case of a monopoly, excess monopoly profit can be earned even in the long-run. In fact, the positive difference between price and MC in a monopoly gives excess profit. In the long run, when the competitive firm gets only normal profit, it operates at the minimum point of the LAC curve. Hence the competitive firm tends to be of optimum size. A monopoly firm, on the other hand, attains equilibrium at the falling path of the AC curve, which means it doesn’t utilize its plant capacity to the full extent. The â€Å"excess capacity† in a monopoly firm thus causes it to be of less than optimum size. Usually, the monopoly price tends to be higher while the monopoly output smaller than that under perfect competition. A direct comparison of price and output under monopoly and competition is however difficult because a competitive firm is just a part of the industry as a whole, while a monopoly firm is an industry itself. MONOPOLY EQULIBRIUM UNDER DIFFERENT COST CONDITIONS Firms under all market condition achieve equilibrium at a point where MC=MR and MC is increasing or MC>MR if an additional unit is produced. Under Perfect competition this is possible only if the firm is operating with increasing cost i.e. marginal cost curve is sloping upward. Equilibrium cannot be determined if the marginal cost is decreasing or constant. Equilibrium is possible only in fig A where both necessary and sufficient conditions are fulfilled, whereas in B only the necessary condition is fulfilled and in C neither necessary nor sufficient conditions are satisfied. Unlike perfect competition, equilibrium of a monopoly is possible under increasing constant and decreasing cost as shown in Figure FIGURE shows equilibrium of a monopoly firm with increasing cost. The firms AC and MC curves are sloping upward. MC cuts MR at E. Here MC=MR and for any additional production MC>MR. Therefore firm A reaches equilibrium at point E. TR=OQ1 TP. TC=OQ1SN. Pie=NSTP Figure B, the firm reaches equilibrium at point E1 under constant cost. At point E1 MC=MR and thereafter MC>MR therefore the firm stops its production. At E1. TR=OQ2T1P1. TC=OQ2E1N1. Therefore Pie=N2S2T2P2 Figure C explains the equilibrium under decreasing cost. Equilibrium output is determined at point E2. Where MC=MR and MC>MR for any additional output. TR=OQ3T2P2. TC=OQ3S2N2 Therefore Pie=N2S2T2P2 The firm however will not be able to decide its output if under decreasing cost its marginal cost is always below the MR curve as shown in the figure. Fig shows the indetermination of Equilibrium under decreasing cost. Here the MC is all the times below MR hence it is not possible to determine the Equilibrium output. However the case shown in the above diagram may not be practical as the marginal cost cannot continuously decline and become zero. CONTROL OF MONOPOLY Evaluating the economic effects of pure monopoly or partial monopoly form the standpoint of society as a whole, on income distribution, price, output, resource allocation, technological advancement, distribution of economic power, it has been commonly observed that there are more evils aspects than benefits in a monopolistic industry as compared to a competitive industry. THE FOLLOWING POINTS MAY BE ENLISTED IN THIS CONTEXT: The monopoly price is generally higher than the competitive price. Evidently, the consumer is exploited under a monopoly. Output under monopoly is restricted with a view to earning the maximum economic profits. Thus, there is inefficient allocation of resources in a monopolistic industry. It entails waste of excess capacity. Only in a competitive industry there can be optimum utilization of existing plant capacity .In short, under a monopoly a higher price is charged, a smaller output is produced the system of allocation of resources is inferior to that under perfect competition. Usually, excess profit is reaped by a monopoly firm even in the long run. A purely competitive firm, on the other hand reaps just a normal profit in the long run. By virtue of their control over market supply, monopolists can export high prices to make substantial economic profits .Excessive price charged by the monopolists is regarded as a â€Å"PRIVATE TAX† on consumers. On account of high profiteering by the monopolists, society’s income distribution tends to be unequal unjust .The owners of monopoly business tend to become richer at the cost of the consumers. Big monopoly houses may acquire concentration of economic power ion their hands which also endangers political democracy in the country. A monopolist is supposed to be very conservative in the matter of innovation technological advancement .Since there is no threat of competition from rivals in a monopoly market, the firm has no impulse to develop new products or introduce new techniques in production. The monopolist is satisfied with the status quo. In fact sometimes monopolists may buy up new scientific inventions patents destroy them so to avoid rivalry. They do so in order to save loss arising from the sudden obsolescence of existing plant machinery. This tactic obviously obstructs technical progress of the country. Monopoly monopolistic competition tend to aggravate the problem of unemployment due to under allocation of resources. The actual production frontier of the country is kept unduly much below its potential level. This results in a low pace of economic growth in creating poverty in the midst of plenty Monopoly firm quite often resort to unfair practices like price discrimination or cut throat competition infringement of trade marks of rivals .etc with a view to eliminating or killing potential rivals in the market. Many big monopoly houses have tended to spread political economic corruption. It has been alleged that some political parties even govt. officials in India always have a soft corner for certain big business houses. METHODS OF CONTROL They are as follows: Restriction on entry of new firms Restriction on output Monopolists hold on price determination MEASURES OF CONTROL They are as follows: Legislative measures Promotion of competition Consumers resistance Publicity drive Control of price output Fiscal measures Nationalization Co-operative movement Misconceptions about Monopoly Pricing Profits It is commonly alleged that a monopolist can charge a very high price and earn high profits because he has the control over market supply and is a price-maker. This is really not so. A monopolist cannot determine price on the basis of his supply alone. He has to consider the demand aspect as well. In fact, the monopoly price is determined by the relative strength of the forces of demand and supply. Again, while determining the equilibrium price and output, the monopolist is interested in maximum sale because he wants to maximise total profits and not unit profits. So if the demand is slack, he will have to set a low price corresponding to profit maximising condition : MC = MR. Again, it is also erroneous p take it for granted that the monopolists price is always higher than the competitive price. It, in fact, depends on various considerations. If the demand is highly inelastic, while the supply is under conditions of increasing costs, ben the monopolist will restrict output in order to produce at a lower cost anchearn a higher profit. Under these circumstances, obviously, the monopoly price will be very high compared to the competitive price. For example, private monopoly is socially harmful in respect of production and sale of essential agricultural commodities like food-grains for which the demand is highly inelastic while the supply is under increasing costs on account of the law of diminishing returns operating on land. If, on the other hand, the demand is highly inelastic, but the supply is under increasing returns or decreasing costs condition, the monopoly price would tend to be nearer the competitive price. In such cases, monopoly can be socially tolerated. For instance, in producing comforts and luxury items, if a private monopolist invests huge capital, thereby enjoying the economies of scale so that he may supply goods at a low price at a competitive rate, then, such monopoly can be tolerated. Again, when there is a very limited market for a product, a monopolist can supply it at a lower price on account of its low cost of production due to large-scale economies than what is feasible in a competitive market by a large number of firms producing the goods on a small-scale. The competitive market price in such a case will tend to be high because though P AC, under competition, the AC itself tends to be high due to lack of economies of scale and the small-scale of production adopted by each firm . If, however, there is a monopoly which has to cater to the entire market, it would resort to a large-scale production. Hence, the output will be produced at a much lower cost, so even if the monopolist sets a higher price than AC for the sake of high profit, it may relatively turn out to be lower than that of the competitive firm. Similarly, it is also incorrect to say that the monopolist can always earn abnormally high monopoly profit due to his advantageous position in the market. In many cases, demand and cost situation may not be very favourable to the monopolist, so that he cannot make profits. In the long run, the monopolist may be under the threat of new entry in his line of production, so that he may resort to price limit which gives him a lower profit but not a high maximum profit. Potential competition thus serves as a significant constraint on the behaviour of the monopolist. Again, in some cases, the demand situation may be such that the demand curve or the average revenue curve in the long run may be just tangent to the LAC curve. In this case, the monopolist would earn only a normal profit (see Fig. to understand the situation). In Fig., the monopolist decides an equilibrium output OM, and charges PM price. Since the AR curve is tangent to the LAC curve at point P, Price = Average Revenue = Average Cost. Hence, the monopolist simply earns a normal profit. The only difference between such normal-profit monopoly equilibrium and competitive equilibrium is that the monopolist is producing at less than optimum size, i.e., at a higher average cost, while a competitive firm, earning normal profit, would be producing at a minimum average cost, i.e., it has an optimum size. In other words, under monopoly, even though there is just a normal profit earned, there is unutilised capacity of the plant and resources, while in a competitive firms equilibrium, the normal capacity is fully utilised. Anyway, it can be concluded from the above discussion that the monopolist cannot always earn high monopoly profits. Again, the monopolist in the long run should earn at least normal profits, otherwise he cannot survive. A monopolist finding the cost situation much above the demand consideration in the long run has no alternative but to wind up his business.

Macbeths Queen Essay -- Macbeth essays

Macbeth's Queen      Ã‚   There are two main characters in William Shakespeare's Macbeth, one of which is Lady Macbeth. Let us in this paper study her character in detail.    Blanche Coles states in Shakespeare's Four Giants evaluates the character of Lady Macbeth:    A woman who could speak as Lady Macbeth does, who could call upon the spirits that tend on mortal thoughts to unsex her and fell her from head to foot full of direct cruelty, who could entreat these same spirits to stop all avenues of remorse so that no compunctions of conscience will interfere with the carrying out of her purpose, who could call upon the night to wrap itself in the murkiest, gloomiest smoke of hell in order to hide, even from the keen knife she would use, the wound she would make when she herself stabs the sleeping King, such a terrible, frightful woman would not scruple at telling a little wife-to-husband lie to accomplish her purpose. (52)    In Everybody's Shakespeare: Reflections Chiefly on the Tragedies, Maynard Mack shows how Lady Macbeth complements her husband:    Her fall is instantaneous, even eager, like Eve's in Paradise Lost; his is gradual and reluctant, like Adam's. She needs only her husband's letter about the weyard sisters' prophecy to precipitate her resolve to kill Duncan. Within an instant she is inviting murderous spirits to unsex her, fill her with cruelty, thicken her blood, convert her mother's milk to gall, and darken the world "That my keen knife see not the wound it makes" (1.5.50). Macbeth, in contrast, vacillates. The images of the deed that possess him simultaneously repel him (1.3.130, 1.7.1) When she proposes Duncan's murder, he temporizes: "We will speak further" (1.5.69). (189)    ... ...Blakemore Evans. Boston: Houghton Mifflin Company, 1972.    Knights, L.C. "Macbeth." Shakespeare: The Tragedies. A Collectiion of Critical Essays. Alfred Harbage, ed. Englewwod Cliffs, NJ: Prentice-Hall, Inc., 1964.    Mack, Maynard. Everybody's Shakespeare: Reflections Chiefly on the Tragedies. Lincoln, NB: University of Nebraska Press, 1993.    Shakespeare, William. The Tragedy of Macbeth. http://chemicool.com/Shakespeare/macbeth/full.html, no lin.    Siddons, Sarah. "Memoranda: Remarks on the Character of Lady Macbeth." The Life of Mrs. Siddons. Thomas Campbell. London: Effingham Wilson, 1834. Rpt. in Women Reading Shakespeare 1660-1900. Ann Thompson and Sasha Roberts, eds. Manchester, UK: Manchester University Press, 1997.    Wilson, H. S. On the Design of Shakespearean Tragedy. Toronto, Canada: University of Toronto Press, 1957.

Summary of the Bell Jar :: essays research papers

Esther Greenwood, a college student from Massachusetts, traveled to New York to work on a magazine for a month as a guest editor. Esther knows she should be having the time of her life, but she feels like she is in a living nightmare. The execution of the Rosenbergs worries her, and this is what triggers the bell jar closing in on Esther and covering her view on life. When she goes home, she finds that she is in more of a nightmare. She tries to cut her wrists, but cannot. She tries to hang herself, but cannot find a place to hang the rope. In a desperate attempt to end her life she takes a large amount of sleeping pills and hides in a crawl space in her basement. But, she survives and awakes in a hospital. She remains uncooperative until Philomena Guinea, a wealthy woman who also gave Esther her college scholarship, pays for Esther to go into a private hospital. Esther improves slowly, she also meets Joan, who is a lesbian. When Esther finds out Joan’s sexuality, she finds Joan to be repulsive. Joan seems to be improving like Esther, but she commits suicide. Esther left the mental hospital in time to start the winter semester at college. She believed that she had regained a grasp on sanity, but knows that the bell jar of her madness could descend again at any time. Esther Greenwood is the protagonist and narrator of The Bell Jar. I find her extremely unique because of her view on life, the way that she thinks of people and how life works is very curious. Esther feels as if no one in the world understands her and is very selfish. During most of the book, no matter where Esther goes, she exists in the hell of her own mind. She seemed trapped inside herself, with no external existence, no matter how new and exciting, nothing could change how she felt.

Nss Phy Book 2 Answer

1 1 2 3 C Motion I 7 (a) From 1 January 2009 to 10 January 2009, the watch runs slower than the actual time by 9 minutes. Therefore, when the actual time is 2:00 pm on 10 January 2009, the time shown on the watch should be 1:51 pm on 10 January 2009. Practice 1. 1 (p. 6) D (a) Possible percentage error 10 ? 6 = ? 100% 24 ? 3600 = 1. 16 ? 10 % 1 (b) = 1 000 000 days 10 ? 6 –9 It would take 1 000 000 days to be in error by 1 s. (b) Percentage error 9 = ? 100% 9 ? 24 ? 60 = 6. 94 ? 10–2% 4 (a) One day = 24 ? 60 ? 60 = 86 400 s Practice 1. 2 (p. 15) 1 2 3 4 5 C B D D (b) One year = 365 ? 86 400 = 31 500 000 s 5 Let t be the period of time recorded by a stop-watch. Percentage error = 0. 4 ? 100% ? 1% t t ? 40 s (a) Total distance she travels 2 ? ? 10 2 ? ? 20 2 ? ? 15 + + = 2 2 2 = 141 m (b) Magnitude of total displacement = 10 ? 2 + 20 ? 2 + 15 ? 2 = 90 m Direction: east Her total displacement is 90 m east. The minimum period of time is 40 s. 6 (a) Percentage error error due to reaction time = ? 100% time measured 0. 3 = ? 100% 10 = 3% 6 7 His total displacement is 0. With the notation in the figure below. (b) From (a), the percentage error of a short time interval (e. g. 10 s) measured by a stop-watch is very large. Since the time intervals of 110-m hurdles are very short in the Olympic Games, stop-watches are not used to avoid large percentage errors. Since ZX = ZY = 1 m, ? = ? = 60 °. Therefore, XY = ZX = ZY = 1 m The magnitude of the displacement of the ball is 1 m.  © 8 (a) The distance travelled by the ball will be longer if it takes a curved path. 7 (a) Length of the path = 0. 8 ? 120 = 96 m (b) No matter which path the ball takes, its displacement remains the same. (b) Length of AB along the dotted line 96 = 30. 6 m = (c) Magnitude of Jack’s average velocity 30. 6 ? 2 = = 0. 51 m s–1 120 Practice 1. 3 (p. 23) 1 B Total time 5000 5000 = + = 9821 s 1. 4 0. 8 5000 + 5000 = 1. 02 m s–1 Average speed = 9821 Practice 1. 4 (p. 31) 1 2 C B Final speed = 1. 5 ? 1 – 0. 2 ? 1 = 1. 3 m s–1 2 C Total time = 9821 + 10 ? 60 =10 421 s 5000 + 5000 Average speed = = 0. 96 m s–1 10 421 3 A By a = 3 D When the spacecraft had just finished 1 revolution, the spacecraft returned to its starting point. Therefore, its displacement was zero and its average velocity was also zero. v ? u , t v = u + at 36 = + ( ? 1. 5) ? 2 3. 6 = 7 m s–1 = 7 ? 3. 6 km h–1 = 25. 2 km h–1 Its speed after 2 s is 25. 2 km h–1. 4 5 D (a) Average speed 100 = = 10. m s–1 9. 69 (b) Yes. This is because the magnitude of the displacement is equal to the distance in this case. 4 B Take the direction of the original path as positive. Average acceleration of the ball ? 10 ? 17 = 0. 8 = –33. 8 m s–2 The magnitude of the average acceleration of the ball is 33. 8 m s– 2. v ? u By a = , t 100 ? 0 v ? u 3. 6 t= = = 4. 27 s a 6. 5 6 (a) Two cars move with the same speed, e. g. 50 km h–1, but in opposite directions. (b) A man runs around a 400-m playground. When we calculate his average speed, we can take 400 m as the distance and his average speed is non-zero. But since his displacement is zero (he returns to his starting point), his average velocity is zero. 5 The shortest time it takes is 4. 27 s.  © 6 Time / s –1 4 0 2 4 6 17 8 22 D Average speed 80 + 60 = 5 = 28 km h–1 Average velocity = Speed / m s 2 7 12 v ? u 22 ? 2 a= = 2. 5 m s–2 = t 8 The acceleration of the car is 2. 5 m s–2. 7 (a) I will choose ‘towards the left’ as the positive direction. 80 2 + 60 2 5 (b) 5 = 20 km h–1 C Total time 10 10 = + 2 3 = 8. 33 s v ? u , t u = v ? at = 9 ? (? 2) ? 3 = 15 m s–1 –1 (c) By a = Average speed 20 = 8. 33 = 2. 4 m s–1 Her average speed for the whole trip is 2. m s–1. The initial velocity of the skater is 15 m s . 8 (a) The object initially moves towards the left and accelerates towards the left. It will speed up. 6 7 8 9 10 C C C B A Magnitude of displacement = 2000 2 + 6000 2 = 6324. 6 m Magnitude of average velocity 6324. 6 = 4 ? 3600 = 0. 439 m s–1 6000 tan ? = 2000 ? = 71. 6 ° His average velocity is 0. 439 m s–1 (S 71. 6 ° E). (b) The object initially moves towards the right and accelerates towards the left. It will slow down. Its velocity will be zero and then increases in the negative direction (moves towards the left). Revision exercise 1 Multiple-choice (p. 5) 1 2 3 C D B  © 11 C Total time = 13 min = 780 s 840 ? 2 = 2. 15 m s ? 1 Average speed = 780 (b) Displacement from Sheung Shui to Lok Ma Chau 1000 = ? 6. 3 1 = 6300 m Magnitude of average velocity 6300 = 359 = 17. 5 m s–1 (1M) (1A) (1M) (1A) 12 13 D (HKCEE 2003 Paper II Q3) Conventional (p. 37) 1 Total time left for the two players = 4 ? 60 + 9 + 5 ? 60 + 16 = 565 s Total time they have been playing = 2 ? 60 ? 60 ? 565 = 6635 s (= 110 min 35 s = 1 h 50 min 35 s) (1A) 5 (a) Total distance = 1500 + 40 ? 1000 + 10 ? 1000 = 51 500 m Total time = 2 ? 3600 + 3 ? 60 + 8 = 7388 s Average speed 51 500 = 7388 = 6. 7 m s–1 (1M) (1A) 2 (a) 50 m (1A) (b) Ma gnitude of average velocity of Kitty 50 = (1M) 1? 60 + 15 = 0. 667 m s ? 1 (1A) (1M) (1A) (c) Average speed of the coach 5 + 50 + 5 = 1? 60 + 15 = 0. 8 m s ? 1 (b) Swimming: Average speed 1500 = 21 ? 60 + 28 = 1. 16 m s–1 Cycling: Average speed 40 000 = 1 ? 3600 + 1 ? 60 + 53 = 10. 8 m s–1 Running: Average speed 10 000 = 39 ? 60 + 47 = 4. 19 m s–1 (1M) His average speed was the highest in cycling. (1A) 3 (a) Since she measures the time interval based on 1 cycle of the pendulum, the error (0. 3 s) in measuring the cycle of the pendulum accumulates. is from 8 to 14 s. 1A) (1A) The range of the time interval (10 cycles) (b) When finding the time for one pendulum cycle, Jenny should time more pendulum cycles (e. g. 20) with the stop-watch and divide the time by the number of cycles. (1A) 4 (a) Time required 7. 4 ? 1000 = 20. 6 = 359 s (5 min 59 s) (1M) (1A)  © (c) Yes. Since the time interval of this competition is quite long, (1A) using stop-watch will not result in large percentage error as the reaction time for an average person is only 0. 2 s. (1A) (1M) (c) Total time = 5 min 45 s ? 1 min 58 s = 3 min 47 s = 3 ? 60 + 47 = 227 s v? u a= (1M) t 431 ? 0 = 3. = 0. 527 m s–2 (1A) 227 The average acceleration of the train is 0. 527 m s–2. 6 (a) v = u + at =0+6? 4 = 24 m s–1 = 86. 4 km h 86. 4 km h . –1 –1 (1A) The maximum speed of the car is 8 (1M) (a) Total distance = 8000 + 4000 + 5000 = 17 000 m Total time = 1 ? 3600 + 30 ? 60 + 45 ? 60 (b) v = u + at = 24 + (–4) ? 2 = 16 m s –1 –1 = 57. 6 km h (1A) –1 = 8100 s Average speed 17 000 = 8100 = 2. 10 m s–1 (1M) (1A) (c) The final speed of the car is 57. 6 km h . v? u a= (1M) t 16 ? 0 = 6 = 2. 67 m s–2 2. 67 m s–2. (1A) The average acceleration of the car is (b) 7 (a) Average speed 30 000 = 8 ? 60 = 62. m s–1 The average speed of the train is 62. 5 m s–1. (1M) (1A) (b) Maximum speed 430 = = 119. 4 m s? 1 > average speed 3. 6 (1A) The average speed must be smaller than the maximum speed because the train needs to speed up from start and slows down to stop during the trip. (1A) Magnitude of displacement = 3000 2 + 4000 2 = 5000 m Magnitude of average velocity 5000 = = 0. 617 m s–1 8100 4000 tan ? = 3000 (1A) ? = 53. 1 ° His average velocity is 0. 617 m s (N 53. 1 ° E).  © –1 (1A) 9 (a) Distance travelled = 10. 5 ? 3 ? 60 = 1890 m (1M) (1A) 10 (a) Total distance = (120 + 50) ? 1000 = 170 000 m (1M) (1A) b) Circumference of the track =2 r = 2 (400) = 2513 m The distance travelled by Marilyn is 3 1890 m which is about of the 4 circumference. (1A) (b) N ?XYZ is a right-angled triangle. Z ? 50 km 30 ° Y 60 ° X ? ? 120 km Magnitude of displacement (from town X to town Z) = 120 000 2 + 50 000 2 = 130 000 m 120 tan ? = 50 ? = 67. 4 ° Magnitude of displacement AB = 400 2 + 400 2 (1A) (1A) ? = 90 ° ? 67. 4 ° = 22. 6 ° ? = 60 ° ? 22. 6 ° = 37. 4 ° The total displacement of the car is 130 000 m (N 37. 4 ° E). = 566 m Magnitude of average velocity 566 = 3 ? 60 = 3. 14 m s 400 tan ? = 400 ? = 45 ° (S 45 ° E). –1 (c) (1A) Total time 170 000 = = 10 200 s 60 3. 6 Magnitude of average velocity 130 000 = 10 200 = 12. 7 m s–1 Its average velocity is 12. 7 m s (N 37. 4 ° E). –1 (1A) (1A) (1M) (1A) Her average velocity is 3. 14 m s–1  © 11 (a) AC = 60 2 + 80 2 = 100 m 80 tan ? = ? = 53. 1 ° 60 (1M) The total displacement of the athlete is 100 m (S53. 1 °W). (1A) 13 (Correct label of velocity with correct direction (towards the left). ) (Correct label of acceleration with correct direction (towards the right). ) (1A) (1A) (a) The coin moves in the following sequence: B A C C A Therefore, it is at A finally. Displacement of the coin = 15 cm (1A) (1M) (1A) (1M) b) Distance travelled by the coin = 15 + 30 + 30 = 75 cm (b) Time / s v / m s–1 0 –6 1 –4 2 –2 3 0 4 +2 5 +4 6 +6 (1A) (1A) (c) (i) Total time = 2 s ? 4 = 8 s Average velocity 15 ? 10 ? 2 = 8 = 0. 0188 m s? 1 (0. 5A ? 6) (1M) (1A) (c) The car will slow down and its speed will drop to zero. After th at the car will move towards the right with increasing speed (uniform acceleration). (1A) (1M) (1A) (1M) (1A) (1M) (1A) A (ii) Average speed 75 ? 10 ? 2 = 8 = 0. 0938 m s? 1 (1M) (1A) 12 (a) Total distance travelled = 60 + 80 + 80 + 60 = 280 m (d) (i) The coin moves in the following sequence: B A C C A B B b) Magnitude of total displacement = 80 + 80 = 160 m 160 m (west). The total displacement of the athlete is Therefore, it is at B finally. zero. the coin is also zero. (1A) (1M) (1A) (1M) (1A) (1M) (1A) (ii) The displacement of the coin is Therefore the average velocity of (c) Total distance travelled = 280 + 60 + 80 = 420 m 14 (a) Total distance = ? r = 5? ? 60 m C = 15. 7 m Total displacement =5+5 = 10 m 80 m  © The total displacement travelled by her is 10 m. (b) Jane’s statement is incorrect. (1A) Since both girls start at X and meet at Y, they have the same displacement. (1A) Betty’s statement is incorrect. 1A) Since both girls return to their starting point, their displacements are zero. (1A) Physics in articles (p. 40) (a) From 19 January 2006 to 28 February 2007, (1A) It takes New Horizons spacecraft a total of 406 days to travel from the Earth to Jupiter. (1A) (b) (i) Average speed total distance travelled = total time of travel (1M) = 8 ? 108 406 ? 24 (1A) (1M) = 8. 21 ? 104 km h? 1 (ii) Average acceleration change in velocity = total time of travel = (8. 23 ? 5. 79)? 10 4 406 ? 24 = 2. 50 ? 104 km h? 2 (1A) (1A) (c) July 2015  © 2 1 2 3 4 5 Motion II 10 (a) The object moves with a constant elocity. Practice 2. 1 (p. 61) D B D D B 30 ? 10 = 10 m s–1 v= 2 (b) The object moves with a uniform acceleration from rest. (c) The object moves with a uniform deceleration, starting with a certain initial velocity. Its velocity becomes zero finally. The velocity of the car at t = 2 s is 10 m s–1. 6 7 C (d) The object first moves with a uniform acceleration from rest, then at a constant velocity, and finally moves with a smaller uniform acceleration again. (a) Total displacement = 4 ? 5 + (? 5) ? (7 ? 5) = 10 m The total displacement from the staircase to her classroom is 10 m. (e) The object moves at a constant velocity and then suddenly moves at constant velocity of same magnitude in the opposite direction. (b) Classroom C 8 (f) The object moves with uniform deceleration from an initial velocity to rest, and continue to move with the uniform acceleration of the same magnitude in opposite direction. 9 (a) The object accelerates. (b) The object first moves with a constant velocity. Then it becomes stationary and finally moves with a higher constant velocity again. 11 (a) The object moves with zero acceleration (with constant velocity of 50 m s–1). (b) The object moves with a uniform cceleration of 5 m s–2. (c) 12 The object moves with uniform deceleration of 5 m s–2. (c) The object decelerates to rest, and then accelerates in opposite direction to return to its starting point. (a) It moves away from the sensor. (d) The object moves with uniform velocity towards the origin (the zero displacement position), passes the origin, and continues to move away from the origin with the same uniform velocity.  © (b) (c) The greatest rate of change in speed 0 ? 3. 5 = 2 = –1. 75 m s–2 (d) Total distance travelled = area under the graph 3. 5 ? 2 2 ? 6 = + 2 2 = 9. 5 m Practice 2. 2 (p. 71) 1 C By v2 = u2 + 2as, 290 3. 6 2 13 (a) =0+2? 1? s s = 3240 m = 3. 24 km < 3. 5 km The minimum length of the runway is 3. 5 km. 2 B Cyclist X is moving at constant speed. Time for cyclist X to reach finish line displacement 150 = = = 30 s time 5 For cyclist Y: u = 5 m s–1, s = 250 m, (b) Total distance travelled = area under the graph (12 + 6) ? 3 = 2 = 27 m a = 2 m s–2 By s = ut + 1 2 at , 2 1 250 = 5 ? t + ? 2 ? t2 2 (c) Average speed total distance travelled = time taken 27 = 3 t = 13. 5 s or t = ? 18. 5 s (rejected) Y needs 13. 5 s to reach finish line. Therefore, cyclist Y will win the race. 3 B Since the bullet start decelerates after fired into the wall, we could just consider the displacement of the bullet in the wall. To prevent the bullet from penetrating the wall, the bullet must stop in the wall. = 9 m s–1 14 (a) She moves towards the motion sensor. (b) The highest speed of the girl in the journey is 3. 5 m s–1.  © By v2 = u2 + 2as, 0 = 500 + 2 ? (? 800 000) ? s 2 8 By v = u + at, 14 = u + 2 ? 5 u = 4 m s–1 s = 0. 156 m = 15. 6 cm < 15. 8 cm The minimum thickness of the wall is 15. 8 m. By v2 = u2 + 2as, 142 = 42 + 2 ? 2 ? s s = 45 m 4 C When the dog catches the thief at t = 5 s, its total displacement is 30 m. The dog is sitting initially, so u = 0. 1 By s = ut + at2, 2 1 30 = 0 + a(5)2 2 The displacement of the girl is 45 m. 9 (a) v = u + at = 0 + 20 ? 0. 3 = 6 m s? 1 The horizontal speed of the ball travelling towards the goalkeeper is 6 m s? 1. a = 2. 4 m s–2 Its acceleration is 2. 4 m s–2. (b) By v2 = u2 + 2as, 02 ? 62 a= = –22. 5 m s? 2 2 ? 0. 8 The deceleration of the football should be 22. 5 m s? 2. 5 6 D 90 36 ? v? u = 3. 6 3. 6 = 1. 5 m s–2 a= t 10 By v = u + 2as, 2 2 10 (a) The reaction time of the cyclist is 0. 5 s. s= v ? u = 2a 2 2 90 3. 6 36 3. 6 2 ? 1. 5 ? 2 2 = 175 m (b) Braking distance (2. ? 0. 5)? 15 = 11. 25 m = 2 Thinking distance = 15 ? 0. 5 = 7. 5 m Stopping distance = 11. 25 + 7. 5 = 18. 75 m child. 20 m The distance travelled by the motorcycle is 175 m and its acceleration is 1. 5 m s . –2 7 (a) Thinking distance = speed ? reaction time 108 = ? 0. 8 = 24 m 3. 6 Therefore, the bicycle would not hit the (b) Since the car decelerates uniformly, braking distance v+u = ? t 2 108 +0 = 3. 6 ? (3 ? 0. 8) 2 = 33 m 11 By v = u2 + 2as, 0 = 32 + 2 ? (–0. 5) ? s s=9m 8m Therefore, the golf ball can reach the hole. 2 12 (a) (i) By v = u + at, 0 = u + (–4)(4. 75) u = 19 m s–1 The initial velocity of the car is 19 m s–1. (c) Stopping distance = thinking distance + braking distance = 24 + 33 = 57 m  © (ii) By v2 = u2 + 2as, 0 = 19 + 2 ? (–4) ? s s = 45. 1 m 2 3 C For option A, apply equation v2 = u2 – 2gs and take s = 0 (the ball returns to the second floor), v = –u = –10 m s–1 (vertically downwards) The displacement of the car before it stops in front of the traffic light is 45. 1 m. This is the same velocity as the initial velocity of option B. Therefore, in both ways the ball has the same vertical speed when it reaches the ground. (b) By v = u + 2as, 17 = 0 + 2 ? 3 ? s s = 48. 2 m 2 2 2 The displacement of the car between starting from rest and moving at 17 m s is 48. 2 m. –1 4 B Take the upward direction as positive. 1 By s = ut + at2, 2 1 0 = u ? 30 + ? (? 10) ? 302 2 u = 150 m s–1 13 (a) By v2 = u2 + 2as, v2 = 0 + 2 ? 0. 1 ? 500 v = 10 m s–1 His speed is 10 m s . –1 (b) Consider the first section. By v = u + at, v? u t= a 10 ? 0 = 0. 1 = 100 s Consider the second section. 1 By s = ut + at2, 2 1 800 = 10t + ? 0. 5t2 2 t = 40 s or t = –80 s (rejected) The speed of the bullet is 150 m s–1 when it is fired. 5 Speed of stone Equation used t=1s t=2s t=3s t=4s v = u + at Distance travelled by the stone 1 s = ut + at 2 2 m 20 m 45 m 80 m 10 m s–1 20 m s 30 m s –1 –1 40 m s–1 Total time taken = 100 + 40 = 140 s It takes 140 s for Jason to travel downhill. 6 1 By s = ut + at2, 2 1 10 = 0 + (10) t2 2 t = 1. 41 s v = u + at Practice 2. 3 (p. 83) 1 2 D D = 0 + 10(1. 41) = 14. 1 m s–1 It takes 1. 41 s for a diver to drop from a 10-m platform. His speed is 14. 1 m s–1 when he enters the water.  © 7 Take the upward direction as positive. By v = u + 2as, 4 = 0 + (2)(–10)s s = 0. 8 m 2 2 2 Besides, since Y spends a shorter time to reach its highest point, it should be fired after X. 10 (a) By s = ut + The highest position reached by the puppy is 0. m above the ground. 8 (a) Consider the boy’s downward journey. Take the downward direction as positive. 1 By s = ut + at2, 2 1 0. 5 = 0 + (10) t2 2 t = 0. 316 s 1 2 at , 2 1 120 = 8t + ? 10 ? t2 2 t = 4. 16 s or t = ? 5. 76 s (rejected) It takes 4. 16 s to reach the ground. (b) v = u + at = 8 + 10 ? 4. 16 = 49. 6 m s–1 Its speed on hitting the ground is 49. 6 m s–1. 11 (a) Distance between the ceiling and her hands = 6 – 2 – 1. 2 = 2. 8 m Hang-time of the boy = 0. 316 ? 2 = 0. 632 s (b) Let s be her vertical displacement when she jumps. As the maximum jumping speed is 8 m s–1, i. e . u = 8 m s–1. By v2 = u2 + 2as, v2 ? 2 s= 2a 2 0 ? 82 = (upwards is positive) 2 ? (? 10) s = 3. 2 m > 2. 8 m Therefore, the indoor playground is not safe for playing trampoline. 1 (a) By s = ut + at2, 2 1 132 = 0 ? t + ? 10 ? t2 2 t = 5. 14 s The vehicle can experience a free fall in the Zero-G facility for 5. 14 s. (b) Take the upward direction as positive. By v = u + 2as, 0 = u + 2 ? (–10) ? 0. 5 u = 3. 16 m s–1 2 2 2 The jumping speed of the boy is 3. 16 m s–1. 9 Take the upward direction as positive. (a) By v2 = u2 + 2as, 0 = u2 + 2(–10)(200) u = 63. 2 m s–1 The velocity of the firework X is 63. 2 m s–1 when it is fired. 12 (b) By v = u + at, = 63. 2 + (–10)t t = 6. 32 s It takes 6. 32 s for the firework X to reach that height. (c) From (a) and (b), for firework Y to explode at 130 m above the ground, the speed of Y should be smaller than that of X. Therefore, Y should be fired at a (b) By v2 = u2 + 2as, v2 = 02 + 2 ? 10 ? 132 v = 51. 4 m s? 1 The speed of the vehicle before it comes to a stop is 51. 4 m s? 1.  © lower speed. (c) Take the upward direction as positive. By v = u + at, –v = v – gt 2v = gt If the stone is projected with a speed of 2v, let the new time of travel be t?. (–2v) = (2v) – gt? v t? = 4 ( ) g = 2t Its new time of travel is 2t. 6 B Take the upward direction as positive. 1 s = ut + at2 2 1 = (10)(4) + (–10)(4)2 2 = –40 m The distance between the sandbag and the ground is 40 m when it leaves the balloon. Revision exercise 2 Multiple-choice (p. 87) 1 D By v2 = u2 + 2as, 0 = 102 + 2a(25 – 10 ? 0. 2) a = –2. 17 m s–2 His minimum deceleration is 2. 17 m s–2. 2 3 D B Consider the rock released from the 2nd floor. By v2 = u2 + 2as, v2 = 2as floor. Note that s2 = 3. 5s. (v2)2 = 2as2 = 3. 5(2as) = 3. 5v2 v2 = 1. 87v (as u = 0) Then consider the rock released from the 7th 7 8 D C Take the downward direction as positive. u = 200 m s–1, v = 5 m s–1, a = ? 0 m s–2 By v = u + at, 5 = 200 + (? 20)t t = 9. 75 s The rockets should be fired for at least 9. 75 s. Both C and D satisfy this requirement. But for D, after firing for 10. 2 s, v = u + at = 200 + (–20)(10. 2) = –4 m s–1 i. e. it flies away from the Moon with 4 m s–1 upwards. It c annot land on the Moon. Therefore, the correct answer is C. 4 5 A C The stone returns to the ground with the same speed (but in opposite direction). 9 10 D D  © 11 12 13 (HKCEE 2006 Paper II Q1) (HKCEE 2007 Paper II Q2) (HKCEE 2007 Paper II Q33) (b) (i) Conventional (p. 89) 1 (a) The reaction time of the driver is 0. 6 s. (b) v a= t = 0 ? 12 3. 6 ? . 6 (1A) (Correct axes with label) from t = 1. 20 s to 1. 25 s) from t = 1. 45 s to 1. 50 s) (1A) (1A) (1A) (A straight line with slope = 0. 35 m s–1 (A straight line with slope = –0. 35 m s–1 (1A) (1M) = –4 m s–2 The acceleration of the car is –4 m s–2. (c) The stopping distance of the car is the area under graph. Stopping distance 12 ? (3. 6 ? 0. 6) =12 ? 0. 6 + 2 = 25. 2 m The stopping distance of the car is shorter than 27 m. The driver will not be charged with driving past a red light. (1A) (1A) (1M) (ii) 2 (a) The object moves away from the motion sensor with uniform velocity at 0. 35 m s–1 from t = 1. 20 s to 1. 25 s. 1A) From t = 1. 25 s to 1. 45 s, the object moves with negative acceleration. (1A) Then, from t = 1. 45 s to 1. 50 s, the object changes its moving direction and moves towards the motion sensor again with a uniform velocity of –0. 35 m s–1. (1A) (Correct axes with labels) (1A) (Correct graph with the acceleration of ? 0. 35 ? 0. 35 about 1. 40 ? 1. 30 = –7 m s–2 at t = 1. 30 s to 1. 40 s) (1A) !  © 3 (a) (b) Total displacement of the car = area bound by the v? t graph and the time axis 1 1 = (5 ? 5) ? (20 ? 3) 2 2 = ? 17. 5 m (1M) (1A) (c) Yes, the car moves 12. 5 m forwards from t = 0 to t = 5 s. Therefore, it hits the roadblock. 1A) 5 Take the upward direction as positive. (a) From point A to the highest point: (Correct axes with labels) (Correct shape of minibus’ graph) (Correct shape of sports car’s graph) (Correct values) (1A) (1A) (1A) (1A) By v2 = u2 + 2as, 0 = 42 + 2 (–10) s s = 0 . 8 m By v = u + at, 0 = 4 + (–10)t t = 0. 4 s (1M) From the highest point to the trampoline: 1 s = ut + at2 (1M) 2 1 = 0 + (–10)(1. 2 – 0. 4)2 2 = –3. 2 m (1A) 3. 2 m above the trampoline. (1A) The maximum height reached by him is (1M) (b) From the graph in (a), the two vehicles have the same velocity at t ? 2. 3 s after passing the traffic light. (1A) (1M) (c) The area under graph is the displacement of the cars. Consider their displacements at t = 3 s, For the sports car: 1 s = ? 15 ? 3 = 22. 5 m 2 For the minibus: 1 s = ? (7 + 13) ? 3 = 30 m 2 The minibus will take the lead 3 s after passing the traffic light. (1A) (b) Height of point A above the trampoline (1A) = 3. 2 – 0. 8 = 2. 4 m (1M) (1A) 6 (a) Initial velocity v = 90 km h–1 90 = m s–1 3. 6 = 25 m s–1 Thinking distance =v? t = 25 ? 0. 2 =5m The thinking distance is 5 m. (1A) (1M) 4 (a) The car moves forward with uniform acceleration at ? 1 m s? 2 from t = 0 s to t = 5 s. (1A) (1A) Then the car changes its moving direction. From t = 5 s to t = 8 s, it moves backwards with a uniform acceleration of ? 6. 67 m s . ?2 Its instantaneous velocity is 0 at t = 5 s. (1A) †  © (b) By v2 = u2 + 2as, v2 ? u2 a= 2s 2 0 ? 25 2 = 2 ? (80 ? 5) = ? 4. 17 m s–2 4. 17 m s–2. (1M) (c) The slope of the graph is the magnitude of the acceleration of the apple. speed / m s? 1 7. 75 (1A) (1A) Hence, the deceleration of the car is (c) By v2 = u2 + 2as, s= v ? u 2a 0 2 ? 25 2 = 2 ? ( ? 4. 17 ? 2) 2 2 (1M) 0 0. 775 time / s (Correct labelled axes) (2A) (1A) (Straight line with a slope of 10 m s? 2) = 37. 5 m Braking distance = 37. 5 m Stopping distance = 37. 5 + 5 = 42. m (1M) (d) The two graphs have no difference. (1A) (1A) 8 (a) Take the downward direction as positive. By v2 = u2 + 2gs, v = u + 2 gs 2 The driver could not stop before the traffic light. Therefore, his claim is incorrect. (1A) (1M) 7 (a) Take the downward direction as positive. 1 By s = ut + gt2, 2 1 3 = 0 ? t + ? 10 ? t2 2 3? 2 t= = 0. 775 s 10 (1M) = 0 2 + 2 ? 10 ? (40 ? 3) = 27. 2 m s–1 cushion is 27. 2 m s? 1. 1 (b) (i) By s = ut + gt2, 2 1 40 – 3 = 0 + ? 10 ? t2 2 t = 2. 72 s (1A) The speed of the residents landing on the (1M) (1A) The apple travels in air for 0. 775 s. (1A) (b) By v2 = u2 + 2as, v = 2 ? 10 ? 3 (1M) 1A) –1 = 7. 75 m s? 1 The speed of the apple is 7. 75 m s when the apple just reaches the ground. The time of travel in air is 2. 72 s. u+v (ii) By s = t, (1M) 2 2s t= u+v 2? 3 = t 27. 2 + 0 = 0. 221 s (1A) The time of contact is 0. 221 s.  © (c) (b) Slope of the graph from t = 0 to t = 0. 28 s 2. 3 ? 0 = 0. 28 ? 0 = 8. 21 m s–2 The acceleration of the ball due to gravity is 8. 21 m s–2. (1M) (1A) (c) (Correct labeled axes) (Correct shape) (Correct values) (1A) (1A) (1A) (i) 9 (a) t = 2 s: Displacement of the trolley = 0. 7 ? 0. 15 = 0. 55 m t = 3. 4 s: (1A) Displacement of the trolley = 1. 175 ? 0. 15 = 1. 025 m t = 4. 9 s: 1A) Displacement of the trolley = 0. 6 ? 0 . 15 = 0. 45 m (1A) (b) It moves away from the motion sensor with a changing speed from t = 2 s to t = 3. 4 s. (Correct sign) (Correct shape) (1A) (1A) (1A) (1A) (1A) (ii) The method does not work Then it rests momentarily at t = 3. 4 s. After that, it moves towards the motion since ultrasound will be reflected by the transparent plastic plate. (1A) (c) sensor with a changing speed. 1 By s = ut + at2, 2 1 ? 0. 1 = 0. 7 ? 2. 9 + ? a ? (2. 9)2 2 a = ? 0. 507 m s? 2 (1A) (1M) 11 (a) (i) The ball is held 0. 15 m from sensor before being released. The ball hits the ground which is 1. m from the sensor. (1A) (1A) Therefore, the ball drops a height of 0. 95 m. which are 0. 45 m, 0. 65 m and 0. 775 m from the sensor in its first 3 rebounds. (1A) The acceleration of the trolley is ? 0. 507 m s? 2. (ii) The ball rebounds to the positions 10 (a) The motion sensor is protruded outside the table to avoid the reflection of ultrasonic signal from table. (1A)  © At the 1st rebound, the ball rises up (1. 1 ? 0. 45) = 0. 65 m. nd The average acceleration is 66. 6 m s–2. (1A) (1A) (1A) (c) v / m s? 1 6. 32 At the 2 rebound, the ball rises up (1. 1 ? 0. 65) = 0. 45 m. rd At the 3 rebound, the ball rises up (1. 1 ? 0. 75) = 0. 325 m. (b) (i) The ball hits the ground with velocities of 3. 9 m s , 3. 25 m s and 2. 75 m s–1 in its first 3 rebounds. (3A) 3. 9 (1M) 0. 95 ? 0. 55 (1A) –1 –1 t3 t1 t2 t4 t5 t/s (ii) Acceleration = slope of graph = = 9. 75 m s–2 ?6. 32 (3 straight lines) (Correct slopes) (1A) (1A) 12 Take the downward direction as positive. 1 (a) By s = ut + gt2, (1M) 2 1 2 = 0 ? t + ? 10 ? t2 2 2? 2 t= = 0. 632 s (1A) 10 It takes 0. 632 s from t1 to t2. (Correct labels of time and velocity)(1A) 13 (a) Speed v = 70 km h–1 70 = m s–1 3. 6 = 19. 4 m s–1 d Reaction time = v 6 = 19. 4 = 0. 309 s The reaction time of the man was 0. 09 s. (1M) (b) At t2, v = u + at (1A) = 0 + 10 ? 0. 632 = 6. 32 m s –1 –1 (1 M) Shirley’s speed is 6. 32 m s when she lands on the trampoline at t2. At t4, she leaves the trampoline at the same speed. Therefore, from t3 to t4, by v2 = u2 + 2as, a= v2 ? u2 2s (? 6. 32) 2 ? 0 2 = 2 ? 0. 3 (b) By v2 = u2 + 2as, v2 ? u2 a= 2s 2 0 ? 19. 4 2 = 2 ? 48 = –3. 92 m s–2 3. 92 m s–2. (1M) (1M) (1A) The average deceleration of the car was (c) (1A) Speed v = 80 km h–1 80 = m s–1 3. 6 = 22. 2 m s–1 = 66. 6 m s–2  © Thinking distance = vt = 22. 2 ? 0. 309 = 6. 86 m By v = u + 2as, braking distance s v2 ? u2 = 2a 2 0 ? 22. 2 2 = 2 ? ? 3. 92) 2 2 (1A) Take the upward direction as positive. 1 s = ut + at2 (1M) 2 1 = 7 ? 1. 75 + ? (–10) ? 1. 752 2 = –3. 06 m (negative means the water is below the spring board) The spring board is 3. 06 m above the water. Alternative method: (1A) = 62. 9 m Therefore, the stopping distance = 6. 86 + 62. 9 = 69. 8 m (1A) Consider the upward motion and downward motion separatel y. For the upward motion, she takes 0. 7 s to reach the highest point from the spring board. Take the upward direction as positive. 1 By s = ut + at2, (1M) 2 1 s1 = 7 ? 0. 7 + ? (–10) ? 0. 72 2 = 2. 45 m For the downward motion, she takes 1. 5 s from the highest point to enter water. Take the downward direction as positive. By s = ut + 1 2 gt , 2 1 s2 = 0 + ? 10 ? 1. 052 = 5. 51 m 2 (1A) This stopping distance is greater than the initial distance between the car and the boy. (1A) Therefore, the car would have knocked down the boy if the car had travelled at 80 km h? 1 or faster. (d) A drunk has a longer reaction time. (1A) This means that the thinking distance, and thus the stopping distance (sum of thinking distance and braking distance), increases. (1A) (1M) (1A) 14 (a) Take the upward direction as positive. By v = u + at, u = 0 ? (? 10) ? 0. 7 = 7 m s–1 board is 7 m s . 1 Therefore the height of the spring board above the water = s2 – s1 = 5. 51 – 2. 4 5 = 3. 06 m (1A) (1M) (1A) The speed of Belinda leaving the spring (b) Total time taken from the spring board to the water = 0. 7 + 1. 05 = 1. 75 s (c) v = u + at = 0 + (? 10) ? 1. 05 = ? 10. 5 m s–1 is 10. 5 m s–1.  © The speed of the diver entering the water (d) Deceleration of car Y = slope of the graph during 0. 5 s? 8. 5 s = 0 ? 19. 4 = –2. 43 m s–2 8. 5 ? 0. 5 (1A) The deceleration of car Y is 2. 43 m s–2. (c) Thinking distance = area under the graph during 0? 0. 5 s = 19. 4 ? 0. 5 = 9. 7 m (1A) (Correct shape) (Correct times) (Correct velocities) 1A) (1A) (1A) Braking distance = area under the graph during 0. 5 s? 8. 5 s 1 = ? 19. 4 ? (8. 5 – 0. 5) 2 = 77. 6 m distance are 9. 7 m and 77. 6 m respectively. (1A) The thinking distance and the braking (e) (See the figure in (d). ) (Correct slope – parallel to that in (d). ) (1A) (Correct position – above that in (d). ) (1A) 15 (a) Speed 70 km h–1 70 = m s–1 3 . 6 = 19. 4 m s –1 (d) The coloured area is equal to the difference in the stopping distances travelled by cars X and Y. (1A) (e) (1M) Stopping distance of car X = area under the graph during 0? 5 s 1 = ? 19. 4 ? 5 = 48. 5 m 2 Coloured area = 9. 7 + 77. 6 – 48. = 38. 8 m < 50 m Since the difference in stopping distances of the cars is smaller than the initial separation of the cars, the two cars do not collide with each other before they stop. (1A) (1M) (1M) Distance travelled by car Y in 2 s = vt = 19. 4 ? 2 = 38. 8 m < 50 m Since the distance between the cars is greater than the distance that car Y can travel in 2 s, the driver of car Y obeys the rule. corresponding v–t graph. Deceleration of car X = slope of the graph during 0? 5 s (1A) (1M) (b) Deceleration of a car is the slope of their 0 ? 19. 4 = 5? 0 = –3. 88 m s–2 The deceleration of car X is 3. 88 m s–2. (1A) 16 a) From t = 0 s to t = 5 s, the car moves with a uniform acceleration of 17 ? 0 = 3. 4 m s–2. 5 (1A)  © From t = 5 s to t = 20 s, the car moves with a constant velocity of 17 m s–1. (1A) From t = 20 s to t = 28 s, the car moves with a uniform acceleration of 0 ? 17 = ? 2. 125 m s–2. 28 ? 20 at rest. (1A) (b) s = ut + 1 2 at 2 1 = 0 + ? 17. 5 ? (8 ? 60)2 2 = 2 016 000 m (2016 km) (1M) (1A) The Shuttle travels 2 016 000 m (2016 km) in the first 8 minutes. From t = 28 s to t = 30 s, the car remains (1A) 19 (a) (i) The cyclist is using first gear when the acceleration is greatest before braking. shortest time. (1A) (1A) (1M) (1M) (1A) b) (ii) The cyclist uses second gear for the (b) Distance travelled = area under straight line PQ (8 + 6) ? 2 = 2 = 14 m The cyclist travels 14 m in second gear. (c) The acceleration during t = 18 s? 20 s 0? 9 = (1M) 20 ? 18 = ? 4. 5 m s–2 The deceleration is 4. 5 m s . –2 (1A) (Correct shape) (Correct time instants) (Correct accelerations) (1A) (1A) (1A) (1A) (1A) 20 21 (c) Yes. (HKCEE 2 005 Paper I Q1) 1 (a) s = ut + at2 2 1 = 0 + ? 10 ? (500 ? 10? 3)2 2 = 1. 25 m Therefore the minimum height the (1M) The car changes direction at t = 30 s. Its velocity changes from positive to negative, showing a change in its travelling direction. 1A) (1M) (1A) (1A) laptop must fall for it to be ‘saved’ is 1. 25 m. (b) v = u + at = 0 + 10 ? (500 ? 10 ) = 5 m s? 1 the ground is 5 m s–1. ?3 (1M) (1A) 17 18 (HKCEE 2002 Paper I Q8) (a) v = u + at = 0 + 17. 5 ? 8 ? 60 = 8400 m s–1 minutes is 8400 m s–1. The speed of the computer when it hits The speed of the Shuttle after the first 8  © (c) Most falls are likely to be from below this height, effect. (1A) (1A) (1A) so the protection will not have taken Physics in articles (p. 96) (a) 2. 45 m (b) (i) By v2 = u2 + 2as, u = v ? 2as u2 = 0 ? 2(? 10)(2. 45 + 0. 07 ? 1. 09) u = 5. 35 m s? 1 2 2 (1A) (1M) Take the upward direction as positive. 22 (a) Any one from: Rate of change of displacement Displacement per unit time (1A) (b) The velocity of a braking car is decreasing (with time) (1A) so the car has negative acceleration. (1A) Its displacement is (still) increasing with time, so its velocity is (still) positive In this case, the acceleration and velocity are in opposite directions. (1A) (1A) (1A) The vertical speed of Javier Sotomayor is 5. 35 m s? 1 when he leaves the ground. (ii) Take the upward direction as positive. Consider the upward journey. By v = u + at, v ? u 0 ? 5. 35 t= = = 0. 54 s a ? 10 (1M) (c) i) Consider the downward journey. 1 By s = ut + at2, (1M) 2 1 ? (2. 45 + 0. 07 ? 0. 71) = 0 + (? 10) t2 2 t = 0. 60 s The time that he stays in the air = (0. 54 + 0. 60) = 1. 14 s Alternative method: (1A) (Correct graph) (1A) Take the upward direction as positive. 1 By s = ut + at2, (1M) 2 (0. 71 ? 1. 09) = 5. 35t + 1 (? 10)t 2 (1M) 2 t = 1. 14 s or t = ? 0. 07 s (rejected) (ii) Vertical distance travelled = area under the graph from 4. 0 s to 10. 0 s (70 + 130)? 6 = 2 (1M) (1A) The time that he stays in the air is 1. 14 s. = 600 m (1A) The vertical distance travelled by the rocket between t = 4. 0 s and t = 10. s is 600 m.  © 3 1 2 3 4 C C Force and Motion 6 (a) The MTR train is accelerating in the forward direction. The man tends to move at his original speed (smaller speed), so he would move backwards relative to the MTR train. (b) The MTR train is slowing down. The man tends to move at his original speed (greater speed), so he would move forwards relative to the MTR train. (c) The MTR train is moving forwards at constant velocity. The man moves forwards with the same constant velocity, so he would remain at rest relative to the MTR train. (d) The MTR train is turning a corner. The Practice 3. 1 (p. 104) (b), (e), (f) 5 a) Stretching a rubber band (b) Standing on the floor (c) Walking time (e) (f) A compass A rubbed plastic ruler attracts small bi ts of paper (d) Exists in every object on the earth at any 7 man tends to move at his original direction, so he would move outwards relative to the MTR train. In space, the gravitational force acts on the spaceship is negligible. When the rockets are shut down, they do not exert a force on the spaceship. Therefore, no net force acts on the spaceship. By Newton’s first law, the spaceship is in uniform motion and can travel far out in space. 8 Joan moves on the ice surface with a constant velocity. Practice 3. 2 (p. 111) 1 2 3 4 5 C C D C (a) No. Athletes would hit the wall of the stadium if it is too close to the finishing line. (b) The mat is used to protect the athletes if they hit the wall after passing the finishing line. Practice 3. 3 (p. 122) 1 2 3 4 5 D A B A D  © 6 (a) 7 (a) Horizontal component = 40 + 30 cos 30 ° = 66. 0 N Vertical component = 30 sin 30 ° = 15 N Resultant = 66 2 + 15 2 = 67. 7 N Let ? be the angle between the resultant Resultant’s magnitude is 67 N and the angle between the resultant and the horizontal is 13 °. (b) and the horizontal. 15 tan = ? = 12. 8 ° 66 Resultant’s magnitude is 67. N and the angle between the resultant and the horizontal is 12. 8 °. (b) Horizontal component = 40 + 30 cos 45 ° = 61. 2 N Vertical component = 30 sin 45 ° = 21. 2 N Resultant’s magnitude is 65 N and the angle between the resultant and the horizontal is 19 °. (c) Resultant = 61. 2 2 + 21. 2 2 = 64. 8 N Let ? be the angle between t he resultant and the horizontal. 21. 2 tan = ? = 19. 1 ° 61. 2 Resultant’s magnitude is 64. 8 N and the angle between the resultant and the horizontal is 19. 1 °. (c) Resultant’s magnitude is 60 N and the angle between the resultant and the horizontal is 25 °. (d) Horizontal component = 40 + 30 cos 60 ° = 55 N Vertical component = 30 sin 60 ° = 26. 0 N Resultant = 55 2 + 26. 0 2 = 60. 8 N Let ? be the angle between the resultant and the horizontal. 26. 0 ? = 25. 3 ° tan = 55 Resultant’s magnitude is 60. 8 N and the angle between the resultant and the Resultant’s magnitude is 50 N and the angle between the resultant and the horizontal is 37 °. horizontal is 25. 3 °.  © (d) Resultant = 40 2 + 30 2 = 50 N Let ? be the angle between the resultant and the horizontal. 30 tan = ? = 36. 9 ° 40 Resultant’s magnitude is 50 N and the angle between the resultant and the horizontal is 36. 9 °. Hence, the angle between the two 5-N forces is 120 °. Alternative method: By tip-to-tail method, the two 5-N forces and the resultant 5-N force form an equilateral triangle. It is known that each angle of an equilateral triangle is 60 °. Therefore, the angle between the two 5-N forces is 120 °. 8 (a) 10 (b) Resultant force = 2 ? 400 = 800 N The resultant force provided by the cable is 800 N. 11 For the 2-kg mass: (c) 9 R = weight ? cos ? = 20 cos 30 ° = 17. 3 N Suppose the two forces act in the direction as shown. T = 20 N Therefore we have: Vertical component Fx = 5 sin ? Horizontal component Fy = 5 ? 5 cos ? = 5 ? 1 ? cos ? ) (magnitude of the resultant)2 = Fx2 + Fy 2 52 = (5 sin ? )2 + [5 ? (1 ? cos ? )]2 1 = sin ? + 1 ? 2 cos ? + cos ? 2 2 2T cos 45 ° = W 2 ? 20 ? cos 45 ° = W cos ? = 0. 5 W = 28. 3 N ? = 60 °  © 12 (a) 2T sin 10 ° = 500 T = 1440 N The tension of the string is 1440 N. 3 4 5 6 B C A Net force = ma = 40 ? 0. 5 = 20 N C By v2 – u2 = 2as, 0 à ¢â‚¬â€œ u2 = 2a(20) ? u2 = 40a u2 a=? 40 Resistance = ma = 12 ? ? u2 = –0. 03u2 40 (b) Component of force = T cos 10 ° = 1440 ? cos 10 ° = 1420 N The component of the force that pulls the car is 1420 N. 13 (a) 7 8 ‘A bag of sugar weighs 10 N. ’ or ‘A bag of sugar has a mass of 1 kg. By F = ma, F 800 000 a= = = 2 m s–2 m 4 ? 10 5 (b) As the mass is stationary, the net force acting on it is zero. When it flies horizontally, its acceleration is 2 m s–2. 100 ( )? 0 v? u (a) a = = 3. 6 = 4. 63 m s–2 t 6 The acceleration of the car is 4. 63 m s–2. (c) (i) y-component of F1 = weight of mass = 10 N 9 y-component of F1 = F1 sin 30 ° F1 sin 30 ° = 10 N F1 = 20 N x-component of F1 = F1 cos 30 ° = 20 cos 30 ° = 17. 3 N (b) F = ma = 1500 ? 4. 63 = 6945 N The force provided by the car engine is 6945 N. 10 (a) (ii) y-component of F2 = 0 x-component of F2 = x-component of F1 = 17. 3 N

Novel Critique: Will Grayson, Will Grayson Essay

One winter night in Chicago, two teenage boys named Will Grayson met by chance in an adult entertainment shop. They were both juniors in high school but they live in different suburbs. The book is told through alternating chapters between the two Will Graysons with John Green writing one and David Levithan the next. Their styles of writing are similar yet different enough which did not make it seem repetitive. John Green’s Will Grayson was written in normal type while David Levithan’s was written in all lowercase, which made the story more interesting. John Green’s Will Grayson believes that all problems and pains in life could be avoided by keeping quiet and not caring. Tiny Cooper, his best friend, is a proof of what heartache can do. Tiny is always in love with different boys and always have his heart broken. Tiny Cooper was a very perfect addition to the story. Him being John’s Will’s best friend and him meeting David’s Will and then them having a thing for each other, really helped the two Will Graysons in figuring out who they are. On the other hand, David Levithan’s Will Grayson is very anxious and does not have any real friends, except for a boy online who he’s in love with. When the two Will Graysons finally met, it seems like it is the most random thing that has ever happened. Though they only interacted a bit, when they did it was both so awkward and at the same time comfortable. The fact that meeting each other reminded â€Å"It’s hard to believe in coincidence but it’s even harder to believe in anything else.† them that just because they are both named Will Grayson does not mean their names define them, and that there are tons of people with the same out in the world, too. John Green’s Will Grayson does not want too much attention drawn to him, which did not exactly work for his best friend, Tiny Cooper. Tiny Cooper, his quest to make the world better though his musical and his and Will’s friendship is really just wonderful. David Levithan’s Will Grayson is angry and sarcastic but as the story progresses, readers would get to see the different sides of him. Also, he did not fit into any typical stereotype of a gay teen. In fact, in the conversation between the authors found at the end of the book, it is very interesting how David Levithan explained how he wanted his Will Grayson to be in the middle of things and also explained that Will writes in all lowercase because he sees himself as a lowercase person. The book blatantly attacks a lot of issues in the modern teenage world, and a lot of it is actually very harsh. At some points, it could be a little bit uncomfortable but it is the authors’ willingness to speak the truth. In a winter full of love, fake IDs, weird band names, two Will Graysons, and an epic musical about love and all the things about Tiny Cooper – Will Grayson, Will Grayson is really a hilarious novel about things that we cannot choose. While some humorous books are just humorous, Will Grayson, Will Grayson also touches on bigger issues that are relevant especially to teenagers nowadays.