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2A: 2.1 #1-25
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4.2 A #1-25THE FIDE LAWS OF CHESS. Introduction FIDE Laws of Chess cover over-the-board play. The Laws of Chess have two parts: 1. Basic Rules of Play and 2. Competitive Rules of Play. The English text is the authentic version of the Laws of Chess (which were adopted at the 93rd FIDE Congress at Chennai, India) coming into force on 1 January 2023. Preface. The Laws of Chess cannot cover all possible situations that may arise during a game, nor can they regulate all administrative questions. Where cases are not precisely regulated by an Article of the Laws, it should be possible to reach a correct decision by studying analogous situations which are regulated in the Laws. The Laws assume that arbiters have the necessary competence, sound judgement and absolute objectivity. Too detailed a rule might deprive the arbiter of his/her freedom of judgement and thus prevent him/her from finding a solution to a problem dictated by fairness, logic and special factors. FIDE appeals to all chess players and federations to accept this view. A necessary condition for a game to be rated by FIDE is that it shall be played according to the FIDE Laws of Chess. It is recommended that competitive games not rated by FIDE be played according to the FIDE Laws of Chess. Member federations may ask FIDE to give a ruling on matters relating to the Laws of Chess. BASIC RULES OF PLAY. Article 1: The Nature and Objectives of the Game of Chess 1.1 1.2 1.3 1.4 The game of chess is played between two opponents who move their pieces on a square board called a âchessboardâ. The player with the light-coloured pieces (White) makes the first move, then the players move alternately, with the player with the dark-coloured pieces (Black) making the next move. A player is said to âhave the moveâ when his/her opponentâs move has been âmadeâ. The objective of each player is to place the opponentâs king âunder attackâ in such a way that the opponent has no legal move. 1.4.1 The player who achieves this goal is said to have âcheckmatedâ the opponentâs king and to have won the game. Leaving oneâs own king under attack, exposing oneâs own king to attack and also âcapturingâ the opponentâs king is not allowed. 1.4.2 The opponent whose king has been checkmated has lost the game. 1.5 If the position is such that neither player can possibly checkmate the opponentâs king, the game is drawn (see Article 5.2.2). Article 2: The Initial Position of the Pieces on the Chessboard 2.1 2.2 The chessboard is composed of an 8 x 8 grid of 64 equal squares alternately light (the âwhiteâ squares) and dark (the âblackâ squares). The chessboard is placed between the players in such a way that the near corner square to the right of the player is white. At the beginning of the game White has 16 light-coloured pieces (the âwhiteâ pieces); Black has 16 dark-coloured pieces (the âblackâ pieces). These pieces are as follows: A white king usually indicated by the symbol K A white queen Two white rooks Two white bishops Two white knights Eight white pawns A black king A black queen Two black rooks Two black bishops Two black knights Eight black pawns usually indicated by the symbol Q usually indicated by the symbol R usually indicated by the symbol B usually indicated by the symbol N usually indicated by the symbol usually indicated by the symbol K usually indicated by the symbol Q usually indicated by the symbol R usually indicated by the symbol B usually indicated by the symbol N usually indicated by the symbol Staunton Pieces p Q K B N R 9 2.3 The initial position of the pieces on the chessboard is as follows: 2.4 The eight vertical columns of squares are called âfilesâ. The eight horizontal rows of squares are called âranksâ. A straight line of squares of the same colour, running from one edge of the board to an adjacent edge, is called a âdiagonalâ. Article 3: The Moves of the Pieces 3.1 It is not permitted to move a piece to a square occupied by a piece of the same colour. 3.1.1 If a piece moves to a square occupied by an opponentâs piece the latter is captured and removed from the chessboard as part of the same move. 3.1.2 A piece is said to attack an opponentâs piece if the piece could make a capture on that square according to Articles 3.2 to 3.8. 3.1.3 A piece is considered to attack a square even if this piece is constrained from moving to that square because it would then leave or place the king of its own colour under attack. 3.2 The bishop may move to any square along a diagonal on which it stands. 3.3 The rook may move to any square along the file or the rank on which it stands. 3.4 The queen may move to any square along the file, the rank or a diagonal on which it stands. 3.5 3.6 3.7 When making these moves, the bishop, rook or queen may not move over any intervening pieces. The knight may move to one of the squares nearest to that on which it stands but not on the same rank, file or diagonal. 3.7 When making these moves, the bishop, rook or queen may not move over any intervening pieces. The knight may move to one of the squares nearest to that on which it stands but not on the same rank, file or diagonal. The pawn: 3.7.1 The pawn may move forward to the square immediately in front of it on the same file, provided that this square is unoccupied, or 3.7.2 on its first move the pawn may move as in 3.7.1 or alternatively it may advance two squares along the same file, provided that both squares are unoccupied, or 3.7.3 the pawn may move to a square occupied by an opponentâs piece diagonally in front of it on an adjacent file, capturing that piece. 3.7.3.1 A pawn occupying a square on the same rank as and on an adjacent file to an opponentâs pawn which has just advanced two squares in one move from its original square may capture this opponentâs pawn as though the latter had been moved only one square. 3.7.3.2 This capture is only legal on the move following this advance and is called an âen passantâ capture. 3.7.3.3 When a player, having the move, plays a pawn to the rank furthest from its starting position, he/she must exchange that pawn as part of the same move for a new queen, rook, bishop or knight of the same colour on the intended square of arrival. This is called the square of âpromotionâ. 3.7.3.4 The player's choice is not restricted to pieces that have been captured previously. 3.7.3.5 This exchange of a pawn for another piece is called promotion, and the effect of the new piece is immediate. 3.8 There are two different ways of moving the king: 3.8.1 by moving to an adjoining square. 3.8.2 by âcastlingâ. This is a move of the king and either rook of the same colour along the playerâs first rank, counting as a single move of the king and executed as follows: the king is transferred from its original square two squares towards the rook on its original square, then that rook is transferred to the square the king has just crossed. 3.8.2.1 The right to castle has been lost: 3.8.2.1.1 If the king has already moved, or 3.8.2.1.2 With a rook that has already moved. 3.8.2.2 Castling is prevented temporarily: 3.8.2.2.1 if the square on which the king stands, or the square which it must cross, or the square which it is to occupy, is attacked by one or more of the opponent's pieces, or 3.8.2.2.2 if there is any piece between the king and the rook with which castling is to be effected. 3.9 The king in check: 3.9.1 The king is said to be 'in check' if it is attacked by one or more of the opponent's pieces, even if such pieces are constrained from moving to the square occupied by the king because they would then leave or place their own king in check. 3.9.2 No piece can be moved that will either expose the king of the same colour to check or leave that king in check. 3.10 Legal and illegal moves; illegal positions: 3.10.1 A move is legal when all the relevant requirements of Articles 3.1 â 3.9 have been fulfilled. 3.10.2 A move is illegal when it fails to meet the relevant requirements of Articles 3.1 â3.9. 3.10.3 A position is illegal when it cannot have been reached by any series of legal moves. Article 4: The Act of Moving the Pieces 4.1 4.2 Each move must be played with one hand only. Adjusting the pieces or other physical contact with a piece: 4.2.1 Only the player having the move may adjust one or more pieces on their squares, provided that he/she first expresses his/her intention (for example by saying âjâadoubeâ or âI adjustâ). 4.2.2 Any other physical contact with a piece, except for clearly accidental contact, shall be considered to be intent. 4.3 Except as provided in Article 4.2.1, if the player having the move touches on the chessboard, with the intention of moving or capturing: 4.3.1 one or more of his/her own pieces, he/she must move the first piece touched that can be moved. 4.3.2 one or more of his/her opponentâs pieces, he/she must capture the first piece touched that can be captured. 4.3.3 one or more pieces of each colour, he/she must capture the first touched opponentâs piece with his/her first touched piece or, if this is illegal, move or capture the first piece touched that can be moved or captured. If it is unclear whether the playerâs own piece or his/her opponentâs was touched first, the playerâs own piece shall be considered to have been touched before his/her opponentâs. 4.4 If a player having the move: 4.4.1 touches his/her king and a rook he/she must castle on that side if it is legal to do so 4.4.2 deliberately touches a rook and then his/her king he/she is not allowed to castle on that side on that move and the situation shall be governed by Article 4.3.1. 4.4.3 intending to castle, touches the king and then a rook, but castling with this rook is illegal, the player must make another legal move with his/her king (which may include castling with the other rook). If the king has no legal move, the player is free to make any legal move. 4.4.4 promotes a pawn, the choice of the piece is finalised when the piece has touched the square of promotion. 4.5 4.6 If none of the pieces touched in accordance with Article 4.3 or Article 4.4 can be moved or captured, the player may make any legal move. The act of promotion may be performed in various ways: 4.6.1 the pawn does not have to be placed on the square of arrival. 4.6.2 removing the pawn and putting the new piece on the square of promotion may occur in any order. 4.6.3 If an opponentâs piece stands on the square of promotion, it must be captured. 4.7 When, as a legal move or part of a legal move, a piece has been released on a square, it cannot be moved to another square on this move. The move is considered to have been made in the case of: 4.7.1 A capture, when the captured piece has been removed from the chessboard and the player, having placed his/her own piece on its new square, has released this capturing piece from his/her hand. 4.7.2 Castling, when the player's hand has released the rook on the square previously crossed by the king. When the player has released the king from his/her hand, the move is not yet made, but the player no longer has the right to make any move other than castling on that side, if this is legal. If castling on this side is illegal, the player must make another legal move with his/her king (which may include castling with the other rook). If the king has no legal move, the player is free to make any legal move. 4.7.3 Promotion, when the player's hand has released the new piece on the square of promotion and the pawn has been removed from the board. 4.8 4.9 A player forfeits his/her right to claim against his/her opponentâs violation of Articles 4.1 â 4.7 once the player touches a piece with the intention of moving or capturing it. 4.8. A player forfeits his/her right to claim against his/her opponentâs violation of Articles 4.1 â 4.7 .4.9. If a player is unable to move the pieces, an assistant, who shall be acceptable to the arbiter, may be provided by the player to perform this operation. Article 5: The Completion of the Game 5.1.1 The game is won by the player who has checkmated his/her opponentâs king. This immediately ends the game, provided that the move producing the checkmate position was in accordance with Article 3 and Articles 4.2 â 4.7. 5.1.2 The game is lost by the player who declares he/she resigns (this immediately ends the game), unless the position is such that the opponent cannot checkmate the playerâs king by any possible series of legal moves. In this case the result of the game is a draw. 5.2.1 The game is drawn when the player to move has no legal move and his/her king is not in check. The game is said to end in âstalemateâ. This immediately ends the game, provided that the move producing the stalemate position was in accordance with Article 3 and Articles 4.2 â 4.7. 5.2.2 The game is drawn when a position has arisen in which neither player can checkmate the opponentâs king with any series of legal moves. The game is said to end in a âdead positionâ. This immediately ends the game, provided that the move producing the position was in accordance with Article 3 and Articles 4.2 â 4.7. 5.2.3 The game is drawn upon agreement between the two players during the game, provided both players have made at least one move. This immediately ends the game. COMPETITIVE RULES OF PLAY Article 6: The Chessclock 6.1 âChessclockâ means a clock with two time displays, connected to each other in such a way that only one of them can run at a time. âClockâ in the Laws of Chess means one of the two time displays. Each time display has a âflagâ. âFlag-fallâ means the expiration of the allotted time for a player. 6.2 Handling the chessclock: 6.2.1 During the game each player, having made his/her move on the chessboard, shall pause his/her own clock and start his/her opponentâs clock (that is to say, he/she shall press his/her clock). This âcompletesâ the move. A move is also completed if: 6.2.1.1 6.2.1.2 the move ends the game (see Articles 5.1.1, 5.2.1, 5.2.2, 9.2.1, 9.6.1 and 9.6.2), or the player has made his/her next move, when his/her previous move was not completed. 6.2.2 A player must be allowed to pause his/her clock after making his/her move, even after the opponent has made his/her next move. The time between making the move on the chessboard and pressing the clock is regarded as part of the time allotted to the player. 6.2.3 A player must press his/her clock with the same hand with which he/she made his/her move. It is forbidden for a player to keep his/her finger on the clock or to âhoverâ over it. 6.2.4 The players must handle the chessclock properly. It is forbidden to press it forcibly, to pick it up, to press the clock before moving or to knock it over. Improper clock handling shall be penalised in accordance with Article 12.9. 6.2.5 6.2.6 Only the player whose clock is running is allowed to adjust the pieces. If a player is unable to use the clock, an assistant, who must be acceptable to the arbiter, may be provided by the player to perform this operation. His/Her clock shall be adjusted by the arbiter in an equitable way. This adjustment of the clock shall not apply to the clock of a player with a disability. 6.3 Allotted time: 6.3.1 When using a chessclock, each player must complete a minimum number of moves or all moves in an allotted period of time including any additional amount of time added with each move. All these must be specified in advance. 6.3.2 The time saved by a player during one period is added to his/her time available for the next period, where applicable. In the time-delay mode both players receive an allotted âmain thinking timeâ. Each player also receives a âfixed extra timeâ with every move. The countdown of the main thinking time only commences after the fixed extra time has expired. Provided the player presses his/her clock before the expiration of the fixed extra time, the main thinking time does not change, irrespective of the proportion of the fixed extra time used. 6.4 Immediately after a flag falls, the requirements of Article 6.3.1 must be checked. 6.5 Before the start of the game the arbiter shall decide where the chessclock is placed. 6.6 At the time determined for the start of the game Whiteâs clock is started.6.7. Default time: 6.7.1 The regulations of an event shall specify a default time in advance. If the default time is not specified, then it is zero. Any player who arrives at the chessboard after the default time shall lose the game unless the arbiter decides otherwise. 6.7.2 If the regulations of an event specify that the default time is not zero and if neither player is present initially, White shall lose all the time that elapses until he/she arrives, unless the regulations of an event specify, or the arbiter decides otherwise. 6.8 A flag is considered to have fallen when the arbiter observes the fact or when either player has made a valid claim to that effect. 6.9 Except where one of Articles 5.1.1, 5.1.2, 5.2.1, 5.2.2, 5.2.3 applies, if a player does not complete the prescribed number of moves in the allotted time, the game is lost by that player. However, the game is drawn if the position is such that the opponent cannot checkmate the playerâs king by any possible series of legal moves. 6.10 Chessclock setting: 6.10.1 Every indication given by the chessclock is considered to be conclusive in the absence of any evident defect. A chessclock with an evident defect shall be replaced by the arbiter, who shall use his/her best judgement when determining the times to be shown on the replacement chessclock. 6.10.2 If during a game it is found that the setting of either or both clocks is incorrect, either player or the arbiter shall pause the chessclock immediately. The arbiter shall install the correct setting and adjust the times and move-counter, if necessary he/she shall use his/her best judgement when determining the clock settings. 6.11.1 If the game needs to be interrupted, the arbiter shall pause the chessclock. 6.11.2 A player may pause the chessclock only in order to seek the arbiterâs assistance, for example when promotion has taken place and the piece required is not available. 6.11.3 The arbiter shall decide when the game restarts. 6.11.4 If a player pauses the chessclock in order to seek the arbiterâs assistance, the arbiter shall determine whether the player had any valid reason for doing so. If the player has no valid reason for pausing the chessclock, the player shall be penalised in accordance with Article 12.9. 6.12.1 Screens, monitors, or demonstration boards showing the current position on the chessboard, the moves and the number of moves made/completed, and clocks which also show the number of moves, are allowed in the playing hall. 6.12.2 The player may not make a claim relying only on information shown in this manner.Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: ⢠Free-falling objects do not encounter air resistance. ⢠All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs ⢠Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 ⢠a ⢠d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 ⢠(-8.00 m/s2) ⢠d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) ⢠d (16.0 m/s2) ⢠d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi ⢠t + ½ ⢠a ⢠t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) ⢠(4.1 s) + ½ ⢠(6.00 m/s2) ⢠(4.10 s)2 d = (0 m) + ½ ⢠(6.00 m/s2) ⢠(16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: ⢠An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. ⢠If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. ⢠If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. ⢠If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi ⢠t + ½ ⢠a ⢠t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) ⢠(t) + ½ ⢠(-9.8 m/s2) ⢠(t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) ⢠(t)2 -8.52 m = (-4.9 m/s2) ⢠(t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 ⢠a ⢠d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 â˘(-9.8m/s2) â˘d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) â˘d (-19.6 m/s2) ⢠d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) ⢠d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.UE B2-2a(2.1-2.32)wordsRIGHT ON 3 2.1 2/2AForces P.2A.1 & 2INCREDIBLE 5-1 2/2AI5 Team 2: 2a.1-2a.12