A Literature Review Regarding Virus Protection

A Literature Review Regarding Virus Protection Although most of the people think that there is nothing to do a research in the area of computer virus protection, there are more and more things to research as everyday more than 200 computer malware are created by the virus creators. In the modern world most of the people are using computers in their day-to-day activities. So it is more important to have knowledge of computer viruses and protecting the computers from those viruses. begin{sloppypar} end{sloppypar} Identifying what the computer viruses are, their types, the threat to the computer from computer viruses, the present situation of computer viruses and prevention mechanisms from the computer viruses are the objectives of this literature review. end{abstract} section{Introduction} As the usage of the computers and creation of computer viruses are increasing all over the world, every computer user began to search about the computer viruses. But there are other various kinds of software like worms and Trojans that can do some harm to the activities of the computer other than the viruses. Although they are different from computer viruses, the computer users are used to call those other types of malicious software viruses. begin{sloppypar} end{sloppypar} Though there is no any real definition for the computer viruses, they can be considered as special kind of software programs that have the ability of self replicating over executable files reside in the computer and do some interruption to the activities of the computer. As the computer viruses are spreading only when the executable files are executing, the viruses can effect only for the executable files in the infected computer. So most of the time the files with .EXE , .COM , .BAT , .SYS extensions are infected. A computer virus can be written with a few lines of programming codes in any programming language. Any person who has a personal computer can write a computer virus and send it to another computer or system far away from the computer which produced the virus through a network or any disk. These viruses can destroy any massive computer system or network easily within few seconds. begin{sloppypar} end{sloppypar} Computer viruses which do less harm to the computers are only spreading over the computers and computer networks. They do not do any dangerous harm to the computers other than just replicating them in the system. The most dangerous type of viruses effect to the computers by changing the content of the files, partially of completely deleting the files reside in the computer. The data stored in the computer can be lost by infecting these kinds of computer viruses. These types of computer viruses cannot be catch by examining the files in the computer. But only the destruction they have done to the computer will remain. So capturing these viruses is the more difficult thing. begin{sloppypar} end{sloppypar} Preventing or protecting from computer viruses not only mean installing an anti virus program and scan the files by getting use of the anti virus program but also awareness of the computer viruses or malicious software and practicing best practices when using a computer. But most of the time most of the computer users trust various anti virus programs to protect their systems against computer viruses. Various anti virus programs use various methods or procedures to capture viruses and other types of malicious software. But with any of the computer protection method, they cannot fully protect the computer from computer viruses or malicious software. The next session of this review is considered about what are the types of Malware and how they infected to computer system. cite{1} newpage section{Types of malware} There is no standard method to categorize viruses into various types. But when we consider current situation of computer viruses in the world we can basically declare types of malware as follows, begin{itemize} item Trojan item Worms item Viruses end{itemize} subsection{Trojan} Trojan viruses do not reproduce in the computer but after a Trojan virus enters into the computer they just allow the outside persons to read the files reside in the computer. Usually Trojans steal passwords and send e-mails to hackers. Then the hacker will get the control of the users account. cite{2} subsection{Worms} Worms are kind of computer viruses copy and spread over the computer networks. It does not need a host to spread. Once they multiplied in a computer, the copied viruses scan the network for further multiplying and spreading via the network.cite{2} subsection{Viruses} Computer viruses are a program, a block of executable code which attaches itself to. It overwrites or replaces some code of computer program without knowing of computer user. A virus always needs a host program to reside. The virus is in its idle state till the host program it resides executes. When the host program executes the bock of code of the virus also executes and searches for another location which it can infect. The computer viruses can be categorized into number of categories like Resident Viruses, Direct Action Viruses, Overwrite Viruses, Boot Viruses, Macro Viruses, Directory Viruses, Polymorphic Viruses, File Infectors, Companion Viruses, FAT Viruses. cite{2} begin{itemize} item Resident Viruses Permanent viruses reside in the RAM item Direct Action Viruses This type of virus spreads and does its work when it is executing. item Overwrite Viruses These viruses delete the content of the files reside in the computer. item Boot Viruses This kind of virus infects to a boot sector of the hard drive or floppy. A boot virus can be infected to the boot sector of the computer by booting the computer from an infected floppy disk. item Directory Viruses These viruses change the path of a file. item Polymorphic Viruses These are encrypting their own code with different algorithms every time they enter into a system. item File Infectors Infect programs or executable files. They infect to a file when the program attached to it executes. item Companion Viruses These are working like resident viruses or direct action viruses. item FAT Viruses These infect to the file allocation table. item Macro Viruses This kind of virus infects to the files that have created using programs that contain macros. Currently most of the times they are affecting to Word 6, WordBasic and Excel as macros are created by WordBasic. In the present situation of the computer viruses, 15 percent of the viruses are macro viruses. On daily basis macro viruses are created by the computer users in their machines. New macro viruses are creating due to corruption, mating and conversion. Macro viruses are the most destructive kind of a virus. Most of the traditional anti virus programs are unable to detect those new macro viruses. cite{2} end{itemize} newpage section{How Viruses affect and infect to the system} begin{figure}[h] par includegraphics[bb =0 0 100 325 ]{virus.png} caption{Malware Detected by Year} cite{10} par end{figure} If the virus generation speed is greater than its death rate, a virus can easily spread within a short period of time. Figure1 shows how Malware spread with time. All the computer viruses do not activate at the time they enter into the computer. But sometimes they activate after some period of entering it into the personal computer or computer system. Some of them will never activate and some will activate and do harm to the files in the system or change the content of the files, format the hard disk, show a picture in the background. begin{sloppypar} end{sloppypar} There are lots of ways which a virus can enter into a computer. Most of the time, they spread and enter into a new computer through a computer network. With a removable media, it is possible to spread a virus. By downloading some games or software through a web site, a virus can enter into a new system. In the past there was a guarantee that the web sites do not contain viruses. But in the present situation, there is no guarantee that the web sites do not contain viruses. begin{sloppypar} end{sloppypar} Trapdoor is another common way of entering a virus into a system. Trapdoors are sometimes created by the programmers who developed the software to avoid going through the security procedure or avoid entering passwords during the period of time the system or software is developed. As a trapdoor is a way to enter into a system without entering a password, a virus can easily enter into a system through a trapdoor. begin{sloppypar} end{sloppypar} If have the attention to the new computer viruses, the code of some newly created computer viruses are encrypted so that the anti virus software cannot catch them. cite{3} section{Protection from computer viruses} To spread a virus from one computer to another, it should have the permission or ability to execute its code and do some modifications or completely delete the files other than the file the virus currently residing. According to those facts, protection from computer viruses means prevent the computer virus from copying it self to another location, the computer virus does not contain or avoid modifying or deleting the other files the computer virus does not reside. begin{sloppypar} end{sloppypar} If the content of a file has modified or edited without knowledge of the user, the user can suspect that a virus has been infected to the computer. Other than that when a virus has attacked a system, sometimes the performance of the computer can be reduced, various error messages are displayed or use some storage space from disk drives unexpectedly. begin{sloppypar} end{sloppypar} Worms normally find addresses to spread and they capture the addresses in three ways. Worms begin{itemize} item Randomly generate addresses item Find addresses in system tables item Find addresses in a program end{itemize} Protection against worms can include, begin{itemize} item Put passwords that cannot easily guess. item Remove some processes which reveal the secured data in the system. item Apply some solutions to the bugs. end{itemize} As worms are rapidly spread over networks and they are trying to overload the networks, protecting from worms include monitoring network activities and do isolation and deactivation of some parts of the network. begin{sloppypar} end{sloppypar} When it comes to protecting computers from viruses, the simplest things the user can do is always backup the data reside in the computer. But it is not a proper solution to deal with the computer viruses. As most of the computer users are now aware of the computer viruses they control write privilege to computer programs. After infecting a virus to a program since it changes the content of the file, there are some kind of software that can be used to check the content for irregular changes in its content. cite{4,12} section{Anti virus software programs} When protecting a computer from computer viruses with the help of an anti virus program, the service providers of those anti virus programs are providing their service to its clients in number of different ways. Some of the vendors or anti virus software are waiting for a request from a user for their product. After the client or the user requests, the service provider provides their service to the user. Another kind of anti virus vendor automatically downloads and installs their product into clients machine without the knowledge of the user. Some of the vendors are sending emails to the computer users mentioning the availability of their product. However some of those above mentioned ways have some ethically not relevant procedures. begin{sloppypar} end{sloppypar} Though thousands of anti virus programs, designed by programmers are there to detect computer viruses, they cannot play a perfect role in detecting computer viruses. More viruses are written for a new platform is the reason for that. To detect those new viruses, new detection technologies should be invented. There are number of computer virus detection methods. begin{sloppypar} end{sloppypar} Over some years ago, only the known viruses could be detected by anti virus programs. What those anti virus programs did were, selecting a string from known viruses and when a scan for viruses is started, anti virus program searches for files include that string and detect that file as an infected file. This method of detecting computer viruses looked only for static characteristics of known computer viruses. But as thousands of computer viruses are creating within a year, the industry people started to create anti virus programs that are detecting unknown computer viruses too. These methods are called heuristic method. begin{sloppypar} end{sloppypar} As computer virus infection has become a huge threats to who owns a computer and who uses computers. Most of the computer users are today getting the service of an anti virus program to detect malicious software or computer viruses. Getting use of an anti virus software is the most secured and popular way to protect the computers from malicious software. Anti virus programs identify the malicious software in two main approaches. begin{itemize} item They use a string matching approach to identify previously known viruses or malicious software. item The second method the anti virus programs use is capturing abnormal behaviors of any computer program running on the computer. end{itemize} begin{sloppypar} end{sloppypar} In string matching approach, the anti virus programs are getting use of a virus dictionary which contains the codes of previously known viruses. When an anti virus program starts to scan a file it refers to the virus dictionary and when it finds a block of code in the scanned file which also included in the virus dictionary, it quickly takes an action against the suspicious file. cite{5} begin{sloppypar} end{sloppypar} If an anti virus program uses capturing abnormal or suspicious behaviors, it monitors the behavior of the file which is scanning and if the file behaves abnormally the anti virus program detect the file as infected file an take an action against that. By using this method, the new viruses can also be detected. cite{5} subsection{The problem in virus dictionary method} As virus creators are now aware of virus detection methods, they are now writing the viruses so that the codes of the virus as encrypted codes when it is in a program. Otherwise they write the virus programs so that the code of itself doesnt look same as the real code of it. Because of those two reasons, the anti virus program cannot detect the infected file by matching the strings using the virus dictionary. The other problem of this method is it cannot detect a new virus which doesnt contain in the virus dictionary. subsection{The problems of detecting a virus by its behavior} As this method is searching for a suspicious behavior of the file that is being scanning, a files that shows abnormal behavior which has not been attacked by a virus or malicious software can also be detected as an infected file by the anti virus program. If the user gets an action against that file sometimes that non infected file can also be deleted. So modern anti virus programs do not use this approach to detect a virus. There are some fake anti virus software which do not clean or protect the computers. These fake anti virus software come with names which are similar to other real anti virus software. When a user sees such a fake anti virus software the user feels that this ant virus software is a real one and then he install than fake anti virus software into the computer. After installing that fake one in the computer, it displays fake messages saying there are some viruses in the computer and to remove it from the computer pay some amount of money. As these kinds of fake anti virus software are there with only the aim of earning money, the computer users must aware of that kind of fake anti virus software too. begin{sloppypar} end{sloppypar} Although there are so many anti virus software to detect computer viruses, the latest computer viruses cannot easily be caught by even the latest anti virus software programs as the code of virus is encrypted. What the virus creators do to hide the source code of the virus is encrypting the code of the virus and sends it to the computers. So the anti virus software cannot easily understand the code and they cannot catch that file as a suspicious file. Some virus creators encrypt the code and keep the key to decrypt in another file to make it more difficult for the anti virus software to find the virus. But good anti virus software, which have used good algorithms, should be able to detect those computer viruses too. cite{6,7} subsection{Most trusted anti virus software} cite{11} begin{itemize} item BitDefender Antivirus 2010 BitDefender is using advanced heuristic detection method and provides protection against online viruses, spyware,phishing scams and more. This provides protection by scanning web, IM and emails and this is capable of encrypting the IM s too. One of the new features BitDefender looking for is Active Virus Control which monitor the behavior of a file continuously. cite{13} item Kaspersky Anti-Virus 2010 Provides protection from viruses, Trojans, bots, worms and spyware. The interfaces and tools used are advanced but this provides agreat protection against most of the real threats. cite{14} item Webroot AntiVirus with SpySweeper 2010 This is a desktop anti virus package and protect the computer from viruses,trojans,worms and other number of malware. This catches the malware before the malware does any harm to the computer. cite{15} item Norton AntiVirus 2010 Uses signature based protection. However provides a new features like the proactive reputation scanning. But uninstalling the software might cause some problems as its partial uninstallation. cite{16} item ESET Nod32 Antivirus 4 Kind of desktop anti virus software. But this doesnt provide a complete security and misses some protection. This is not in the competition with other anti virus software. cite{17} item AVG Anti-Virus 9 Includes antivirus and antispyware protection. provides complette protection from harmful downloads and web cites. cite{18} item F-Secure Anti-Virus 2010 Great desktop anti virus. Has one of the most effective scan procedure and test results are shown to prove that. When installing this anti virus software, it has been automatically configured to remove the other anti virus software installed to the computer. cite{19} item G DATA AntiVirus 2011 Uses two distinct antivirus scanning engines, behavioral/heuristic protection, and even self-learning fingerprinting.This provides protection against malware spreading via emails and IM. The types of malware which are detected by this anti virus software are,phishing scams, dialers, adware, malicious scripts, Trojans, rootkits and worms. cite{20} end{itemize} section{Getting use of a firewall} The firewall is a kind of a program design to protect the computer from harmful things that are coming from the internet. Firewalls are divided into two categories as hardware firewall and software firewall. begin{sloppypar} end{sloppypar} Hardware firewalls are some kind of small hardware which can control the data coming from multiple computer systems. Software firewalls are kind of software that have the ability to block suspicious data coming to the computer from the internet. So to protect the computer from viruses and other kind of malicious software a software firewall and hardware firewall can be use. cite{5} section{Best practices to protect the computer from computer viruses} Though there are thousands of anti virus software are there, the computer users are also having the responsibility of protecting the computer when using the computer. They have to follow some best practices when they use a computer. begin{itemize} item Although the computer user is the owner of the computer he should not always log into the computer with the administrator privileges. If they log into the computer as a normal user, some kinds of viruses will not be able to enter into the computer. item A computer user should avoid from installing some anti virus software or some other software in some other persons computer. item As some viruses are coming with email attachments, when opening emails it is good to not to open emails from unknown addresses. item When downloading and installing anti virus software, download a recommended anti virus software. end{itemize} newpage section{Conclusion} According to the researchers, each and everyday over 200 computer viruses which can destroy a whole computer system within a few seconds are released by the computer virus creators. The worst thing that result an inflection is loosing data the reside in the computer. begin{sloppypar} end{sloppypar} Most of the time, these kind of destructions happen because of the lack of awareness of the computer users about the computer viruses. To mitigate the risk of infecting a computer virus to a users computer, the main thing we can do is make the computer users aware about the computer viruses, risks of infecting a computer virus and avoiding from computer viruses. begin{sloppypar} end{sloppypar} The people who have an idea about computer viruses most of the time trust anti virus programs. But just installing an anti virus software is not enough to protect the computer from computer viruses. The computer users also have the responsibility to protect the computer from computer viruses. As most of the time the computer viruses can come with the e-mail attachments, the e-mails from unknown addresses should not be opened. When downloading something from a web site the users should use only the trusted web sites and recommended software. But now there are nothing called trusted web sites. Even the software downloaded from Microsofts web site may contain viruses. Some viruses cannot enter into the computer if the user is logging into the computer with user privilege. So it is good to normally log into the computer with the user privilege. As viruses attack only the executable files, the write permission to those files can be restricted. begin{sloppypar} end{sloppypar} After installing an anti virus software program in a computer,to get the best protection from that anti virus, keeping it up to date is more important. But though there is an anti virus program installed in the computer, when plugging external removable devices into the computer, it should be scanned. begin{sloppypar} end{sloppypar} But the final conclusion which can come through this review is that though there are many protection methods,that are available in the world, a computer user cannot be completely safe from the computer viruses. That means any of the computer protection methods is not perfect in protecting computers from computer malware. newpage begin{thebibliography}{widest entry} bibitem{1} Markus Hanhisalo,emph{computer Viruses},Available at:http://www.tml.tkk.fi/Opinnot/Tik-110.501/1997/viruses.html# 1.Introduction% 20to% 20Computer% 20Viruses bibitem{2} Top Bits,2010, {http://www.topbits.com/types-of-computer-viruses.html} bibitem{3} McAfee,emph{An Introduction to Computer Viruses and other Destructive Programs},Available at: http://www.mcafee.com/common/media/vil/pdf/av_white.pdf bibitem{4} Stanley A. Kurzban, emph{Defending against viruses and worms},Available at:http://portal.acm.org/citation.cfm?id=68697 bibitem{5} emph{How AntiVirus Works},Available at:http://www.antivirusware.com/articles/how-anti-virus-works.htm bibitem{6} emph{How does anti-virus software work?},Available at:http://www.antivirusworld.com/articles/antivirus.php bibitem{7} emph{How Antivirus Software Detect Computer Viruses},Available at:http://security-antivirus-software.suite101.com/article.cfm/how-antivirus-software-dete bibitem{8} emph{What Is A Computer Virus? An Introduction To Computer Viruses},Available at:http://www.computertipsfree.com/computer-tips/security-tips/what-is-a-computer-virus-an-introduction-to-computer-viruses/ bibitem{9} http://www.washington.edu/itconnect/security/tools/ bibitem{10} Tech Bitz, http://tech-bitz.com/2008/04/05/virus-and-spyware-threat-is-larger-than-ever-before-anti-virus-companies-struggle-to-keep-up-with-flood-of-spyware/ bibitem{11} emph{AntiVirus Software Review },http://anti-virus-software-review.toptenreviews.com/ bibitem{12} Charles P. Pfleeger,Shari Lawrence Pfleeger emph{Security in Computing (4th Edition) } bibitem{13} emph{BitDefender AntiVirus Review },http://anti-virus-software-review.toptenreviews.com/bitdefender-review.html bibitem{14} emph{Kaspersky Anti-Virus 2010 },http://anti-virus-software-review.toptenreviews.com/kaspersky-review.html bibitem{15} emph{Webroot AntiVirus with SpySweeper Review },http://anti-virus-software-review.toptenreviews.com/webroot-antivirus-review.html bibitem{16} emph{Norton AntiVirus 2010 },http://anti-virus-software-review.toptenreviews.com/norton-review.html bibitem{17} emph{Trend Micro AntiVirus + AntiSpyware },http://anti-virus-software-review.toptenreviews.com/pc-cillin-review.html bibitem{18} emph{AVG Anti-Virus 9 },http://anti-virus-software-review.toptenreviews.com/avg-review.html bibitem{19} emph{F-Secure Anti-Virus Review },http://anti-virus-software-review.toptenreviews.com/f-secure-review.html bibitem{20} emph{G DATA AntiVirus 2011 },http://anti-virus-software-review.toptenreviews.com/antiviruskit-review.html end{thebibliography} end{document}

The Data Communication Networking Computer Science Essay

The Data Communication Networking Computer Science Essay Data Communication can mean many things to different people, but when industry with computers and their communication with them. So, people are usually discussing equipment that wan designed to provide or provide or gather information needs to communicate. Data Communication provides the tools, product and equipment to make it. The distance over which data moves within a computer may vary from a few thousandths of an inch. The amount of Data Communications builds from this point on, because there are many factors such as distance, topology, protocol, signaling, and security. Data Communications will continue to develop and change considerably for the probable future.  In the companies unlike manufacturers are not biased toward solution and the applications. Question 1 A group of business employees would like to set up a small networking office. Explain the meaning of topology. Discuss and draw the different types of network topology that are available. Answer of Question 1 Nowadays, networkings are very popular. So, a network consists of multiple computers connected with some type of interface, each have one or more interface devices such as a Network Interface Card (NIC). Each computer is support by network software that provides the server or client functionality. Network can be of any of the following three types such as Local Area Network (LAN), Metropolitan Area Network (MAN) and Wide Area Network (WAN). Hence, the local area networks (LANs) interconnect data processing devices that serve communities of users. In the first industrial in the context of the minicomputer world, than the LAN come into its to a personal computers (PCs) develop into the norm in the most networking environments. Seen the LAN networks were as the solution to the research problem. Then the computer could be fixed with a new I/O device and directly linked to one another. In the beginning, LAN is organizing to support shared printer access and to assist the movement of files between systems. The system would be taken long to harness the platform to support other application, such as email. The components of LAN are requiring the incorporation of many different components that determine how the devices are connected. Second is what the format of the data will be transm itted in. Third is how to ensure that multiple stations can transmit at the same time. By the way, this is the major elements of the LAN and some options commonly available to network designers. Metropolitan Area Network is the connection of devices that covers a geographical area of region that is larger than Local Area Network (LAN) but smaller than Wide Area Network. It implies the interconnection of network in the city into a larger network. Then the Wide Area Network (WAN) is the interconnection if devices across a geographical are. The connection spans from country to country. What is a topology? In a LAN, the organization can be described by the physical topology and the logical topology. The physical topology of network refers to the arrangement of cable, computers and other peripherals. Therefore, physical topology should not exist confused with logical topology which is the method how data actually transfers in a network as different to its design. So, the logical topology of a LAN is attaching devices and the flow of signals between attaches devices. Topology can be measured as a virtual shape or arrangement of a network. However, the shape actually does not match to the actual physical design of the devices on the computer network. The different topologies available to LAN, while discuss in the following. There are five types of topology network including star, bus, ring, tree, and mesh. Bus Topology Figure 1: Bus topology (Florida Center for Instructional Technology College of Education) (University of South Florida, 2009) Bus topology is the simplest ways a network can be organize. In bus topology, all computers are connecting to the same transmission line by using a cable. It is coaxial. Bus topology is easy to handle and put into action and is best suit for small networks. So, the advantages of bus topology are easy to use and understand. Second is requires least quantity of cable to connect the computers together. Therefore, it is less expensive than extra cabling arrangements. Ring Topology Figure 2: Ring Topology (  Network topology, kioskeas Creative Commons License Deed V2.0, 2007 ) (http://en.kioskea.net/contents/initiation/topologi.php3) In this type of ring topology, each computer is connect to the next computer with the last one connect to first. Consequently, each retransmits what it receives from the previous computer. Then the message flows around the ring in one direction. Ring topology does not subject to signal loss problem as a bus network experiences. By the way, there is no execution because there is no end to the ring. Ring topology advantages are each node has equal access and capable of high speed data transfer. Mesh Topology Figure 3: Mesh topology Mesh topology is a unique network design in each computer on the network connects to other. It is creating a point-to-point connection between each device on the network. The function of mesh design is to provide a high level of redundancy. If one network cable fails, the data always have an alternative path to get its destination. The advantages of mesh topology are provides redundant paths between devices and the network can expand without interruption to current users. Star Topology Figure 1: Star topology (Florida Center for Instructional Technology College of Education) (University of South Florida, 2009) A star topology is design with each file server, workstation, and peripherals. All of them are connect to a central network hub, switch, or concentrator. So, data on a star network passes through hub, switch or concentrator before ongoing to its destination. The common arrangement is use twisted pair cable. It also is use coaxial cable or fiber optic cable. The advantages of star topology including easy to add a new computer system to the network, crash of one workstation does not affect the entire network, uses a single access protocols and very fast Tree Topology Figure 5: Tree Topology (Florida Center for Instructional Technology College of Education) (University of South Florida, 2009) At the last tree topology is also known as a hierarchical topology and a central root node that is connecting to one or more nodes of a lower hierarchy. In each node the network has a exact fix number of nodes connect to a lower level. A tree topology combines individuality of linear bus and star topology. Hence, it consists of groups of star-configure workstations connect to a linear bus backbone cable in figure 5. Tree topology allow for the increase of an existing network, and enables schools to configure a network to meet their needs. Finally of advantages of a tree topology is point-to-point cabling for individual segments and support by numerous hardware and software venders. A group of business employees would like to set up a small networking office. Before set up the topology, office must consider when selecting a topology such as cost, flexibility and reliability. Cost of selecting that is selected for a Local Area Network has be install and perhaps a lengthy process including the installation cables and raceways. Another way for a network to be cost effective one would attempt to minimize installation cast. This may be achieved by using the suitable hardware linking cables, good modems, cost effective computers to reduce cost. Second is flexibility is one of the main advantage of a Local Area Network. It is ability to have the data processing and peripheral nodes distributes around a given area. Next be consistency is select for the network can help through allowing the location of the responsibility to be detect and to present come means of isolating the responsibility. The best of topology for small networking office is bus topology. It is the easy method of networking computers. So, this consists of a single cable as a trunk, backbone or segment that connects all the computers in the network. However, each system is directly attached to be common communication channel. Then signal is transmitter over the channel makes up the messages. While each message is passes along the guide each system receives it. After receiving the message each system scan the destination address contain in the message. On a bus topology signals are sending to all the computers in the network to keep the signal from active back and forth along the cable a terminator is place at the end of the cable. A bus topology only can one computer send data at a time, therefore the more computers in the bus slower data transmission in the network. Usually, bus topology is easy for small office use (example in figure 1). Question 2 The purpose of data link control is to provide functions like flow control, error detection and error control. Explain each function in detail. Answer of Question 2 The data communications have much more needed to control and manage to substitute. So, the list some of the requirements and objectives for effective of data communication between two directly connected transmitting receiving station such as frame synchronization, flow control, error control, addressing, error detection and recovery, control and data on same link and link management. So, line access controls determine which station can pass on next. This is easy for two stations on a full-duplex link. When more than two stations are in use on a full-duplex link such as multipoint or any number of stations is in use on a half-duplex line. However, transmission needs to be not acceptable suspiciously. The service of frame synchronization is the data link layer is responsible for providing synchronization at the frame level. This is determines the beginning and end of each frame. Therefore, the physical layer is usually responsible for maintaining bit synchronization. Flow control is so metimes of the receiving station must be able to cut off the transmitter, so the receiver may be too busy to accept of new frame. By the way, at the data link layer, flow control enables the receiver to tell the transmitter it is not ready, and to later identify its motivation to accept more frames. Another is error control is bits errors introduce by the transmission system should be correct. In the addressing on two station connections, addresses can be use to separate commands from responses. Then, addresses are necessary on multipoint links with more than two stations to denote the future receiver and sometimes to classify the sender as well. Error detection and recovery is using a grouping of order numbers and an error detecting or correcting code, so the Data Link layer protocol ensure that frame with error are accepted and not deliver to higher layers. Then, the recovery is by means of retransmission for error-detecting codes. Timers are use to ensure with the purpose of all transmit frames are receive. Maintain of control and data on same link. It is usually not attractive to have a physically divide communications pathway for control information. As a result, the receiver must be able to distinguish control information from the data being transmitted. The link management of initiation, maintenance, and termination of a sustained data exchange requires a fair amount of coordination and assistance among station. It actions for the management of this exchange are involve. This requirement is content by the physical interfacing techniques. A data link protocol that satisfies these requirements is a rather complex of issue. Begin to looking at three key of mechanisms that are part of data link control such as flow control, error detection, and error control. 2.1 Flow Control First part of data control link is flow control. Flow control is mechanisms are necessary in order to stop the transmitter form overwhelming a receiving entity with data. So, this can be achieved by letting the receiver control all data flow from the sender. Accordingly, the popular of flow control mechanisms allow the receiver to dens credit to the sender in terms of how much data can be transmitted. That function credit exist at the receiver call the window size. Flow control might be needed due to many reasons. At first is link the capacity. However, if the link is share to many transmitter-receiver pairs, the total amount of data on the link may exceed its capacity at some time. By the way, in second reason can be unavailability of sufficient memory resources at the receiver station. The link is possible is not busy and enough memory available to process or store the usual the data and still can congestion. So, congestion means a condition in which packets queue to be process inc rease a line above a certain threshold. The queue can simply be because of the receiving station having to forward in each packet on a slower link. At the same time, flow control also can be necessary and implement at all layers. When implement on DLC layers, the results in rule of data flow across a single link. Thus, of implementation scheme is when the widow size is in one packet. Finally, flow controls have two examples to processes. There are stop-and-wait (SnW) flow control and sliding window flow control. 2.1.1 Stop-and-Wait (SnW) Flow Control In this system, stop-and-wait is an entity transmits a packet. After, the destination entity receives the packet, and then it indicates its motivation to accept another packet by sending back an acknowledgement to the packet just receive. So, this small packet is call as Acknowledgement Packet (ACK). At figure 6 illustrates a timing diagram. Figure 6: Stop-and-Wait Flow Control (Data Communication Principles for Fixed and Wireless Networks, 2002) The transmission time is due to a limited capacity of a link. So, propagation time is due to a limited velocity of signal propagation. Then, the processing time is due to limited processing capacity of the receiving station. It depends on a number of factors, including but not limited to, processor type, queue size and protocol type use after receiving a data packet. 2.1.2 Sliding-window (SW) Flow Control Figure 7: Sliding-window (SW) Flow Control  ¼Ã‹â€ Data Communication Principles for Fixed and Wireless Networks. ¼Ã¢â‚¬ ° Sliding window is generality of Stop and Wait to more than one packet. This for system to receiving allows the sender to send up a sure maximum number of packets without getting further of ACK. Therefore allow to be transmitter without receiving an ACK is to be maximum window size. Usually each ACK allows to expanding the number of packets to the maximum window size. In the example, the receiver specifies a maximum window size of 4. The packets can be sequence number from 0 through 7 a maximum of four packets are allows to be transmit without getting further credit. Thus, suppose that packets numbers 0,1,2 and 3 have been transmit and then the receiver has not accept them. After the processing of receive packets is success then the receiver an ACK for the entire four packets. Finally, on receiving the ACK packet, the transmitter is acceptable to send packet numbers 4,5,6 and 7. 2.2 Error Detection Error detection is about communication impairments and the effect of data rate and signal to noise ratio on bit error rate. This system will be error, resulting in the change of one or more bits in transmitted packet. At example, there have two copies of data. The receiver compares copies equal then no error. So, the probability of same bits corrupted low. The parity is value bit. That character has even or odd number ones. The even number bit error goes understand. 2.2.1 Cyclic Redundancy Check (CRC) CRC is an extension of the parity block concept. It is nothing degree polynomial. The value of each bit is a coefficient. It is given a k bit block of bits, or message, the transmitter generates an n bit sequence. So, the resulting packet consisting transmit k+n bits which is exactly divisible by some number. The receivers then divide the incoming packet by that number and if no remainder, assume was no error. 2.3 Error Control Error controls are transmission impairments random and affect bits at random locations. So, describe a link with respect to its impairment effect is by probability of error. That call variously link error probability, bit error probability, bit error rate, or frame error rate. In addition, there have two type of errors are lost frame and damaged frame. Lost frame is a frame fails to arrive at the other side and damaged frame is a recognizable frame does arrive, but some of the bits are in error. In the most common technique for error control are bases on some or the entire following element. There are error detection, positive acknowledgment, retransmission after timeout, and negative acknowledgement and retransmission. 2.3.1 Stop and Wait ARQ This is sending station keeps a copy of every packet transmitted. Once transmission it waits for an ACK for each packet before sending the next packet. If an ACK received before the timeout, the stored copy of transmitted packet is discarded. So, an expiring the timer then the transmitter sends the copy of the packet again. If ACK damaged, transmitter will not recognize it. The transmitter will retransmit the same packet on timer expiry. There will start from sequence number; the receiver will know that this was duplicated packet. Example in figure 8. Figure 8: Stop and Wait ARQ 2.3.2 Go back N ARQ Go and back ARQ are the transmitter sends packets as allowed by current window size. If there are no errors in the packets, then the normal flow control operation continues as discussed above. In the way of error, the receiver discards the packet and does not increment its receiver window pointer. When it receives the packet with next sequence number, it may send a negative acknowledge (NAK). Usually, NAK implemented by sending the ACK packet asking for the discarded packet. If transmitter is already expecting an ACK for this packet, it will know that the packet in question was never received. Then go back by resetting its window passion at the discarded packet number and restart transmission of this packet. 2.3.3 Selective reject ARQ The only packets retransmitted are those that receive a negative acknowledge. It calls as SREJ, means time out. Protocols provide for a selective reject mechanism in which retransmission is sought for only the packet that was in error instead of a while black of packets. This can be accomplished by sending a NAK as soon as a packet is received in error. After the packet in error is successfully receive, all the packets can be marked as received and processes by the receiver. This performance measure of the ARQ schemes, then the selective reject tops. Finally, the performance of go back N ARQ is much better than stop and wait ARQ. Conclusion Finally I have to finish this assignment. What I have know about networking and Data link control. In first question is easy for me to do. Networkings have three types such as Local Area Network (LAN), Metropolitan Area Network (MAN) and Wide Area Network (WAN). The main for question asking is local area networking. Local Area Networking for personal computers and back end network and storage area network. Then, spare out five type topology. There are bus, star, ring, mesh and tree. All of them have advantages and disadvantages. Because, it can give users to choose which want is better for their home or company. In question two, I have explained all of them like flow control, error detection and error control. Three type of them are discussing transmit or retransmit. In conclusion, I have learned about networking how to set up their system and the mechanism how to process the packets.

An Analysis of the Fast Moving Consumer Product Industry and a Review of Kao Corporation :: Business Marketing Japan Essays

An Analysis of the Fast Moving Consumer Product Industry and a Review of Kao Corporation Executive Summary In this project, I have chosen the Fast Moving Consumer Product industry as the topic of study. First of all we will take a brief look at how the industry started in the late 19th century as soap making companies and slowly evolving into some of the most successful multidomestic company of today. Following we will have insight on the industry’s prominent characteristics and highlight some of the major players. We will also get an idea of the attractiveness of the industry through the use of Porter’s 5 forces industrial analysis. Included in this project is an in-depth review of Kao Corporation, Japan. Kao Corporation is one of the major players in the industry. Here we will take a look at how the Japanese based company employs strategies to reduce cost and at the same time differentiate its product from its competitors to gain competitive advantage. We will also examine some of the key financial ratios to aid us in identifying some of the company’s strength and weaknesses. Then a SWOT analysis is carried out on the company. From the SWOT analysis we can formulate suitable strategies in order to improve the performance of the company. By closely examining the company’s internal environment to better understand the company’s capabilities and limitations and then analysing the changes in the external environment that could affect the company favourably or adversely, appropriate strategies can be formed in order to ensure high performance of the company. Then finally we will look at other po ssible recommendation, which I believe would help improve the company’s performance in the competitive fast moving consumer products industry. Fast Moving Consumer Product Industrial Brief Fast Moving Consumer Product are products that consumer would use regularly. The product line of Fast Moving Consumer Products encompasses a wide range of products such as shampoo, body foam and facial wash. These products are classified as fast moving due to the nature of its usage and durability. While shampoos are non-perishables, the consumer would eventually finish utilizing it and would require to purchase another bottle of shampoo. Therefore, unlike products like television and radios which consumers would only buy once in a blue moon, Fast Moving Consumer Products are bought constantly from time to time by consumers. The Fast Moving Consumer Product Industry has evolving since the 19th century.

Technology is a Friend of Humankind Essay -- Argumentative Persuasive

Technology is a Friend of Humankind Technology is important in our world today. Terry Tempest William’s â€Å"The Clan of One-breasted Women† is about women having breast cancer because of bomb testing from 1952- 1961 in Utah. In this case technology has a negative effect on the human race. On the other hand, â€Å"The Technology of Medicine† by Lewis Thomas is about money and the technology of medicine. There are three different levels of technology in medicine according to Thomas and they are â€Å"nontechnology,† â€Å"halfway technology† (582), and â€Å"technology of modern medicine† (583). It is obvious that technology benefits humankind with cures for diseases and prevention of sickness. Without technology, medicine would not be as advanced and there would not be as many medical procedures as there are today. By using technology, doctors have found out how to prevent certain diseases or viruses by immunizations. In fact, the human race needs to have technology to advan ce the medical field, as disclosed in Richard Selzer’s â€Å"Sarcophagus† about surgery, from the doctor’s point of view, showing all of the technology he uses during procedure. All these essays have examples on how technology benefits human kind. Therefore, technology is a friend of humankind, when used with restrictions, because it has made improvements in medicine and has proved its usefulness in hospitals. There is no doubt that technology has to be used with restrictions because of what has happened with the bomb testing in Utah. Between the 1950s and 1960s the government decided to test nuclear bombs in a deserted area in Utah. Because of testing this technology out it has cost a whole clan of women to be diagnosed with breast cancer. As Williams puts it, â€Å"Children grow... ...es. Though technology has gotten humanity far, there is still a lot more information to be discovered. Without medical technology, many people would have died of diseases because of inadequate treatment. Therefore, technology has been a friend of humankind and will continue to be even a greater friend in the future. Works Cited Klass, Perri. â€Å"Macho†. The McGraw Hill Reader. 8th ed. Ed. Gilbert H. Muller. New York: McGraw Hill, 2003. 577-580. Selzer, Richard. â€Å"Sarcophagus†. The McGraw Hill Reader. 8th ed. Ed. Gilbert H. Muller. New York: McGraw Hill, 2003. 564-570. Thomas, Lewis. â€Å"The Technology of Medicine†. The McGraw Hill Reader. 8th ed. Ed. Gilbert H. Muller. New York: McGraw Hill, 2003. 581-585. Willams, Terry Tempest. â€Å"The Clan of One-breasted Women†. The McGraw Hill Reader. 8th ed. Ed. Gilbert H. Muller. New York: McGraw Hill, 2003. 598-604.

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

Difficult Teachers: Recent Development on how they can be dealt with

Competition as a factor plays an important role in shaping up nearly all sectors in the present operational environment. Competition is rife in all sectors and education is swimming within these tides. Success of the teaching staff is largely guided by results and having even two ineffective teachers could lead to a bad name to a principal and his administration (Brock, & Grady, 2003). Inefficiency of the teaching staff can lead to loss of accountability and even place a school at risk of losing its reputation or failing to develop any.Like in the result oriented business world the ability of the teaching staff to set standards of peak performance or be close to the top performers is important in ensuring that parents and hard working teachers remain motivated and always seeks the best for students. Such levels of motivation are bound to trickle down to students and the result may be improvement in the levels of performance and development of a culture of success (Hopkins, 2009).It i s thus upon administrators especially principals and head of department to ensure they develop creative, humane, supportive, tough and timely approaches to deal with teachers who are not performing for one reason or the other. This is further complicated if the teachers being referred to are difficult to deal with. Appreciation of the Problem In practical teaching, many principals confess of having dealt with difficult teachers in more than one occasion. It is generally believed that each institution has what can be referred to as troubled teacher (Hopkins, 2009).The fact that the teachers are difficult to deal with must not be assumed to be directly correlated to their performance. In some cases the best teachers can prove to be hard to deal with which presents a larger problem considering the impact they could have on a school if they were to quit or their problem addressed badly. It is generally true that troubled, exhausted and even confused teachers have multiple negative impac ts on morale and school environment.Such troubled teachers have the ability to single handedly break the team spirit that is critical to staff success which results in fragmentation that is a breeding ground to failure in issues relating to school improvement initiatives. Difficult teachers have been termed by some researchers as a proverbial elephant in the staff. All in the staff are often aware of the existence of such a character but none is willing to confront for the fear of losing or what most refer to as being ‘trampled'.Many low performing and even high performing (based on result) institutions are overrun by such teachers who appear to be operating within their own code. It has been observed that some principals are even scared of such troubled teachers and though they are aware of their existence they do little to address the situation (Wilmore, 2007). The reality is that ignoring the existence of such teacher does little to address the situation which is let to man ifest within the teaching environment and with time the negative effects may even be observed in the levels of performance that can be attained by a staff (Brock, & Grady, 2003).Failure in performance may not only result to the ousting of a principal but also affect the lives of students who may have had better futures had it not been for the failure of the administrators in addressing difficult teachers. Principals must be appreciative of the fact that their roles as leaders is worth the risk because the goals seek more than material gain or advancement. This should also involve appreciation of the fact that the lives of the people within the school community and even outside the school community thus the society is dependent on the school system to provide meaning and purpose.This appreciation must be reinforced by personal assessment of the administrators to determine if they have the ability to effectively handle difficult teachers. Principals as managers and leaders have the ro le to ensure that the school community is motivated by dealing with the challenges to high levels of motivation and ensuring that impediment or threats to achievement of educational goals are addressed (Brock, & Grady, 2003). Administrators have the ability to misjudge a straight or good teacher for a trouble maker. This is especially true for administrators who are still getting the feel of being in a new environment.Depending on the existing administration for support is cited as a possible avenue to ensuring that new administrators get a feel of the environment and therefore develop objective assessment of the staff including their own ability to effectively manage the challenges presented by the environment they are in (Wilmore, 2007). Difficult teachers can come in hordes or could be unique in a staff that is highly cooperative. Even in a dysfunctional school community, there are teachers who are considered difficult. Researchers have come up with values that they view as being important in dealing with difficult teacher in varied conditions.The values that must be inherent of administrators can be developed and play a role in ensuring that difficult teachers are dealt with in both functional and dysfunctional environments. Assertive administration is cited as one of the critical success factors in dealing with difficult teachers. The term difficult in difficult teachers is not out of their knowledge of martial arts or spiritual ability rather is assertiveness that manifests negatively. The level of assertiveness that such teachers display has been cited by some as being manifested in difficult to deal with but successful teachers in class.Disregard for the existing systems and rule on a regular basis that defines difficult teachers is a manifestation of a negatively developed assertive nature (Brock, & Grady, 2003). It is only an assertive administrator that can effectively manage such a teacher and even reform his ways into those accepted within the est ablished systems. Character building is an important quality that administrators should possess if they are to effectively deal with difficult teachers. Character building is the ability to mould the perception and thereafter actions of a teacher in a manner that leads to their entry or fit into an existing set of values.Character affects perception and therefore actions and should thus be developed in a manner that is positive if the actions are to lead to generation of value to both students and other teaching staff (Wilmore, 2007). Many researchers have come up with findings that blame the behavior of difficult teacher on the nature of their cognition. Such difficult teachers display their negative character irrespective of the nature of administration thus addressing the character issues appear to be the best approach to dealing with difficult teachers.Communication is considered one of the important tools that managers and leaders have in ensuring that goals are transmitted thr ough out a system and people are motivated towards achieving set goals. Under constrained teaching environment, teachers may take on repulsive behavior to gaining the attention of the administration to issues that may be affecting their efficiency in teaching. While a negative teaching environment is not a precondition for difficult teachers it is a possible cause of negative teachers (Wilmore, 2007).Communication is an avenue through which the nature of difficult teachers can be discerned and it is only from this understanding that suitable intervention measures should be developed. Administrators must study and personalize the art of affective communication if difficult teachers are not to arise from a teaching environment and to also ensure a proper understanding of the negative teaching behavior. Developing a positive school culture has come up as one of the critical success factors in reducing the prevalence of difficult teaching.Poorly performing schools have been recorded as being a breeding ground for difficult teachers (Whitaker, 2002). A poor school culture develops a negative picture of what is expected of a teacher and develops a breeding ground for negative perception and energy that could result in difficult teachers. Some researchers have tried to develop a theory in a bid of have a clear image of difficult teacher formulation in an environment which postulate that a negative culture direct the otherwise positive energy that could have been channeled into positive development to poor interaction with students and other members of staff (Wilmore, 2007).The role of developing a positive culture has been studied in business and involves interaction with positive people and promotion of positive norms and values. The role of managers and leaders is prominent in this phase and could be the defining factor between failure and success. Contribution of the administrators to teaching and addressing issues that teachers and students are faced with plays a n important role in ensuring that administrators are appreciated as part of the school systems.By contributing to issues relating to staff development administrators are placed in a position where they can effectively monitor events within their environment (Whitaker, 2002). This is in line with proactive approaches to issues where reporting systems are a formal or documentation systems and not an avenue through which leaders gain insight of problems. By actively contributing to the school community and being at the fore in addressing issues, administrators are put in a position where they can be effective in ensuring overall school development.Conducting assertive interventions and timeliness are the other important values that administrators must display to be able to effectively manage difficult teachers. It is generally believed that the most effective way to deal with students is addressing the teachers. A school in its basic definition takes on a hierarchical structure was the teachers act as a bridge between students and the administration (Whitaker, 2002). Assertive intervention systems are important in ensuring that other teachers whom are often aware of the existence of a problem become informed of strategies that are being taken to address them.Timeliness is a value that is of critical importance in minimizing the negative effects of difficult teachers in a school community. Procrastination has been cited as one of the avenues through which administrators let negative effects of difficult teachers affect existing systems with negative repercussion (Whitaker, 2002). Addressing procrastination is therefore a critical success factor and can only be attained if timeliness as a factor is ingrained within systems seeking solutions. Manifestation of Difficult TeachingThe ability to identify problematic teacher is important in ensuring that the effect that he has on other members of the teachings staff are mitigated. Difficult teachers come in different for ms according to a recent survey that seeks to develop a clear understanding of difficult teachers (Whitaker, 2002). Understanding the exact manifestation of difficult behavior in teachers and interaction with other staff members and administration is important in devising strategy that can be used in dealing with threats that they pose.Complaining and negative teachers have been cited as the most difficult to deal with. It is advisable that such teachers be directly and confidentially addressed (Waterman, & Waterman, 2006). Ensuring that teachers are aware of the effects that their behavior has on existing system and achievement of the educational goals and presenting a platform for them to raise issues that may be affecting them in a confidential manner is important in ensuring the interventions do not manifest negatively. Research shows that some principals however choose to ignore the negative comments from difficult teachers.Though success of the interaction in such a case is de pendent on the degree with which the administrators can focus on positive staff members, ignoring the negative staff members is misplaced and could be counter-productive. Stating expectations and offering assistance have for a long time been considered vital steps in dealing with difficult teacher; however, the change in environment and the need for leaders and managers to be proactively involved in day to day running of their systems has led to increased requirement on administrators (Whitaker, 2002).Noise makers and anarchist are considered the second most common characteristic of difficult teachers. Directly addressing their behavior groups is considered the first step to addressing issues they may be faced with. Stating what is expected of them including policies, behavior and expectation while monitoring progress are also considered success factors (McEwan, 2005). Research shows that if the initial intervention framework fails in developing observable change in the teachers pla cing them on an improvement plan is considered the next phase by many.In most cases, the intervention system for these behavior group end with a teacher being asked to leave a faculty after all measures fail in developing positive behavior change. Difficult teachers can also be defined by high propensity to gossip. While gossip may be considered a form of communication in any social setting if it continues to grow it could prove difficult for administrators to replace it with the truth. A factor that is widely appreciated as being a stumbling block to effectively addressing issues that an organization is faced with is gossip which affects the level of efficiency that can be attained in communication.Administrators must ensure that such members of staff are made aware of the negative effects that gossip could have to the attainment of school goals. Most leadership experts point to the fact that openly showing disgust and disapproval of gossip could lead to positive results (McEwan, 2 005). Institutions that are aware of the negative effects that continual gossips has on communication efficiency have in the past asked difficult teachers to consider leaving a faculty if they could not deal with their love for gossiping.Backstabbers are another problem group where the direct approach is cited as being most effective. In fact a more direct approach than in all other cases has to be used in confronting culprits with questions on the why, what, where and when regarding a case. Depending on the magnitude of a case the intervention systems may either involve letting the culprit be aware of the fact that his actions have been brought into light and citing insubordination which may also involve restating expectation of behavior and initiation of an improvement plan (McEwan, 2005).Research shows that most principals are aware of the existence of such backstabber but consider their actions less influential on attainment of goals. Backstabbing is viewed by principals as diff ering opinions rather than lack of appreciation of the input of other members. This is a negative perception that is reflective of the differences that exist between practice and research. Discussion There appears to be a wide appreciation of the extent of difficult teachers. Researchers have tried to dissect the problem from different dimensions to ensure that its emergence, manifestation and even approaches to its management are well understood.Behavior development; nature of experiences that a teacher has undergone; the nature of the operational environment and objectives of the teachers are factors cited as being vital in defining the extent and effect of difficult teachers. There is no doubt on the effects that such teachers could have on the levels of morale and even attainment of administrative goals. One of the most important developments in recent research is an effort to develop a clear understanding of the different behavior groups and how each group can be addressed.Dire ct measures and restatement of the vision have been stated as being vital in ensuring effective management of different cases. The role of the administrators and the critical success values are applicable in the measures that have been developed for different behavior groups. It is apparent that there is little that can be done by researchers to address individual cases however administrators can pick from the general guidelines that have been developed to come up with measures in management and leadership that can ensure difficult teachers are dealt with in a manner that leads to professional development.Critical review of success factors and steps involved in mitigating and addressing specific behavior problems points to the fact that effective leadership and management are important in identification and address of difficult teachers as a key problem in achievement of schooling goals. The art and science aspect of management have to be reinforced with appreciation of the potentia l effect that problem teachers have on attainment of a school's objectives and facilitation of communication and transmission of positive values in attaining efficiency.There is no doubt on the role played by innovation, creativity, skills, experience, coordination and overall strategic management in ensuring difficult teachers are managed and their effects addressed. Conclusion Difficult teachers have the potential of disrupt learning and lead to loss of morale in the teaching staff. Understanding the teachers which involve ensuring high levels of interaction with them and application of strategic management principles in addressing the challenge they present have widely been discussed by researchers.There however appears to be a gap between research and practice in that some principals and administrators despite the effects that difficult teachers have on attainment of school goals ignore their existence. Researches on how this appreciation can be developed are lacking and are one of the key areas that have to be addressed if the current state of research is to be helpful to practicing administrators. Little has been done with regards to difficult administrators who may in fact pose a greater threat to attainment of schooling goals.A further understanding of the cause of difficult behaviors among teachers must be developed to create a good platform for administrators to base their interventions. In a nutshell, the current researches provide a suitable platform for definition of strategic directions that can be taken in addressing difficult teachers; it is however upon administrators and the entire school community to develop specific approaches that are relevant to their cases which differ different owing to different values and expectations that characterize schools.