The Market Research Analysts Approach to the Decision Making Process †Business Research Paper

The Market Research Analysts Approach to the Decision Making Process – Business Research Paper Free Online Research Papers The Market Research Analysts Approach to the Decision Making Process Business Research Paper As a market research analyst, the first step in my decision making process would be to analyze the situation. The problem is that my company has just created a new type of packaging product that improves the shelf life of food products, but the product development department does not know what to do with this novel product. The product could potentially be marketed to consumers and businesses, in which case it would be a convenience product (consumer) or a packaging/supply product (business). The problem is not only what to do with this new product, but who to market to and how to market it. Before performing any research on the potential market, I would test the product as extensively as possible to make sure that it is a quality product. The most important thing when selling a product is quality, and if they product is not up to company standards, I would throw the idea out, or perform research to make the product the best it could be. If the product was up to the company’s standards of quality, I would then move on to researching the market. After testing the product’s quality, I would also identify what the potential uses for the product are. Once I have identified my problem and thoroughly tested my product, I would perform as much research as possible. First, I would research my competition’s product development to determine if anyone else in the industry has created this new form of packaging as well. If they had created the product, I would research their product development and marketing strategies. If they had not created the product, I would research for other new and innovative technologies that could be potential competitors with my new product. After performing research on my competition, I would focus on the potential consumers for this product. For this particular type of product, I would use focus group research as my primary research tool for usage of this product. I would use the product in several different ways and test the reactions of the consumers to determine which would be best to market. In order to select people to participate in the focus group, I would choose people from many diverse demographic groups to determine whether the responses to the product would be similar in people from different age groups, education levels or ethnicities. After the focus groups are completed, I would follow up with a consumer survey to get the participants reactions to the product. Using this information gained from performing the focus groups and consumer surveys, I would be able to determine the potential consumer markets for this product. I would also have to take into consideration my potential customers, including existin g customers and unrealized customers. Once the consumer research has been completed, I would focus on the business market. Is there a need/desire for a product like this to be sold to other companies? Since the demand for industrial products is derived from the demand for consumer products, I would assume that there is a desire for business to use this product as well (given that based on consumer research, this product is something that consumers would be interested in using in their homes). When all the testing and researching is completed, it still must be decided what to do with this product, and in order to do so, the options must be considered. First, using my research on consumer and business responses to this new product, I could sell the product. If I am a producer of food products, it would make sense to market the new packaging product as well as sell more of my own food products. For example, if I sell dairy products, I would market my food products as usual, but I would emphasize the fact that my company is using new and advanced packaging technology that will keep the consumer’s food fresher for longer. If I do not sell food products, one of my options is to license my new product to permit another company to use the new technology. If I choose to license the packaging technology, it would limit my involvement greatly. Of course, I would determine the advantages and disadvantages of this form of business and weigh them against my other options. Another option for my new product is to form a joint venture with another company (regardless of whether I sell food products). In this case, I have a very useful product that my company would be contributing to the venture, so I would have to choose carefully before determining that this is where I want my new product to go. There are so many potential uses for packaging technology, and surely there would be great potential from food producers to get their hands on this product. I think this would be my last choice in deciding what to do with this product, but I need to weigh my advantages and disadvantages and determine what the risks are in going into this form of business. Once I have determined what my options are, the risks and advantages of each must be compared. Using this information, in addition to my research gathered, it will be easier to decide what to do with this product. In order to develop successful new products, there are three activities that must be completed: first, to uncover unmet needs and problems, second, to develop a competitively advantaged product, and third, to shepherd products through the firm. The first step in this process has already been identified: the new packaging technology increases the shelf life of food products, and that is the â€Å"unmet need or problem.† Developing a competitively advantaged product is much more complex. This step of the process involves: (1) input from customers, (2) input from marketing about what competition is doing about the need, (3) manufacturing input about what is being done to satisfy the need, (4) engineering input about additional technology that is available, (5) R and D input about new ways of addressing the need, and (6) financial input regarding costs. I have already received customer input by performing the focus groups and consumer surveys, but I do need to complete every ot her step before I can move forward in deciding what to do with this technology. It is important to consult with every department so that I can once again identify my risks and my advantages in each step of this process, before I can decide what to do with my product. Once i have consulted with each department and decided how to market this product, the objectives for selling the product must be established. Research Papers on The Market Research Analyst's Approach to the Decision Making Process - Business Research PaperMarketing of Lifeboy Soap A Unilever ProductRiordan Manufacturing Production PlanBionic Assembly System: A New Concept of SelfAnalysis of Ebay Expanding into AsiaResearch Process Part OneOpen Architechture a white paperThe Project Managment Office SystemMoral and Ethical Issues in Hiring New EmployeesGenetic EngineeringDefinition of Export Quotas

Societal Functions of Religion According to Emile Durkheim and the Essay

Societal Functions of Religion According to Emile Durkheim and the Decline of Religion in the United States - Essay Example The wealth that the U.S. has enjoyed over the past decades, the freedom of expression entrenched in the U.S. Constitution, and the acceptance of other religions into the American society have diluted the place of Christianity in the U.S. The decline of Christianity in the U.S. at the expense of other religions and secularism risks to threaten the American society if it fails to provide a compromise to people from other faiths or sharing other beliefs. Such risk exists because there could be clashes between people from different communities/ faiths (e.g. Northern Ireland). The situation, however, is not alarming because religion has been replaced in the U.S. by education and to some extent a sense of patriotism. To revert back to Emile Durkheim's theory of religion, patriotism, education, the "American dream", Hollywood, etc. now ensure some form of social cohesion in the U.S. through the sharing of common values. With regard to social control, the law enforcement system and the judic iary system are now recognized as legitimate institutions to deal with conflicts. In this modern U.S. society, religion has been relegated to the personal realm and people living in the U.S. now enjoy the freedom to choose the religion that will answer their existential thirst.

Children and Technology Research Paper Example | Topics and Well Written Essays - 2250 words

Children and Technology - Research Paper Example On taking Matthew’s history, the doctor learnt that Matthew has been in the habit of spending more than 7 hours watching the television and playing video games on the computer for the past 9 years or so. Matthew told, â€Å"I started gaining weight quickly from a very young age, and this, along with my poor academic performance, made me an easy target for bullies. I found it difficult to settle in school and to make friends. Excess playing of fighting games often got me into fights with the bullies of the class, often believing that I would be able to imitate the kung fu moves of my favourite video game character. With such a discouraging environment at school, my only resort was my mum’s computer and the television- something that has been my pastime as far as I can remember.† Matthew’s example shows a more deep-rooted problem. Years of computer and television use have taken their toll on him. This gives rise to the question that is use of technology just ified. Use of technology is common both at homes and schools. The primary element that the schools instil in children through the use of technology is the ability to adapt to change. For this purpose, technology has emerged as a saviour, affecting both the efficiency of pedagogical tools and the individualization of education for the students. However, the varied use of technology in various fields has led to negative impacts too- a topic that is the hotbed of debate for the past few years. The government and the media have strongly been favouring the use of technology in education. Most educationists are of the perspective that learning computer skills at an early age result in better performance in the workplace. However these claims are challenged by skeptics who believe that technology has little contribution towards the enhancement of learning and tends to affect not only the minds but also the bodies of the children. It impacts the imagination of children and initiates a proce ss of degenerative changes in the body. Also, increased use of technology takes time away for physical activities. According to Jane Healy, computers are not an essential part of the lives of young children and have a deleterious effect on the motivation and imaginations of the children. She argues that institutions should wait for children to attain the age of 7 before exposing them to computer use. Despite the benefits technology provides, such as being an interactive interface for learning, use of non-technological ways of learning should be encouraged for young children in schools. Excess use of technology adversely affects the bodies and imaginations of the children; thereby making it a viable solution to discourage the use of technology for children below the age of 7 and use non-technologies based methods to learn basic knowledge such as promoting play, using pedagogical tools to actively engage children in physical activities and mobilizing the media to promote reduced use o f technology in schools for children aged less than 7. In the early years of life, the child is in the process of learning and writing. He is undergoing a series of linguistic and verbal changes with the process of identifying letters, analyzing them and their sounds and then spelling them out being achieved by the age of 7. What would happen if this complex process is disrupted at some point by use of technological means, interfering with the natural order of learning. If a child is exposed to the computers even before they are able to write, it may have significant effects of his health and development. It is seen that gaming is the most popular use of computers for children aged

Auditing case assignment Study Example | Topics and Well Written Essays - 500 words

Auditing assignment - Case Study Example The balance sheet accounts will substantiate or given ample evidence that the revenues reported in the income statements are valid and relevant. For, an increase in cash in the balance would indicate that a cash sales transaction occurred. However, an increase in cash could also be attributed to collection of accounts receivable and not a new sales transaction. In addition, an increase in receivables account would indicate a possible sales on account transaction. On the other hand, an increase in receivables account could be due to an reinstatement or reversal of an accounts receivable that had been written off. Accounts receivable are written off if there is a probability that the a certain amount of receivables can no longer be collected because the customer may be absconded or has filed for bankruptcy in the courts2. Yes, the substantive approach appropriately was applied in EM's audit of Maxall. First, EM determined that a violation of the auditing principle that segregation of incompatible functions occurred. Auditing principles state that incompatible functions of recording, keeping and approving should not be placed in the shoulders of one person.

A Cup of Hot Tea Essay Example for Free

A Cup of Hot Tea Essay A cup Of Hot tea Tea is a kind of drink. We get It from leaves. Now-a-days it Is popular drink in the world. where grows: Tea grows well in the hilly places where rain water can not stand at all. It grows in Bangladesh, India, China and Japan. In Bangladesh it grows well in the hilly areas Sylhet and Chittagong. How grown: Tree plants grow seeds. Seeds are shown in March; seedlings are planted In rows five feet apart. They are regularly prunced and allowed to grow only four or five feet high. How gathered: When tree lants are four years of age, leaves are plucked three or four times a year. The leaves are first dried in the sun and roested in fire. Thus the leaves are ready for use and sale. How prepared: At firest water Is boiled in a pot. Tea leaves are put Into boiling water. After three or four minutes it Is poured into cups through a sieve. The suger and milk is mixed with it. Thus it becomes a good drink. usefulness: Tea is a useful drink to us. Refreshes body and mind. Gives us energy to work. It also helps us o keep awake. Taking tea three times a day Is healthy for healt While gossiping with our friends and relatives we can not heardly think without a cup of hot tea. It brings a new mood of gossiping And In our country it is an important crop also. Demerites: Tea is not always good for healt. Too much of it is bad for health. It kills our hunger. The Importance of tea In our national economy Is very great. It Conjunslon: brings a good deal of foreign money every year. So we should take care of better production of tea.

Dayton Hudson Corporation Case Analysis :: GCSE Business Marketing Coursework

Analysis of Dayton Hudson Corporation Case In the case of Dayton Hudson Corporation, the company fell into a situation of a hostile takeover attempted by the Dart Group in 1987. At that time, Kenneth Macke was the CEO of the Dayton Hudson Corporation and sternly disagreed with letting the company fall into the hands of the Haft’s. Macke’s decision on what could be done to terminate the takeover turned the circumstances over to the hands of the state of Minnesota where Dayton Hudson’s headquarters resided. Macke requested a special session of the legislature to revisit the Minnesota corporate takeovers statute. This proved to work in Dayton Hudson’s favor and a statute was enacted that left the decision of a takeover up to the Board of Directors of the company. The actions that were taken by Kenneth Macke to assist the company with the their takeover situation were an example of how business relies on the government to be responsible for social issues. When pleading their (DHC) case to the governor, Dayton Hudson made sure to make know all of the contributions that they have made to the community and how they have provided safe and secure jobs for its residents. They claimed that if they were to be taken over, the Haft’s would probably break up the company and sell it off to pay for the expenses incurred from the takeover. Jobs would be lost and there would most likely be no social contributions made by the Hafts. Out of the thinkers that we have studied regarding business’ responsibility, I think that the two that would agree with Macke’s decision to look to the government would be both Freeman and Friedman. In the readings that we have covered, we have seen that they are believers in that the government is who should be responsible for social issues in some way or another. I think that Freeman may feel a lot more strongly about Macke’s decision than Friedman because he argues that government is the sole caretaker when it comes to taking responsibility for social issues while Freeman argues that it is mainly government’s job, but that business plays a huge role in that responsibility. Friedman’s views are that business’ main responsibility is to maximize shareholders wealth and that in doing so, they are being socially responsible. He also contends that corporations are not people and therefore they cannot be responsible for social issues.

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