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sol... II - cap11-dynamics - f beer & e russel - 5th edition solution bo, Manuais, Projetos, Pesquisas de Engenharia Mecânica

Resoluções do livro Mecânica Vetorial para engenheiros; Cinemática e Dinâmica, Beer 5ª Edição

Tipologia: Manuais, Projetos, Pesquisas

2012

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Baixe sol... II - cap11-dynamics - f beer & e russel - 5th edition solution bo e outras Manuais, Projetos, Pesquisas em PDF para Engenharia Mecânica, somente na Docsity! » » Instructor's Manual to Accompany ; Vector Mechanics “for Engineers ii ao oa * Ferdinand P, Beer DIES E. Russell Johnston, gr. º º e º e º Instructor's Manual to Accompany Vector Mechanics for Engineers | pynamics Fifth Edition FERDINAND P. BEER Lenigh University E. RUSSELL JOHNSTON, gr. University of Connecticut McGraw-Hill Book Company NewYork St Louis San Francisco Auckiand Bogotá Caracas Colorado Springs Hamburg Lisbon London Madrid Mexico Milan Montreal NewDeihi Oklahoma City Panama Paris San Juan São Paulo Singapore Sydney Tokyo Toronto ET ção x = sr» DESCRIPTION OF THE MATERIAL CONTAINED IN 'VECTOR MECHANICS FOR ENGINEERS: DYNAMICS", Fifth Edition 11. Kinematics of Particles In this chapter, the motion of bodies is studied without regard to their size; all bodies are assumed to reduce to single particles. The analysis of the effect of the size of a body and the study of the relative motion of the various particles forming a given body are postponed until Chapter 15. In order to present the simpler topics first, Chapter 11 has been divided into two parts: rectilinear motion of particles, and curvilinear motion of particles. In Section 11.2, position, velocity anã acceleration are defined for a particle in rectilinear motion. They are defined as quantities which may be either positive or negative and students should be wvarned not to confuse position coordinate and distance traveled, or velocity and speed. The significance of positive and negative acceleration should be stressed. Negative ácceleration may indicate a loss in speed in the positive direction or a gain in speed in the negative direction. As they begin the study of dynamics, many students are under the belief that the motion of a particle must be either uniform or uniformly accelerated. To destroy this misconception, the motion of a particle is first described under very general conditions, assuming a variable acceleration which may depend upon the time, the position, or the velocity of the particle (Sec. 11.3). To facili- tate the handling of the initial conditions, definite integrals, rather than in- definite integrals, are used in the integration of the equations of motion. The special equations releting to uniform anã uniformly accelerated motion are derived in Sections 11.4 anã 11.5. The students should be warned to check carefully, before using these equations, that the motion under consideration is actually a uniform or a uniformly accelerated motion. Two important concepts are introduced in Section 11.6: (1) the concept of. relative motion, which will be developed further in Sections 11.12 and 15.5, (2) the concept of dependent motions anã of degrees of freedom. The first part of Chapter 11 ends with the presentation of several graph- ical methoês of solution of rectilinear-motion problems (Secs. 12.7 and 11.8). This material is optional and may be omitted. Several problems in which the data are given in graphical form have been included (e.g., Probs. 21.72 - 11.75). The second part of the chapter begins with the introduction of the vectors defining the position, velocity and aeceleration of a particle in curvilinear motion. The derivative of a vector function is defined anã introduced st this point (Sec. 11.10). The motion of a particle is first studied in terms of rec- tangular components (Sec. 11.11); it is shown that in many cases (for example, projectiles) the study of curvilinear motion may be reduced to that of two in- dependent rectilinear motions. The concept of fixed anã moving frames of ref- erence is introduceã in Section 11.12 and immediately used to trest the relative motion of perticles. Sections 11.13 anã 11.1) introduce respectively the use of tangential and normal components, and of radial anã transverse components. Both systems of coordinates are first introduced in two dimensions and then extended to include three-dimensional space. iii 12. Kinetics of Particles: Newton's Second Law As indicateã earlier, this chapter and the following two are concerned only with the kinetics of particles and systems of particles. They neglect the effect of the size of the bodies considered and ignore the rotation of the bodies about their mass center. The effect of size will be taken into account in Chapters 16 through 18, which deal with the kinetics of rigid bodies. Section 12.2 presents Newton's second law of motion and introduces the con- cept of a newtonian frame of reference. In Section 12.3 the concept of linear momentum of e particle is introduced, anã Newton's second law is expressed in its alternate form, which states that the resultant of the forces acting on & perticle is equal to the rate of change of the linear momentum of the particle. Section 12.4 reviews the two systems of units used in this text and previously discussed in Sec. 1.3, namely the SI metric units and the U.S. customary units. This section also emphasizes the difference between an absolute and a gravitational system of units. A number of problems with two degrees of freedom have been included (Problems 12.26 through 12.33), some of which require a careful analysis of the accelerations involved (see Sample Problem 12.1). Section 12.5 applies Newton's second law to the study of the motion of a particle in terms of rectangular components end tangential and normal components. In Section 12.6, dynamic equilibrium is presented as an alternate way of express- ing Newton's second law of motion, although it will not be used in any of the Sample Problems in this text. Note that the term inertia vector is used in pref- erence to inertia force or reversed effective force to avoid any possible con- fusion with actual forces. In Section 12.7 the concept of angular momentum of a particle is introduced, and Newton's second lew is used to show that the sum of the moments about a point O of the forces acting on a particle is equal to the rate of change of the an- gular momentum of the particle about O. Section 12.8 analyzes the motion of a particle in terms of radial anã transverse components, and Section 12.9 considers the particular case of the motion of a particle under a central force. The early introduction of the concept of angular momentum greatly facilitates the discussion of this motion. Section 12.10 presents Newton's law of gravitation aná its appli- cation to the study of the motion of earth satellites. Sections 12.11 through 12.13 are optional. Section 12.11 derives the differ- ential equation of the trajectory of a particle under a central force, while Section 12.12 discusses the trajectories of satellites and other space vehicles under the gravitational attraction of the earth. While the general equation of orbital motion is derived (Eg. 12.39), its application is restricted to launchings in which the velocity at burnout is parallel to the surface of the earth. (Oblique launchings are considered in Section 13.9.) The periodic time is founã directly from the fundamental definition of areal velocity rather than by formulas requiring a previous knowledge of the properties of conic sections. The instruct- or will note that he may omit Sections 12.11 through 12.13 and yet assign a number of interesting space mechanics problems to his students after they have reached Section 13.9. iv 13. Kinetics of Particles: Energy aná Momentum Methods After a brief introduction designeã to give to the students some motivation for the study of this chapter, the concept of work of a force is introduced in Section 13.2. Attention is called to the fact that the term work is always used in connection with a well-defined force. Three examples considered are the work of a weight (i.e., the work of the force exerted by the earth on a given body), the work of the force exerteã by a spring on a given body, anã the work of a gravitational force. Confusing statements such as the work done on a body, or the work done on a spring are avoided. The concept of kinetic energy is introduced in Section 13.3 aná the prin- ciple of work and energy is derived by integration of Newton's equation of motion. In aspplying the principle of work and energy, the students should te encoursgeã to âraw separate sketches representing the initial and final positions of the body (Sec. 13.4). Section 13.5 introduces the concepts of power and efficiency. Sections 13.6 through 13.8 are Gevoted to the concepts of conservative forces and potential energy, and to the principle of conservation of energy. Potential energy shoulã always be associated with a given conservative force act- ing on a body. By avoiding statements such as "the energy contained in a spring" a clearer presentation of the subject is obtained, which will not confiiet with the more advanced concepts that the students may encounter in later courses, In applying the principle of conservation of energy, the students should again be en- couraged to draw separate sketches representing the initial and final positions of the body considered. In Section 13.9, the principles of conservation of energy and of conser- vation of angular momentum are applied jointly to the solution of problems in- volving conservative central forces. A large number of problems of this type, dealing with the motion of satellites and other space vehicles, are available for nomevork assignment. As noted earlier, these problems (except the last two, indicated by asterisks) may be solved even if Sections 12.11 through 12.13 have teen omitted, The second part of Chapter 13 is devoted to the principle of impulse anã momentum and to its application to the study of the motion of a particle. Section 13.10 introduces the concept of linear impulse and derives the principle of impulse and momentum from Newton's second law. The instructor should emphe- size the fect that impulses and momenta are vector quantities. He should en- courage his students to draw three separate sketches when applying the principle of impulse and momentum and show clearly the vectors representing respectively the initial momentum, the impulses, and the final momentum. It is only after the concept of impulsive force hes been presented that the students will begin to appreciate the effectiveness of the method of impulse and momentum (Section 13.11). Direct central impact and oblique central impact are studied in Sections 13.12 through 13.14. Note that the coefficient of restitution is defined as the ratio of the impulses during the period of restitution and the period of defor- mation. This more basic approach will make it possible in Section 17.12 to ex- tend to the case of eccentric impact the results obtained here for central im- pact. Emphasis should be placeã on the fact that, except for perfectly elastic impact, energy is not conserved. Note that the discussion of oblique central impact in Section 13.1) has been expanded to cover the case when one or both of the colliding bodies is constrained in its motion. x6. Plane Motion of Rigid Bodies: Forces and Accelerations This chapter is devoted to the plane motion of rigid bodies which consist of Plane slabs or which are symmetrical with respect to the reference plane. Cases involving the plane motion of nonsymmetrical bodies and, more generally, the motion of rigid bodies in three dimensions are considered in Chapter 18. If the determination of mass moments of inertia has not been covered in the previous stetics course, the instructor should include material from Sections 9.11 through 9.15 of Appendix B (or from the second part of Chap. 9) at this point. In Section 16.2 the fundsmental relations derived in Chapter 1h for a system of perticles are used to show that the external forces acting on a rigid tody are equipollent to the vector ma attached at the mass center G of the body and the couple of moment Ha. This result, which is illustrateã in Fig. 16.3, is valiã in the most general case of motion of a rigid body (three-dimensional as well as plane motion). It is shown in Section 16.3 that, in the case of the plane motion of a slab or symmetrical body, the angular momentum E, reduces to Tu and its rate of change to Ta. Section 16.) is devoted to D'Alembest's principle. It is shown that the external forces acting on a rigid body are aetualiy equivalent to the effective forces representeã by the vector mã and the couple ia. As noted in Section 16.5, this result is obtained independently of the principle of transmissibility (Sec. 3,18) anê may be used to derive this principle from the other axioms of mechanics. At this point the students will have reacheà the climax of the study of rigid-body motion in two dimensions. Indeed they mey solve any problem by draw- ing two sketches - one showing the external forces, the other the vector mã and the couple la - and then expressing that the two systems of vectors shown are equivalent. To avoid drawing two separate sketches, the method of dynamic equi- librium may be useã, with a single sketch showing the external forces, the in- ertia vector - mã and the inertisa couple - Ty (Sec. 16.6). However, to facilitate the transition to the study of three-dimensionel motion (Chap. 18), the two-sketch method showing separately the external forces and the effective forces will de used in all sample problems. The various types of plane-motion problems have been grouped according to their kinematic characteristics. The translation, the centroidal rotation, and the plane motion consisting of a translation and of an unrelateã centroidal ro- tation are considered first, since they are the simplest ones to analyze, They are followed by plane motions with various kinematic constraints: non-centroidal rotation, rolling motion, aná other types of plane motion. Problems involving systems of rigid bodies have been included at the enã of this chapter, with either one degree of freedom (Problems 16.14! through 16.117) or two degrees of freedom (Problems 16.148 through 16.155). The instructor should stress the fact that, in spite of the different kinematic characteristics of these various motions, the approach to the Kinetics of the motion is consistently the same: all problems are solved by drawing two sketches - one showing the ex- ternal forces, the other the vector mã and the couple io - anã then expressing that the two systems of vectors shown are equivalent. viii Since the approach used in this text differs from others by the emphasis placeã on the direct application of D'Alembert's principle, rather than on specialized formulas, it might be appropriate at this point to sumarize the ad- vantages derived from this approach. (1) A single methoã is useã, which applies to all cases of plane motion, regardless of their kinematic characteristics, and may be used safely under any conditions. This is in contrast with the equation EM = Ia which is limited in its applications, as pointeá out in Problem 16.105. (2) By stressing the use of the free-body diagram, this method provides a better understanding of the kinetics of the motion. There will be little danger, for example, in the solution of a problem of non-centroidal rotation, that the students forget the effect of the acceleration of the mass center on the reaction at the fixed point, a mistake which oceurs frequently when the specialized formula EM, = Toa is used. (3) The method used divides the solution of a problem into two main parts, one in which the Kinematic and kinetic characteristics of the problem are con- sidered (separately if necessary), anã the other in which the methods of statics are used. In this way the techniques of each separate field may be used most efficiently. For example, moment equations may pe written to eliminate unvanted reactions, just as it was done in statics; this may pe done independently of the kKinetice characteristics of the problem. (bh) By resolving every plane motion (even a non-centroidal rotation) into a translstion and a centroidsl rotation, a unified approach is obtained, which will aiso be used in Chapter 17 with the method of work end energy and with the method of impulse and momentum, and which will be extended in Chapter 18 to the study of the three-dimensional motion of a rigia body. This approach is a pasic one, which may be applied effectively throughout the study of mechanics, in ad- vanced courses as well as in elementary ones. 17. Plane Motion of Rigid Bodies: Energy and Momentum Methods The first portion of the chapter extends the method of work and energy, already used in Chapter 13, to the study of the plane motion of rigid bodies. The expressions for the work of a couple and for the ginetic energy of a rigid body are derived in Secs. 17.3 and 17.4. Using the results obtained in Sec. 14,7, the kinetic energy of a rigiã body is separated into a translational part and a rotational part (about the mass center). The authors believe that, while it may lead to slightly longer solutions, this method is more fundamental and should be used in preference to special formulas. Indeed, it follows the basic idea of re- solving every plane motion into a translation and a centroidal rotation. It is shown in Section 17.5 that the method of work and energy is especially effective in the case of systems of rigid bodies connected by pins, inextensible corãs, and meshed gears. In Section 17.6 the principle of conservation of energy is used to analyze the plane motion of rigid bodies. ix In the second part of Chapter 17, the method of impulse and momentum is ex- tended to the study of the motion of rigid bodies. The approach used is different from that of most elementary textbooks. Ready-to-use formulas are evoided; in- stead, the students are taught to express the general principle of impulse and momentum by means of free-body diagrams and to write the equations most appro- priate to the solution of the problem considered. The results obtaineá in Section 1h.9 for a system Of particles are directly applicable to the system of particles forming a rigiã body anã may be used to analyze the plane motion of rigiã bodies. It is shom in Section 17.8 that the momenta of the various particles forming a rigid body reduce to a vector nv and a couple To in the most general case of plane motion. While the principle of conservation of angular momentum is discussed in Section 27,10 because of its physical and historical significance, it is not actually used in the solution of problems, To solve any problem, regardless of the type of motion, and whether it involves constant forces oí finite magnitude applied for & finite time, or impulsive forces applied for a very short time in- terval, the students are told to draw three separate sketches showing respectively the initial momenta, the impulses of the external forces, and the final momenta. The momenta of a rigid body are represented in the most general case by a momentum vector my attached at the mass center and a mgmentum couple Iw. If the students consider then the components of the vectors involved, they obtain relations be- tween linear impulses and linear momenta. If they consider the moments of the seme vectors, they obtain angular impulses and angular momenta. If, by equating moments about a point such as a pivot, they obtain an equation which does not in- volve any of the external forces, they will have automatically established conser- vation of angular momentum about that point. The advantages derived from this approach may be summarized as follows: (1) The students leern only one method of solution, a method based directly on a fundamental principle, a methoá which may be used safely under any conditions. This is in contrast with the equation EMt = Twp-u,), which is limited in its applications (see Problem 17.66). (2) The methoa stresses the use of free-body disgrams and thus provides a better understanding of the kinetics of the motion. It is unlikely, for example, that the students will forget an impulsive reaction at a fixeà support. (3) The students use the basic tools they learned in Statics: reduction of a system of vectors to a vector and a couple, equations relating the components or the moments of these vectors. (4) Again, the same unified approach is used: every plane motion is re- solveê into a translation and a centroidal rotation. And in Chapter 18 this approach will be extended to the solution of problems involving the three-dimen- sional motion of rigid bodies. Some teachers may fear that the inclusion of momentum vectors and momentum couples in the same diagrams may lead to confusion. This will not be the case, however, if the students are instructed to write separate equstions involviag either components or moments, as they did in Statics. The first equations will contain linear impulses and linear momenta expressed in Nes or lbºs, anã the latter angular impulses and. angular momenta expressed in Nemes or lbeftes. Section 17.12 is devoted to the eccentric impact of tvo rigid bodies, a topic seldom included in an elementary text, No special difficulty will be encountered, however, if separate sketches are useê as indicated above, 18. Kinetics of Rigid Bodies in Three Dimensions In this chapter the restrictions imposed in preceding chapters (e-g., plane motion, syrmmetrical todies) are lifted and the student proceeãs to the analysis of more general (and more difficult) problems, such as the rotation of nonsymmetrical bodies about fixed axes and the motion of gyroscopes. In Section 18,1, the general result obtaineã in Section 16.2 is recalled, namely, that the external forces acting on a rigid body are equipollent to the vector mã attached at the mass center G of the body and the couple of moment Kg. It is also pointed out that the main feature of the impulse-momentum method, namely, the reduction of the momenta of the particles of a rigid body to a linear momentum vector my attached-at G and an angular momentum E, remains valiã, and that the vork-energy principle and the principle of conserYation of energy still apply in the case of the motion of a rigid body in three dimensions. The dif- ficulties encountereã in the study of the three-dimensional motion of a rigiã body are relateã to the determination of the angular momentum Hg, of its rate of change Hg» and of the kinetic energy of the body. The determination of the angular momentum H,. of a rigid body from its an- gular velocity w is discussed in Section 18.2. Since this requires the use of mass products of inertia, as well as of mass- moments of inertia, the instructor should cover Sections 9.16 and 9.17 from Appendix B (or from the seconã part of Chapter 9) if this material has not been included in the previous statics course. Section 18.3 is devoted to the application of the impulse-momentum principle to the three-dimensional motion of a rigid body, and Section 18.4 to the deter- mination of its Kinetic energy. In Sections 18.5 and 18.6, the rate of change of the angular momentum Hg is computed and the equations of motion for a rigid body in three dimensions are de- rived. D'Alembert's principle is extended to the case of three-dimensional motion ty showing that the external forces are actually equivalent to the effective forces represented by the vector má and the couple Hç. Sections 18.7 and 18.8 are de- voted, respectively, to the particular cases of the motion of a rigid body about a fixed point and of the rotation of a rigid body about a fixed axis, with appli- cations to the balancing of rotating shafts. While Euler's equations of motion have been derived on page 907, it should te noted that the more fundamental vector relations represented by Equetions (18.22), (18.23), and (18.28) are used in the actual solution of problems. xi The remaining portion of this Chepter (Secs. 18,9 through 18.11) is de- signed for advanced students and, in general, should be omitted for ordinary classes. In Sections 18.9 and 18.10 the motion of a gyroscope is considered. At this point Eulerian angles are introduced. It should be carefully noted that the rotating system of axes Oxyz is attacheã to the inner gimbal; these axes are prin- cipal axes of inertia and they follow the precession and nutation of the gyro= scope; they do not however spin with the gyroscope. The special case of steady precession is considered in Section 18,10. Several problems dealing with the steady precession of a top are included and, in one of them (Prob. 18.89, it is shown that the formula usually given in introduetory texts is only approximate. Two problems dealing with the general motion of a top have also been included (Probs. 18,108 end 18.109). The motion cf an axisymmetrical body under no force (Sec. 18.11) introduces the student to one of the most interesting aspects of classical dynamics - an aspect which nes gained widespread attention in recent years due to the interest in space vehicles and artificial satellites, In this connection, it shoulãá be pointed out that the instructor may cover the Poinsot theory of the motion of a nonsymmetrical body under no force by assigning three problems (Probs. 18.11h through 18,116), An adãitionsl problem (Prob. 18.117) relates to the stability of the rotation of a nonsymuetrical body about a príncipal axis. 19. Mechanical Vibrations Tnis chapter provides an introduction to the study of mechanical vibrations. While only one-degree-of-freedom systems are included, all the basic principles are presented. The various topics covered are as follows: (a) Free, undamped vibrations of a particle (Sec. 19.2). The differential equation characterizing simple harmonic motion is derived and 822 basic terms, such as period, frequency, anê amplitude, are defineã, Both the analytical and the geometrical methods of solution are described. It is shown in Section 19.3 that the motion of a simple pendulum may be approximated by a simple harmonic motion. Section 19.4, which is optional, shows how an exact solution may be ob- tained for the period of oscillations of a simple pendulum. (b) Free, undamped vibrations of a rigid body. The principle of equiv- alence of the systems of applied and effective forces is first used to determine the frequency and the period of oscillations of a rigid tody (Sec. 19.5). Note that the same positive sense is assumed for the angular acceleration and dis- placement; this results in an apparently unrealistic assumption for the sense of the vector mã and the couple Ia. The principle of conservation of energy is then used to solve the same type of problems (Sec. 19.6). (c) Forced, undamped vibrations of a particle (Sec. 19.7). This section introduces the students to the concepts of natural and forced frequencies, tran- sient and steady-state vibrations, and resonance. While all students will be able to understand this section, those with a knowledge of elementary differential equations will derive a greater benefit from it since it provides a direct application of the solution of linear nonhomogeneous equations with constant co- efficients. xii TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS Problem Number? SI Unite U.S. Units Problem Description CHAPTER 11: KINHMATICS OF PARTICLES RECTILINEAR MOTION OR PARTICLES Analyze motion of a particle, given 11.3,4 position = f(t) also find total distance traveled 11.11,8 acceleration = f(t) 11.15,1h acceleration = f(x) 11.19,18 seceleration = flv) 11.20 net 11.25 velocity = f(x) 11.26 1.27 accelerstion v L/r2 21.29 1.28 Simple hermonic motion Uniformly accelerated motion 11.31,30 2.33,32 motion of one particle 11.35,34 11.39,36 12.37,38 motion of two particles Relative motion of particles systems with one degree qr. freedom and 1.k5 ko 11.h2 constant velocities 11.b1,bh ai.h3 constant uccelerations 1h h6 21.h9,18 systems with tun degrees of Ireedon 11.51,*50 Graphical methods 11.55,54 12.53.52 analyze motion using a given motion curve and inílial condilions 11.57,56 11.592,58 motion of one particle 11.61,60 aL.6p 11.6h,63 motion of two partícies 11,67,66 11.65 11.68 211.69 problems involving average velocity nn 12.70 and rate of change of Acceleration 11.73,74 n.7 problems solveã by numerical integration 11.75 problem solved by use of v-x curve 1.77,78 11.79,76 nroblems solveã by moment-area method ” CIMVILTIHEAR MOTION OF PARTICLES 11.83,82 11.81,80 Anstyze motion of a particle given 4 1.85 x and y as functions of t 11.86 position vecter in two dimensions position vector in three dimensions Projectiles 1.91,88 11.89.90 with horizontal Initiul velocity 11.95,9h 11.93 with initial velocity of given mugnitude and direction 12.97,96 11.99,92 riven direction, find magnitude 07 initial velocity 11.98 11.103,*102 *11.101,100 | given magnitude, find direction of initial velocity (solution of 11.105,104 quadratir equation required) Relative motion of partácles with 11.107,108 11.109,106 constant velovities 11.111,110 veloecities and accelerations 11.133,112 given relative ncccleration 12.115,114 constent accelerations 2.N7,116 q1.119,218 two given observatious of relative motion *Problems which do not involve any specific system cf units rave been indicated ty underlining their number. TABLE TI: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED) Problem Numbert SI Units U.S. Units Problem Description Wethod of impulse and momentum applied to problems involving 13.113,1H 13.115,112 single body 13.116 13.127,120 13.119,28 two or more connected bodies 13.121,122 13,123,124 13.125,126 force defined by a graph 13.127,130 13.329,128 impulsive motion of A single body 13.131,132 13.135,136 impulsive motion of two »edies 13.133,134 . 13.137 13.138 impulsive motion vith different supports 13.1h1,3h2 13.139,10 Direct central impact, 13.143 2h 47,148 33.145,146 Oblique central impact 51,152 13.149,150 Impact against Fixed surfaçes 13.155,156 13.153,15 13.159,160 13.157,158 Oblique central impact with one body constrained to move in a 13.161 given direction 13.165,16: 13.163,162 Problems involvíng conservation of energy, conservation of 13.167,166 momentum and impact 13.171,170 13.169,168 13.172 13.173 and oue body constraineã to move in s given direction “13.175 13.174 — Epecial problems IHITLITO 13.179,180 Review problems 13.181,178 13.183,18) 13.187,182 13.185,186 13.01,03 13,ce,eh Computer problems CHAPTER 1h: SYSTEMS OF PARTICLES 14.3,4 1h.1,2 Conservation of linear momentim (one dimension) 1h.5,6 14.9,10 14.7,8 Computation of angular momentum 1h.12,14 14,13,12 Motion of mass center 1h,15 1:.19,18 1h,17,16 Conservation c? linear momentum (2 or 3 dinensions! 14.23.,20 14,25 24 1h.23,22 — Nerivations and proots 11,27 11.26 Conservatiou vT linear momentum and computations of cnange 14.29,28 24.31,30 in kinetic energy a 1h.37,32 Conservation of linear momentum und conservation of energy 135,3 1i.39,38 TEhl ho aWh3,h0 Conservation of linesr and anguler momentum 15.h5,h 1h.h9,h8 Conservation of linesr momentun, angular momentum, and energy 21.47,16 +Problems which do not involve any specific system of units xviii have been Indicated by underlining their number. TABLE IT: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED) Problem Hunber? SI Units U.S. Units Problem Description 1h.51,50 1Y.52 Thrust Causcd by diverted flou 2h,53,5h 1b.55,56 14,59,56 ab.sT,6o | Reactions at supports of vane or conveyor 14.61,62 - 14.632,66 1h.65,6h Thrust developed by a fan, propeller, or jet engine. 1).69,68 1h.67,70 Wo Ui 73,7) Power and efficiency *4.75,M76; TT Theery problems from fluid mechanica 1.78 — Motion of sprinkler 14.79,80 14.81,82 Motion of chains or of cars of variable mass 11,85,86 10.03,8h 15-89,90 1h.87,88 Thrust and acceleration of rocketa 1h.91,92 14.93,94 Yelacity of rocket for given fuel usage 14.97.98 14.95.96 Distance traveled by rocket 14.99 14.100 Derivetion of efficiency of jet engine or rocket 14.107,106 24,2101,102 Review problems 14. 109,108 1.103,10) 30.711,110 15.105,112 15.ca cl 1h,C2,03 Computer problems CHAPTER 15: KINEMATICS OF RIGID BODIES Rotation about a Tixed axis 15.3, uniformly accelerated rotation 15.11,8 rotation about akew axis 15.13,12 motion of the earth 15.19,18 motion ín two dimensions 15.021,20 rolling contact vith no slipping linear and angular motion 15.29,28 disks brought into contact, slípping occurs *25.30 special problem General planc motion - velacíties 15.31,3h 15.33,32 motion of single rigid body 15.35.38 Da application of vector algebra to motion of a plate. 15.39,h0 15.42 planetary gear systems 15.43, motion of 8 rod connected to: 15.145,18 15.47 ,h6 a crank and a sliding block 15.149,50 15.51,52 two cranks 15.53,5h & rolling vheel anã a horizontal surface “15.55 special problem Instantaneous center Gf rotation 15.59,58 15.57,56 problems involving parallel velocities biido 15.63,62 single rod: angle between controlling velocities = 90º 15.65, 15.67,66 single rod: angle between controlling velocities É 90º 15.69,68 single plate guidcá by pins fitted into slots 15.70 15.71 problems involving tuo instantaneous centers 15.173,72 15.75 15.7h space and body centrodes 15.79,76 15.717,78 previous problems solved using instantaneous center *Problems which do not involve any specific system of units have beer indicuted by underlining their number. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED) Problem Number ST Units U.8. Units Problem Description General plane motion - accelerations 15.821,80 find reletions among aecelerations (w = 0) 15.863,82 15.85,8h 15.89,88 rolling motion 15.87,86 motion of a body controlled by 15,93,90 15.91,92 one crank vith constent engular velocity 15.95,9h 15.97,96 two eranks 15.99,98 15.101,100 a crank and & sliding block 15.102 special problem *15.103,%104 *15.105,*106 Analysis of plane motion in terms of a parameter »15.107,*108 *15.109,*110 +15 131,9112 415.113,%11h Plane motion of a particle relstive to a rotating frame 25.117,16 15.115,118 velocity of a partícle with respect to a rotatirg frame 15.119,120 Coriolis ecceleration 15.121,122 15.123,124 accelcration of particle moving ut a constant speed with 15.125 ,126 respect to a frame rotating with a constant anguler velocily 15.127,128 15,129,130 15.131,132 15.133,134 accelerstion of « particle moving with respert to a frame 15.137.138 15.135,136 rotating with an angular acceleratioa *15.139,41h0 415.1h1,*1L2 special problems Motion «bout a fixed point. General motion 15.1h45,246 15.153 ,14k relation betveen linear aná angular velocity for a rigiã body with a fixed point 15.119,15€. angular scceleration of body rotating sirmltaneously about two axes vetocity and/or acceleration of a point on a buly rotating about a fixed point 15.153,15h 15.155,156 angular velocities ang rate of cuasge o? angular velccitics given 15.159 157,158 15.161,160 163,162 velocities of points on tody specified 15,164 65 cone rolling on fixcd surface 15.169,165 15.167,166 rod attacheã to moving collers: find velccities of coliara 15.171,170 (angular velocity is indeterminate) 15.173,172 universal Jointo 15.175 15.174 rod with clevis attached at one end *15,179,ML78 R1S.17T,*ITO rod attached to moving collars: “ind acceieretions o” cotLaro (angular velocity anã acecieration are indeterminato) ivo to E rotaling ramo. Thres dimensional! motion ol a particle re : forjolis accelerati 15.183,18 Q=0andã=0 15.185 Bfosna ufa 15.187,186 15.189,188 previous problems to be solveá usink rotating frames 15.193,192 25.191,190 axis of º in plane of mechanism 15.195,19 15.197,198 15.199,196 frame of reterence in gencral motion 15,203,20P 15.202,200 «208 15,205,204 Review problems 5.213,212 15.209,206 .215,21b 11,210 «C1,Ch 3,02 Computer problems *Problems which do not involve any specific system of units nave D ed ly underlini: xx g their nutber. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED) Problem Number* . SI Units U.S. Units Problem Description 17.131,130 17.133,132 Review problems 17.137,136 17.135,13 17.139,138 AT.JuL,iho 17.C1,03,Ch 1T.c2 Computer problems CHAPTER 18: KINETICS OF RIGID BODIES IN THREE DIMENSTONS 18.1,2 18,3,4 Computation of angular momentum (no product of inertia) 18.7,6 18.5 .. 18.9,8 18.211,10 tise of equation HE = PXTV+H 18.12 18.13,14 Computation of angular momentum (using products of inertia) 18.15 18,19,18 Impulse and momentur in three dimensions 19.25 ,24 18.727,26. 15.29,30 Impact o? meteorite on saLelliLe 18.35,38 Computation of kinetic energy 18.37,36, 18.h3,h2, 18.h1,44 Change in kinetic energy due Lo impact 18.15 ,46 1á.L7,L8 Computaticn of derivative of angular momentum 18.51,50 Dynamic reactions on roteting shaft for 18.55,56. 18.53,5h constant w 18.59,58 18.57,60 m=Oandazfo 18.61 18,62 wand az o 18,63,6h 18.65,66 Simple gyroscopic cffect (axes of rotation perpendicular) 12.67,68 28.69 Relative equilibrium under rotation about fixed axis Iynamic reactions 18.73,72 28.712,74 for combincã rotations (constant rete) 28.75.76 +18 for combined rotations with angular acceleration *18.79 *18,90 Advanced problems 18.851,82 18.853,04 Steady precession. Simple problems 18.85,86 18,87,88 Steady precession. More advanced problems 18,90 18.89 18.91 Spevial problem (ayrocompass) 18.95,94 18.093,92 Frecession under no force (theory) 18.599,98 18.971,96) 96 Precession under no force (applications) 28.103,102 78.101,100 . 18.10h Conservation of energy and angular momentum for bodics *18.105,*106 *18,107 witr precession sná nutation *20.115,4110 *18.109, with precession, nutation, and apin +18.112 418.113 *18.115,*1]h “18.117,16 Poinsot analysis 18.121,120 18.119,118 Review prótlems 18.125,122 2B.123,124 . 26.127,126 18.129,128 18.C1,Ch 18.03,c2 Computer proble-s “Problems which do not invulve any specific em of units have been indicated by underlicing their number. xxiii TABLE IT: CLASSIFICATION AND DESCRIPTION OF PROBLEMS ( CONTINURD) Problem Humber* 81 Units U.S. Units ' Problem Description CHAPTER 29: MECHANICAL VIBRATIONS VIBRATIONS WITHOUT DAMPING Free víbrations of a particle 19.1,2 19.544 problems involving amplitude, frequency, maximum velocity 19.3,6 x anã aeceleration 19,9,8 39.11,10 problems invclving velocity anã acceleration at q given tnstant 19.215,14 19.13,12 aprings in parallel or series 19.19,18 19.17,16 change mass or stiffness of system 19.21,22 19.23,20 special problems *19.27,426 *19,25,%ah simple pendulum with large amplitude Free vibrations of rigid bodies systems with elastic restoring forces 19.31,28 19.29,30 rolling or noncentroidsl rotation, also find vg 19.33,32 19.35,36. 19.37,3h special problems systems vith restoring forces due to gravity 19.38 19.39,40 derivetions for motion of compounã perdulua 19. hz 19.43,h4 find period or frequeney of compound pendulum 19.h5,h6 19.47 19.49.18 19.51,50 19.53,52 19.57,56 torsional vibrations 19.55,5h Conservation of energy used Lo analyze the notion of 19,61,60 particles 19.65,6h single rigid body: pendulum 19.67.58 19.69,70 single rigid body; elastic restorins forces 19.73,76 comected rígid bodies 15.78 19.83,80 special problems 19, BL Forccd vibrations . 19.89,88 problems involving periodic applied furve cr periadic 19.90 support motion 19.914,92 19.93,94 problems involving rotating machinery 19.95,96 . 19.97,98 19.99,100 special problems 19.103,10 19.101,102 19.105 10.197,106 *Problemns vhich do not involve ary axiv specific system of units have been indicated by underlining their number. TABLE IT: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED) Problem Number * 51 Units U.S. Unite Problem Description DAMPED VIBRATIONS Damped fres vibrations 19.109,108 19.111,110 proofs concerning damped free vibrations 19.112 19.113,11 problems involving logarithmic decrement 19.115,116 eritically damped system Demped forced vibrationa 19.117 19.118 proofs concerning damped forced vibration 19-.121,122 19.119,120 problems involving rotating machinery 19.123,12h 19.125,126 problems involving periodic applied force “19.127 *19.128 special problems 19.131,132 slectrical analogues 19.134 19.135,138 19.137,136 Review problems 29.2h1,1h4 29.139,2h0 19.143,146 19.145,1h2 19.C1,C2 19€3,Ch Computer problems “Problems which do not involve any specific system of? units have been indicated by underlining their number. 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Up MES Tm Some = toa Voe” omt A 7 H 4 I me crcasE Pestriver resertoy Loures Mes ces MOON dE comer £ Go Za? w som» Lato) + Ay 7 OBM ayro ta (E) Imce GutIE COAEMOT, MITO, AO App ão Tee: que-ome-(osm), arotmt «4 Riso, Ge rRBETCONSIANT, EADÍEO pro a t24,=0 Des: aez-20, AC oras) -+ idaho; esti ya Cb) secoerry 0F LtbuA TUA MFTER Es, Walt AE= OTIS A) =— Gm) to tmrt a O RIstm Gem 58 A EEJ6 io 2ug-2u, 0 IÍUB Me) 27 O Ati mk=Rr ij (b) come ECRO IS OF constant LEMETA XetED-L FE CONST Ve + p-2Mro (elê indo) + 0o Ala rindo (e) vetoeiry Or A RELATIVE To 8º aja SUR RT ih IO dolo = EP ddr E) petociry or E 15 SAME AS tetoaITy or Es Ur tstide = Sé od 4 P ul 8 LEU A8 din do Vos (18 dinda) -(36 tmb )==/Bindo px 18 indo INE CHOOSE THE POSITIVE DIR TMN TO THE EdFT 0 bt sum E D RENSERES LENTA CF COLE: Alta t tg E) =consTava %a MEX consraANT ; G +24 "O Ve E UA ap t+ 2Zagros REATINE MO TOM: For VA o THE LEFT, Age delas EU «aguas LDL Meca À e e AR Êo a WE MTE THAT gpa MUST BE To ANT, CM, Ugo — 950 meosls mos unem É=ÉS: -R5O mmlso-ZUa Mnsã00 conto AEE UpE AD medo (o) SprcE RECEEMRADIA ARE CONSTA, Mp = Qt! 4300 mujrza,(és, 2 z SO mente Lg Smnde "a «q 450 m/s = Ag (és) Ag =-RSamfs” A Emo! mA 28 an mf 8 )=-2000mmlr : ag=800 pls + Slcasmnfe Bs) == 800 mm Kr80 om a MMS] SEE ErECRE DF Sorwzmom or Pros, Ate, MOTES POSITIVE DIRETO 4$ Ta LET. (a) LemeTU DE chetE, PM Ya Pe %p) = comszans Xp +2Hg = Consinra Mteug=o Q 36 E 400 oem [en = ECO mim de lp 2100 ommlsj= Zoo mmis Lat) Mtebge Ae 200 mal» 8 Ea MHZ | FE PIEURE Cr Socuror or Pres (4. DT SME moves Dump LEE E Also Moses LET Block 8: ()rO agrconsamy E ruzo: VgrdBinls toner Mg (Valps = Dido, [o Tuusi gl tab *209ve - (vob) (rain po o + 2agi-amin) agr Bufo LEBRE US DF CONSTANT LEA 2Eg-2%p = conszanç Pdg-Z% Do Sag-20, 20 3 gun fe?) -Zapzo apS AS info! = Sto fe A (DD) ame A sr8ers From RE, (ttal=2, aee fra] zo AT É=-S8i Ag binho E Oesp ts) Nom 22 E dgh “p= cesto “4 Cata tOaot td age co rorkfrs metsa)E Ey= SUL im Ryo SÉrBri (6) sEmeTA OF Perry GE CpgeE CRE, KARL eim dora Vg= LO ommfr ipr2 20 muli)-o 212400 mm js emsTasT Eotr00 mora A be) Ritative vetar op O Wir CESPET To E We Ugo Ly Wa emos Ugo +00 twin af = 400 omoaf Ut S00 mom; — 108 mn = + 300 mms 1 300 iemnlp am « 8 DPOCODOCOCOOCCOOOCUCOCOCOCODOCCCCC 0000000000 00000% 11.54 BIÉ Cove E haLUE OF.O = Stopa OF Ul coma SF] SPECO: GE Amp = Leco É Ze É=/05 ua. “>BOs: aro Érhos po estas: o fls-a)=roSms” E=/85 Po tros: ac dls-a)o= 4 ms” Aimio) 9! os La So ds) 56 = Potts) . 51 6) MÉ CURVE: CHANGE IJ AREA UNA É CURVE, Es dé mm Erro EMOS To E=188' %ygWu7 Hemp tdm bofs)= Sem rt 84m Kay ge Elm k 65) = 18 Etta Naa Xmy EE mishes) = Am Lt d=2os To Erbos! Kyo ans CEmÁsNIO 5) = "BOM Kg 2% SINCE AN aura Pagverina 12S ms UItL ARENE AZ. Siena. 8 JUST AS 47 TORNE EEE, THE TIME Cen Sétiil E JUANES BREEM 45 É = DistâNtE dg 25 TO IRS ek 12.5 mfs SLGPE = CONSTANT ELLIRATICOS e LS Onde uma RÉ LS E - 263 rênt) EÉxI2bs Ay ta 325 m HOUSE E HSM ME) = 324 -D,75 4) +37E —zas do (E =-Ws8s Joe telotes MEX SPEED: = HSÉ, EhSÍIO4Z) = 1:63 20; GSE hm A MOTDEIST EBOULO MCCELICATE FOR MO GES My THEM Co TIME AT SRED OF SEI fomh PISTANCE TREE OA DERAM BATOM OE DG) FROM Srrrun TRANS Eita) NHSBl ato Acersgntom Ponce avo fes » dé taeto) WO aoas / , Loss “2 “3500 injs tome, “ee ij A(Rjo + (a) wrem É =245, MAX VALUE OF FOSITIOM COML =45Pm 270 += pm == a trromb ace) (b) Posso Coon Sem IpHEN É=I8S quo b=308 4 ps! Dna 26) Estroin, É ta) , & MES Pesre Pias MSF For az et quad cons, Vutã) TO mn açtas É=O vo dePis: DySPecE TEAVEÇED = ME LEF=Z NOM TATO main. Elanoind)e É-RisTo É=308! DiSTAME afvEMM) = LP-3E = 48% Va Oh, Eros «4 f=0 Tu É=%os TUR LISTÍME TEGVELEO =88m » Fe Lise) th) Dezreminma o é: Eno suma rramereS Deca nove Pins q -270 = 978 20007] Dis Or IO (amo Nim emos) Bs A “e 02% -Ó0R” 0-0 Rj ELLBIM HS? WE CAGE POBITIVE TO THE IEtaMT, THU rare e tO-ês do pés a o For x=cyt=Soró = 365 4 16 | (SOUTO aprs=(06 mb)E=iZnho 5 6=185 õ8 ze) mw L+& Za LA voto) ue Ag SHam)i sm : E aa — am TO DISTANCE = P0O Mm = A, Hits 300 media mitos ) ria nisfE,)+42om Los /4s PLAMER AMoves = 30 in, fo RAT: FIO FE H6A,AS bes Tera TIME = E ebgrlço Seitas fo = TES | Ponce muesosoin orem! caos Dto E ROSS ia ETIML TIME = ISPESHAS E LS FOSSA ASEIOES Em E ES COMECE OF PeRMSTÇLE UBACITAS (5 ARtM)s E We Era Simet pecstemamom 15 - BinÃO, or 260. [5% THE SLOHE THE É COEVE MUST ALSO RE Eri ls) DALbin/S afiado) & e) mês “rea lusj& ts Las (|él2s ' | poa t 00) 6 SO) Mto 1 Aee) o Lol ee) E, ' ; e Eras) : anã afim) j AT a som. e 8 DT RP TT vulinis) É H 160 Depw é crove mira il = Etbefoe 18) ps tag ÁREA UNDER WE curve = Z80m. For sie ACCELERATION AN DECELSRATIOM CHOOSE CONSTANT ACCLLEIATIA AMZ CONSTANT DECELERATIAM HE CURE AMENITURE, Almjs?) sa o Zu artomte) nr A 5 eye fe) Archos APAE RA, 2280 Mm : 5/40 m Br É[1 mt + (45) = 4840 om Comin TRA nin 18 Ani mia A = sor or vt corvi as Sms os sas a-tousm 4 Pew v.E cor vira GS mosca AREA VDER VIÉ CURVE = J80m. WE CHOOSE COMSIAMT DECELERATIOS AMO CONSTANT RECELER A TIDAS, Nozê: PéceemaTos + RitLttAIaM OE O 6 mist palmos) +a= 0,6 msn ô z “o UR Zês vii ms, + Aga io DANI AM / Aa “49 É 1 E : k RBS Aa beso m: - US MI Tay VZ8 5) = 280 m oi Eae BF 18 toh A For fereunanos Fans : aro np A AS + OE mito sinfe pr PE comete 4 e mÊs 96 po Fatos e É ME.667S É HE 86Is = 284 z =// 3385 É 4É,228s 5 for Decertepren Base: Lo= SOME CE V-E CORVE as Pam] h&mis Some — pSrmde o & ZA 2325 ap - 0 880m/e* 4 MEZ] ren. Da= Et dmho 4Em]s , aço Avzo: Cas Pi Bmlh = 25 of 3 tanto) auto Rs 1 ») “o so Ee For Hevck Yg=!5 E £or Avto: (t>i0s) ugralzs kar ZE (é -t0) ASt, > 25 + 26 (p-d0) tirasse 4 Xe “8AS Á tomem Pct, Tg %g? Ku Kg Um ANAL) = 187 m (&) AuiO STBRTS 35 AFTER PeVCk PMESES AUTO atom) AUTO AS|--— — 48 ç 1 . ls) Dig E = « For us: “gt %a= HASÂE-S)A AE (E 13) NE tp= IRS HASTE = 08) Es Zos %="300 m For AvTO! fE>eZs) We dot, Ra VE AA Vr= LE Us mbtos) 1.63 | PRAZIVE 70 TRUE, COR MusT move & DISTANCE! furl rHOrSa Ino [PE EE ALLOIARIE INCIEATE 1H SEED: AN, 250 “BC ads mido= 2RTES cae (ré pe fd zo fe ise Cm [ci mM 5 rpuere 46) t— E —+& +44 Aecerenanos Punstr Er He = as As Dececegamem funse : Ego" Bhonh!S A (22N4%) = 88.946 fakas) = 424 fé BET: AVEM AA LHE TEMP HIDE SIL) EP 387S Compal UPE HE AME PETAAS IPI tras 4 LES | RELATIVE TO TRUCK, CAR MUST move a Distance: ax JátyosSatro =/ vá PE vfiets) car ANE O a de [DO (rata Try T a “o Preto ets) ps —>—+ 4 A 4 SE = 20h53 = FE, Are Ada Srt ft = lim tsto) sete Gisa NE Ade) Ba HR Zy= é. 8885 Esorad Etta 6 RICAS Abr SE ES(ESM)= SHIB PET RES mito Somee Das BSmilo, EI IZE SM; HIS ao att 12 = [1765 ml Rat A | Memes Rena er. CoRLISIOM 15 AUST... AVMDER. THU WevEM PISTANCE BE TwEEM. ma ANO TRUCK IS ZERO TUE , o) TRIM AMD AUTO MUST HAVE SAME VELGUTL Ve | Spade MEGA = 4x = 0% É lua tefsK os) + É (18 fio) £ = 30 FE AY Creu Net EJA USAS HA s22n)= SO Ftjs Bog asas Ava (E) -30ft- Agyml1 3335) Agyro (2) ab=somi/h = Be/t% Genasss =-27,5 ft =SE ls we Dra vt curve (assine Bog is ese) LET commnuro “vo - Vímio cena sm Passemem suewrr OVER TAES FRENHT REUITOR nem fracas UNDER THE É Curi RE ESURL, Du7 6, Vime SIADGO MEIA (uuttcm PEPRECENTS OIPPRENCE / JO MESAS) 5 Bo, mos Ab> 88, Axo | TIME REG UBEO IS | E-sratsees W 1a Feranr; gr as at Disrames ranvemo s a=Bet(Sakkiso) d' = 705 eae FAN = TEEM - 30 Aka E) = HE anil =é6 th 66 Venal Cotar dE Vtls Both = 26 Fis for Bora) pres) TRE Eme peleciry 45 PassTrê AS WE MINE ASSUME, THES For gera te) ame DECEERA TOM = LES FU? Amit) dio FP REEMT LULEVATOR MEVES UPWARO AT CONSTANT VEL CITY Y =E mf Gp=o Poscemem EsivATOR STPEA Fem Ee ano SINO vue rarzt = 42%? THREE SECAS MITER IT AS FISSO EF Fauna Emivrras. (Voa = Os. WE Now Def Tue ai aro gt colour (correm) mee | aos agro “| ob fronea) 5 sm) | de 1 AREA = Mora sue | jnrfm/s) eres 4 PELATIV Marim “so br) 0244 5 irei Guto 40 cómçtm Lo 4 OL jo md caes EMITE =, O —— 1 E “ “ - é AREA = Sem CEDMMETO OF GrL-SUAPED Poe mono Tt curve Hm om; 6» doxue=s tre2s pa É apro rOME.. His pts “zo roasça) — A 1 o < + a id E) 2 (O [ | bsmls Gs Ap 10%, dn [026 E NR TUE MARA BOL smnores Mpweep y M ARE of / y Passo ESA ARBA, THUS, Terdc AREA OE SmAprO Poe Tom dl rectas DE NÉ Dinsmaa IS: Judas - =dxa ga FúGee LA Zen Cross) tons dfo NS adsipao) = +0XCanje e(s) A fmh?) b é, 39) tos) os 0) 25 sã Tess) [ too) | (a) ugeu = ne: SD «| Ka (60mte K 200) (suor Area) a 200% 8%» Cpo 8 (b) tram É AX= (60 muy s— Es)” (smapão 6208) = fevendehas)- mz P8Om Áx caapã Cavernas” de” Ss ema E PAS ms 13 UPR | ue aroma THE or eve or rue SERIES OF HORIBONTAL EOSMEU LINES Srta - PY THE FIBORE, FOR ENCH INTERVAL THE ACC ELERATIOS à PSSUMED Comsrin Ava a af or abr Ya axe varrza(né)? = 90 Bm/h= 25 mes Reta 1= PS quis TO S= 201, ASA ms AVE ZO-M ="5 mft = AY Aé= axa mstatalas) (ES 285) HEINZ zo) ESSRSPm SP m Áx= SES A, FiZoMA S= 20 m/s TO = domêbs: A PE mf AT=/O-20 = — dom st= SE. das AS Ax= sé +dalat)= Comtio) Hasta)" dx= = 80%» 70m dom FR = Sms Tp EzO A Sms Axe bate do dafatj= mto Dá (e Fm fik 2.828 x = DEI3m gm AT AEE m For ENTIRE MT» VeRE m/s 79 vo = 1 HE. TE [o Re mtnitA(é- 2) al y . = Bt Ut dat NÉ) 3 e EA £ Vnnitepa st? « amis 3) - A 1 so E E 6733 h OT Z/4..0.8) 0) ga) 10 E ses RE) since Ar=0 WeEm É=0 Aro lira É=d, THE CHANGE JM AT RETUEEM É=O AMO É=-A 45 ER, TÉLS NEEM Umgan a-f cveve ds ZERO Aria tão Leste dlritolos)+ E (-rofa-os)=0 9-2-Sg p40 g 285 4 (8) Postzom unem tet= 248 EK, A,/E- 02) P Ap(E,-0:733) + Aa (EXE -o) Oro dakas- aaja dns genas as] Ztay-8) = Bm LEMA ps 766 m HIS] sec soro or Peg (184 poe at core Etmey tnios , ds (a)é=F0t= 2885 +4s+2235 A=Qys afmis?) tes dos RE, (8) x CEM= SELIM táom tibs xt CA EIITA tre 2 ato) of » B UTE | ta) AT E Por Compésporame ro fa So, X=3ip, we DES TRMSENT DO mx cueve (A) Fosrmem Wuin Esdos -Hels “otindo) a WE UEM mens: X= Kore + TREE) = AEB mis =-tm + femtotzos so eso fr litasUs) zo dr. 3 ade =emasom 424% = 2m RE Hom e A (6) Minumoura Posrmis Cmmemmare occups Wam ds26 Ev: arde SEE É Corvê AMO NOTE THT Vga, OéCUAI We COPAS =lisens), me) sos =-14M am GA ij" | tm io) lar A bin, (8) Br que Bermr Corr estorno To V=ó infos INE Dfw THE TANSENZ TO THE VÊ cuers, » Vinho) WE THEN MEnSORE: mma YEm gm -/a%m Ena SPO ' . das Sh mA 1.78 | E as ”/ Bur &s E 8! E do O 2 7 -48 4% “pokes) 2 ah Ne Or) na ZA to » els) o a=- SE tus a 2] K i VA o. AO. IM ORQER To SE TRE LET Soovem tm Fars (ÉS TRE SAME STALE MUST BE MED ON TÉ A pp x PIES, BY LISA THE BOSE RRLATRrIS AA ATL NÉ MAS MEASURE 41 Ar Ala CUANDO OF TRE SEALES USED Bo Ema Yoga E Ko! mt TA(Eç E) Roi 16 Entendemos) male) foe) Lypes duiys: = Rr FA) = cc tisrenyinaa to restos) ze) + le tutekas)os) +fsamfag ze) EA LB m + IE m LED LO Xu, = 1/04 40 16 MOO | nu 283 g4* =22s24 Br So Et oe la = se 12 A Su rIZt Je Ape Ha = é (0) pmem E=is: [184] qo fee po : = gem Ap rip é to % no “Ape pe vsma & me mar toras é <E xi oá = ly Epesçé o as = Npc pls ção Bon. tysé Weldtih zo 6 4 yo o pEfA cosa )o pise gs Le de + agaré As TU mA ES - cf P(A css pe) = as" y (6) mdep tos? “8 Jr nrês Ê CB apa O 18 int « Eae Dis atous rum de A 18 A a, E IES «4 “Ss OrgrersD Alone OP. Homer, fe) me “= 3 9 a , > AsxPRESSIoAS FOR Gy GO AG Shots em Lx 25% > = Há vga 44 Testes Per a “al qse a o aero agrté esmo” AB) afcaiçaz = eras calor Eostab) O 7487 é nero 67 bo Pag a(A Cntot + EE sris pt) o h mer Foriro que Enocuer Ke Orverne E er (D, mengo gr memger =97 (e) Lite) Te 4 4 «a GEMA MIS A Hrtrega « rar emplr us amr 2 1,85 =s6lm- e TE) g= Esse g=(é+) g- HtaerE nda = ture Uj = 1257 cas STÉ Be 24) WE-B(E mi 4 ae = pre agr OW = -samsnvré mA Ag: até” (2) Wpem Eixos Epa fo: º = +47 =+07º U=82lntediao A Wet ty Ve sab quo | az agro Er VS dn do ca -« e+2 au rap A CR CEA 882” E] (2) mem toorers= ps: (bi pm Eros, Dz tre MO = 939% mr - ts d,=-2.37 82 fa TIE “ E mam eastema)= O Vo ssdindo « It? AGU Bam A EU? “| a cute stages ir +, as ssintirms A = norteia) == 03 JR O ML82 | ge sigam grosttaze-s És va glretra =2+2 11 86 = [E sis péli retj HReospo) ae Sigde = =Cya-1 W=/Epcopll + cj -(Rponpi)B van = (ze Rena Be (Rent) — (potes pa) a MARGIITINE CP VEl oerry deem A1 do drnerno, 115 mes peniana; Trus, mê aprno | AS tg nO Lespeesgi) 4 + Re pesin poe) O dE) afae- 23) nMtra=2(88-4 442) =O te aaa Piceeç a =2(tt-mj-o E- 2.805 « dudértdE 00 Gees st ted Om fecrc evito 2 tetos = 28 drarrajra;- Cp ls) ol pasa) = Re x=(2ataea)+? Z--28% «a amostras se Sm 4/1487 87] r= até AGE res Ga atas) - -8 WE Age) cr deasº A URL, quase) peee? ESP ess o 2 “yr Z8+2 Micrnao S Veserad « Ve corram Taro qr (agoe)- -agg? 183 z For B=| 82: ati ratgr ze pe, XE ve ci fre, =45é 7 sas oe b = ce = amar cas dt roda » soé a sej + & =. =“ =, Car ciaõom de Mu vão at ege Dada? w=f/a sed Paises im tete AR alyne eis ve som! Rs) pum 4 velmto = 4ms* «4 ae veem o AQ (6) item EA ri ER eo] 3 feresado o axPrsuia)Tem = 7a 3 onde (8) sue tio 25 + + Fer se 1 SDEUTIM OF Pesa MRE, WE compre 453 44 cos = 247 Te Muy hs] Sra => A vga Ep 22 dy” dofa)= do - aê r” é tee Vis aa EEE a rosa 2= so mafl | me Ez! “ro «1 ay= de bre b=2s* de routers) 1 tt opopmpt d "2 santo ? IZ 1188 | (2) veericas mega (um, Accer. Morrer) joe, Caro ge fat 20 = flop * das Hotrzantae porto (ur Fetm x (dt HOm [NAZI E) tc dert mts (6) For he Zêm amo fbs bl Pomnis, me mave no geést? monHONa)E trrors Cefult= (1) ImisN2OR 5) x"527m EP 7 / Ass Lg 2m E la A: -57 tm Ea o amomípe 1 mui o y GAME HITS CS LRRIEST VALUE SF, Ka ESOM de= 1om Dom Elos me? ES Da = VU HEIS 6) Exp usas s ys 294” G= 587 uts Ke WE SEND HITS 83 SMALLEST vaLVE OE Vo Rar Am Gg E 4350 tu Je 7a” AI SLB) Eco sam 5 me mê Lmeblesms) age 28tmts <d FOR SAND To ENTER CHUTES 281 m/s SUL EBIm/; A (a) Bau grs Cr: Ee/S E, ge=3ft Bair durs i qrógtl art=fiszame)t RE Wt ME tt=W(om8r7s) 0) Bmemes 8: Ya SME, Gac 28, (romena) Get apeilzrane! Pro Bsoss ug bÊ NS fé = 05 (0.3826) = Hab Ft); BAIL HITS DE KpT US /= Le gp2 EEE from) Gorági? art=Álmarrei E=04217 5 Yo GÊ mer = fonte) apr 32% HE] FoR BRL FO STOME COBUPR RCD3 BLA CUL fr, Ret =o.78901, (1, Ja = Yo smzo o, 220 dy 1492 x sk -— a 3& Eros 1sR 5 xa 66 Enzezs fis ATE: AnsO VaLvs or Us jade 3 FE st PE DES E=ESS/ a = Ly z stt = 0 ao é +Elsrade (asas SENTE For És Bafo 300) SESI À FELH3+ ES) menti A vas RPE ATE: sra est VALUE Or Dente, CRE=OSLES E rERES, dg ts astro ar mtv (sec? et “asse fo gem pras) 16 E)" Ls essttts FOR SOMEeE TO EnTER PIRES JERTEIA SIG E MS A 93 HI0 | SEE swerem or soturmon or Prog 4) 89 WE Noto Have dz RE (a) eat miTs E: Hs Pitt, ge SR gesto afacb(gazp)o Ecos Tb 28ft-Wlomers) esto 6) BALL ITS ES Rg RPE “as 28% Yo Ego? ate blaza po d=oatães %g" Ut as tem ilassass 70.9 AAA] Ve) BR MITS DE MASI up st getos am-dlempe” Leonsos Rel Hft=Blov3s) Ur ssErt) <T DRA BUU Ta STR Cone BED! SEL USUI set rt] Vs 28 ride OS fr )z0d hr Cosgtnaos tele WATER ETIOXES RB: SMMEEST HETOBT b Kg PO Fé gb Meslida DO fetal )a Ge” VoHu)t dgs* : o=h ilinzsricosme o)-d(22:2 210 NO PE <) 0=h+/40025-/5.2553 h=4359 +E Érova6és WATER STekES CS toReET HEHE by Ve RPE Goch RU RIfE-Boratak)t é=2/095 Yes Goat Sae 2 D= halimesreapne)-2(222 Jin s) o» venia) * We gal de For WATER To EnTER BET A3SIt cheese 18 5 E 102] rstomj oiêess (ajetbaina [MOS PR 3 HoRiItoNTaL Motrem Eidedad (Exportar Méier 4G=/6 E a B - EU = (8 cosa Je Soe/6 nbs 2 ATE: Yp=/8m = (18 case t E Sgt « Qrt233 1) A E Í Cosa heltça sem To VER TRA. Mo Troy =, gude dg é? pn => 9-8 soa) - -A(asa * Far partimos Gg, BALL RENCHES CELINE WITY ATE: dg =IRm=(18 sir DAI sigo (OE ops A Honsromzal vecocery. ; VERTICAL LAO Tras Ré UNE Omi AEERATED) SUBSTITUTE cos E secim = /4 bon? (uy +2ay 12= 15 Cano — Py0 (Je bata ) ATE: vo, q am? 02 fu) -2(cbrmf ia Fara = phor Land + 4$73= O é (UT 989 cais Lana = 23 tanus/ 434 Flom A oC Es Cia EE - comes a = Taz PARE 9 Sretma MTS RB FOR a=dtPor apr MÁ] pra sy =) elo Vito] vom) Celossra tg C)aemss mto HORIBONTA MO TIass hotsm =, = (1235 mo, 90985) = 12, A BS mohSm a 2om KoF (Ed, Exa é mio. 95005) em E gm Z= 49% E =4BM-S240m = S66 m = From E TOR? t os /6 mf; (UE tb cosa Cad 6 sina £ e X - Som Ss pas sm = 43 ce fl VAI mk Lemon torne m=(ULE=(H cost) xja AT RI YaF18m= (focos) £= LES (1) Ven qreõe taozom - Zago , MEZTGA AvOT: g-lnt-tgê g=29 tie alga minho sau a) > = 0:36 m ALR: Ugo Pm=(/e sinat- ELA! FRoM SETE WE MTE Zur Garbo : p Als - - use Eol): 2= (e sino fito) acl E h= Gai = ESPPm asa ! EA suasmrom Expj SECAS Iara (h106 | Mata = 4% 4 Uatulideas 130º 2 dB rana 62079 (rttana) p= RE Pais Enio PAPO ond + Lageo= DR Sin tão! am O ABS Land + hIZar= o Ses = Sosa and = 7,3320 ÉanA TONE a = ég80" e =224+º B=19.7 , LHES enETuer Bate Mo7S Cpremkg = Puro tis IRD oe onde” fm) = Mg sin dio? = ep 204 fel 4, vg (Ur st=/406 PRE = de Wi= ss 2(sks) cos v208 Ve gt emoe cps posts dE SE calota) cos FOR ha* MEET Va O = 18,20 =p.até E=hrses dos 777, = 2 339 bs 2 ” “Aa Ta t-2gé ah= = t-ég À Va é fede E HEIs PAES éys = mm jé = Strst spo duto Etr bases 9a E UR A ass)” PZzo0 yr ty Rr 270º AM Nisa E MO27%m 6) í . x 2. sm À Seneé Crreneê E 47 rum AtovEd, ig SE rtn0) 2(chsoro Fel A=UAN Brel HITS Ceni. E VC = Mto For à=298%º (ae VE sin 2USH = 2E8ES Fe A ao E? Soy sinto! “ado gt=2888-p7/é 8=189 “BIO ftys SLB fts Fem MAY item g TESES RUE E O FOU)S a =s2s" g- Cut -Lget= tee (e). loane , à | Dtd=do!; LESTdedo fue! Mato Bd <«“ mer ESTE ABES: Jomeç? 2MEI(O BOM) alrmfo NA) | (e) pm dos pnrervars Lee Lap 000 é em yes Goma 3/72 om so ADD E SUA Tio SME csprenmê v5 AP = hm AEOME A, ms Teo) maia EA eira Bate wtee giro Porn B For AESA] TREE Desa E PO e Vo tda Dee ema em! Va £! NORTE LAW OF cospar 1pã e (520) (são) 2lsakczojeess" Vain = E HO era dh LAW DE sines E sda SK persas Ve” Sofa, 82" A “VE 00 Rao/h, x Ê + o p= Sl .— - Le 227 ta df = 904% cratera" 1.108 Uestobn/h Lpw GE cos puts nas? = aoot+700"- 2 [imo 700) cos B cosa = asas p= 382 vaso = A 48.7º Va Dhaceror or Amrre Bis GE we or scwm M UE More pHAT Ye 14,107 dos dé A Va Wmiln= é FG vd ad 7 30 meia = “pttls tu 0 cosnEs E te A Up do! 9" a(s Naglcos do* MPis & Vaga PO. 99 Life Lita DF simes BU, "Loja Sep o EM gua ss 20.39 rm Pop ndo! gas “202? (2) RELATIVA VELoeiTr: Ea o figa 8 (6) cnguee Ip Posirom cor At=as, AY ajp” Veg dE = [ORAGO Cap Sss e PO (c) DosmaMcE GENTEIN AUTOS 2 Secad AFTER Auto À NOS MESED INTERSEE Troty BuTo A Temes For As. As 66 ij 4! 20º, For UE =/6 Erk)(os)= 3218 SBR FE do DO Lg te - “" 7 Auzo 8 =p deh ter / To GE s(otilolss)= 200 dé À Tm di CL Costner . Cao” (132220) = afpsziesalcos tro Ca? 292280 Je lat aa Liomecg provem Guros= 293 EAR uito LEMETH QE CABLE = CONSTANE L= RH consraer BAU to Ag tZcap -0 (a Dt Para: = AS indo? me Pk) cast)mata) 272% gj e ap 45 =-Zga 7 =-2 Aga Cear IE ineo agr PE dado Lj VE dnfo? Ca GOO E 4 vin fg Pes 0º fa)lggocr E 8 Ag Vem LB CF CosmES 2 “ toa tas 2000) cos 50 = 7013 m/s Ler OF Sins E se Ea “203 f à = 90 pasa qa! = dorsét = 0) info 406 (b) peceteRAmoo GER: Ag MAL IB couro er Usinê FIMÂLYSIS SIMINEAR Ta THAT USED GROVE pros Lo AN MTERNRTE MAGIA (7 Rg Eu * Lema = dE int) 4 DELE! Dom dr =-15 9 «(25005 sei +(1s mta = g = EME É MEUS a TE E “0.19j a É AL a Iê Y 1 1 40.027 | = ME bj pés A ZA Hm CC) For E=205 gr Pis qe ME Às So mis ree decr: & o Mat Vea tau Of cCosmes, “ A, a(s)" UXAS)cosdo! po Não Gs ="gporass . Sto LA or Eures É SE EIS qu 7? 25 go fé Vos HOY infs bag a CB) Fon = 120, Cor VS tdo 730” Law SE Cossnes PADUA) cos te uz 10.892 tm LOW SE Eiras Sbg sitios aroma! 25 70857 sas Acdop-sar as = = 00h Vós A - 2 La = Emenda . Emar 420 on mprwl do TED EI7 77 Le“ Ga "Qu CE) = Cimo) comeram o é t8fs = Sec smo enris EFE, Core pec, Mazur) Cu Cs mo Fes» =Fataaz fps) = 1294 Etts À CrLoTvE VegoesTy Do Como EF Tm 15º Ve = “o Usgo Vo 8 LE E Tape CEPE E / g/8m* = =p -2"/848 “ to), Mori qejo-g A Y= US ! =P) my om = 18.926 Sp £LAv OF costs z Es “small mmXB)eos (15º ms) Vo= 70,08 Pg Zi SE Sus Sa WHEN be ZE: = Agtas)= (826% mmleNas) Vo E ASAS conhe ADE ASA mf; Cap Lap como At LAw of comes sol) sntent) lara! io [IE UAI Zoe ot 2 a 2 É Coca + Cm “OAB. ue ai-(se) tiro) eloelicerasão! | SRA 20.08 File Er 20.08 Ri aço SELL mn fe . = 243º Low DE Comes $ Us fetiã 2? se 120" . Vo pasame cm ADA = = [ZE - . 20 me EREE rente (E8.6 . Vsk = 20186 N 88) «q A = 180 -v286 = Sp Poa rAsoçã a (0) Pecmemamers E! ESPE mf A SAS A (3) peecesry of & KE Ag= comzar? RO Nh=6 WMEnÊcO NS men d= 85 K= toa) ALeSMÃa5)=182. 6 aonko DO | rum Poor HS pro Herr mus cstrm luz 1826 nho Et! A - 15 Perito 2748 te (ed Per Soy Te 86 veria MS “ag LE Var Mus] x = Bom” ro! E es, nda « Zoe Lap” !Romnlos —m /s a pá Dad Ag” 24 tLgya g + 15 ta ps Som. Lt 05 Costner Ha Se am PET unteço Es age (20) ao ala eçzos see pre Ma =a2r2tm É . Anjos nET Aga 52.82 mens * raso Gnt6s9 . ZA OF sims . Med A CSBEO sirms = 2448 6) For Mirra Vajg, WE mate Go Ng, Dom” SPLL Nà g . o pe cinza" (0) heceutprav ar E Cp LES men? GEE” aj % : E mera em 3.484 (6) tecoerry 0E E; me Ageconsaaro tro NG=O tones É =, = LIS Ff amo 1$º= 249º K persa” e dose tie mis! E 23 PHLIEG CONTINDED W Acer manos WE MA ver” Ba sbS vols? —s Ag = 03232 00/68 Tem PULO 2 * Rafa Su — 2% agr An E 4 am = = ar Sms “a peafç 4 2 4 au (Bond; =230m ow ce cosmes apyg(o3r22) + (he) doresalisjeasfas"-12) anos E vainto «(lord fo 8 Aitg7 338 eomts + Hh 132 due ee a aros Une Prodbernd is Fraso ) “e ams 9 Bltos 308) = rio an/e x Usontscação” Eu ASS Om AT: (rmangarorm desta) apa 90 RB Law OF Sines Sink Sins us a Vossas mp)" di MST” 2338 v= es a-8º + 1 RUmpi= Pa =>2480m 4 2ajge IES À fe 1,130 E e bo) festa jose cones | 1) /33 | a) ser some peos dA ATA S Fela A A= 372 ride mes a-g= 82% de ans (SAR sin 88" e 26.829 «58 N n uÊ E Pa (6) Lotizonça sro Tom 15 umiform: constam? Wocrry A a a e = Vence E o RU = oe : assa z eva ms 2 2.39 eds (ari ane E ; e R a um qe 227 8 LC istecsa)E 2 AT A= 3 so cs = 3 A 5 + (Pose - BTAIA=O, cosd=1 E A usa a tar 4 Pon” 3 p= foi Ga) « = TE HS] cana: acer é fm Ri . tr Dib — lE)= 08%; eme ele, ate (E) “E m Ba ES From Plog ne: a-Rp?, U=[eisRão piso Geo eo, som (RpPco (Case fa) 47 ds 3» =2%» y armado eee en É « aço A Ê 7 Rp? a Ef a uÊ 3 z Ed 4 HP) = geo Kb /25) 20.139 asas Fm em ts 2 N RV = pre smcg €, Len olof$z PE) ) No p= 16573 fe Na ES mf AM Gram - See soco 0E Fur 487 Rot uatues OF à, 5 fo Pilde era sl ) (02) mute tros am ponl S=tm)s, ouldéso : co Bros. = — a? TB: UC testo A3RSP anfe Eot): famoso o + [gar] ag tar) a = sed prot é trmfeio BZ)” favas mag) (lihen deas: a se nfs VT nd, Pta s sato: (sm)* (encsjº | gal É SU «(STEPS VERAS 3 peletIm af Ré = EMO mê — ZO3P x10t 4: 322 eds * ooo rugas a sãos rd abre rot VE 2$ 501 life =/ 739706 Ra Fe IT To milho, 4 = [BI6O mé 4 MO mi )= Hbowê = AddBNIO ft 136 | ar= 16000 mi tiigo r= 6ot% qoel = milh=23.%4, ilh=? or . EA É estos - god? o fasso x 5280) Peso -sgot am E ara ao Bu tudu É=ds: p= dotortoim. P=sta-tos Bomje : 252 Feszo-moro É) vecowrr org: ” r= 322 tensas = AS sua not fe E Verê (23,46 )* q r=gerz me mos, «=. $ Here agove Susrace = VEW.3 330 = PBL3 mí depris dom “tt? hs tm 4 . Prapeo ne nos id M.137 | Na = (tombo vlitomiles MM Ds (4) Recermeprom or &: 2 gr ape Pré O (toi Kt radfe em do dn) eo Ag 2 TB raTE = (von ferrado) e 2bomiforade) = +45 1/8) « SO fa) me mou: a, = PA creio (e) Aececmeamon E E RELATIVE “PO g0b a=éraVE = POSA wide * Y= sa eas = vaso a=-leroinde +oinide, A algo 1) soe E In" O « SETIME Ae? =. er Aq 5 - 508) 3 (9 Hs r= nasttoge Duos vs PROBOLTANA Po . SED, =2.5p-274" . , Fozs-cré lb) temem 68 cre omez: = 27 same Wuem P=/67 q =)28-09=0,38m Premo (26 Tirati For eme org, ÉS (2) f 7=-09mjf fot: ves E v=R/E Fones -arms % (2) uesociry qr 8: . fofa) dez AEr am LA “Ozmls Aga Veras dead Rap R ga hs =A10ô Í MEGLECIME MIMOSINERE, WE cimose Va = Tp E=/85+ta; ie =2529º . 27 4 A mma gu La ra E = LB m/s - - é o R= 4370-Rm = suo - p Vas LS mfo, pros? « Emnrenã ente, exi agy = Suas Emir Etr minides | (W)pccmensro» or &: [1.139 “A Phi) (2800 irradh) = ceu8stemf6E Ago TB +2Ê = (002 Cm rindo ati 2 ms redhs) x =-RL9P+ALTEO = 2047 ampe * Ser Sotoram or Frog 4/3) Fe OR past 20 Petro, o Est) , Y=(80Wtas! Eat = sora (a S7x16m) —————— =(RBh, a 4 ss 2 a ção am Celtas leo c sta ae posmério? aT= 4008 m/s Es cá AT = SUS Rmfh nr= SEO Lnjo Ago ET mk, ums, A — NOTE SincH DiAMOpIER OF MDOM *Pr$Am, TH (O) neceeratios 0E E REBTNE TP TREROD MOD MOVES DELATIVE TO THE EltT4 A Drag sm PEbo7 ema movr Tejo RIO nfs Y- 80; E RT asrôsró o sabotar) = 407 bo; Velimmider (0) men tro: ds 0a ke, ADO, Aga HOTELES De (J2S7 im fo des AA Per Ramo E see Pres igidom, (6) mms bos Vos 4D di is e ró=beibJar= sm 48.7 nb puro ras EM = 46 5 VE = JB, dg Va jo mise, +USVindolãa, a ="tor (a)a or E - anjo! = 207 tar SÃO = 218º as (287 mile +28 md eo Aga 0OT= 1857 44, Anadia = UE ndo? Hrmê Po 26 cout Saut Hg r=2&li+cos jrê) rtmebat ra FonTê ade (2) tetoerry P= fm etre Vo Pe cAbw sinal Gar6s Ibicaswê geoerr Um Pao-Db safra q atado (280) Fcis usPe (cos co) = foéu) marés Mó (r+ecs £re) «a dcesuraros ae téc geerecamy a =Y-r6* = F- rd casacos ut (bes at Xi)? Lengua - do Blrrcos fre) Ce rates ata) Cro cg TI + 2 cas fr) a sebitsia ut : ae rôsará eatrabo (pa?) cadi + sinui)= (44) = 0 +2( 28 sintré£S) ar sbu* «q as -mEsintré <] (U) snee Ve fbw- constmir, ageo (a) waiy d=o: r=28(n)=H8 0 Fes: Ag A= vam sado vezury Wo Gs bl) fia az Ês (5a, disse er bo p=b uezrber 4 P Foe TUE FRIA p= comeram | Rerum as FZ o e THUS, PRIR IS A CIRCLE “q b 43 fmcmpazocam SPA CElmien este: vozêlmos Ejaz4 A jo e=ané setoery = -DbsinTo-p& ie É ram = Dera) DE lemocrry rcÊ=to mis ape rô= (ot)n)= 2ont V=- DBer + TÉ Es « Aeegreamom astro o esta) =-s07% fectresôpor q Lnglages if) agr e PL AT -28 « a -L9BeVEo MIBE | see soro or Pane lJAS Poe 50, O,10s fodpdem t=dss p=2birasE)samib, aus nbs EDITA Vas meureos 2)= s702 0h «a a -omme e +10M8£, A he zaã) a 5 Selic szonrt as Pibeint= 030074 + 2 aA=-1207 ze EDIT ÊL -« (E)wmem é=2s vao e=7 =u8* . P=rvass s-Mbsero =nblitesn)co — V=O « = Begins J fecfs ram e-77 as fiBtiszen= Los agr -PAsnT=o o ' AÍ a-inbeo M Srisga' —— Z = SRF mms” ? Sigo | Aos RE dj geo! D fé=e, ro, or0p a Boto ve; (820 R8 HIS8 =8 S=2nt E smABnê ê zo &=27 Ê 20» 8 cos 2rnt Ê-o 6-0 E 97W'6 ain amnt Vercerry (em dhr) Vc Ret ROC, + ER V= HALO Es tam nE cos 2né R v=grfa*+ ntelcosdemnt E) Peteenmoy (EG $0) a=(Ê-RSTE pres abb est Eb 2- sine MW din crnt A -« az sra q nº et sintemme 1159 contivuro qe zrfnirs cosa dE VAIIRTaRO é n2s nã cado since AT, 9=0 Zuus smce nz (BD, Ga fy 1.159 | From Frog hIS3 Fóre n=! aro Es É Ve ADA E tim Bes6A= a0[ A E, + Bens E] Eine £] A urAto- 7 Esme bi = 47 [her Diptera or ElhormaL 45 MERLLAL TO EYQ From Fr MR, WE OBSWEE THAT Core Cr B=€p Brte= Ea " traz em[iforêcs e 6] sato Eme £| = emfai “AB cosO E, AR Sine el (1) WE SMAIL EXPRESS PrT vsime Tue M9,8 Commona ra Ep" CosE E + Ina G -smer +cosod SUBSATLTE InTe gel): ara=str [AR -ABf-cos o sino i + cos 3) dr 97 Esin6 cego visar) acara o gé E A a= o Prim 5 fu as SINCE JHt=0, WE MAY Were assa ante Aa xira PrgrUS DE Cpavarpes cecues arD dar P(A ve) fel SMA [Mem À MiMIMMvM SEADIDS DE CLRVATURE Oceves n7 BD ; E tesao a=trVateg? do. Tavsr st e ut smaE asas 5 E” mm latçer (ui vãs e « 1.160 | Pohtto gre ache = bh temp 6-47 é=h fo é-o Z-o Ve Re rRÓR HER = h dans Corhifimplere, rh ve hy tas € +47%44 daria + ve ham AÍ I77 7) = as-(Ê-csjar(rã ab ER =(-ht Emekem de, + (2h bana)ae, = Com" bt tenJta (uh tema E « -AB(sinGcaçã E + sinta á)] = sr [66 ma (todo voimto);] va a amS [RR - Ag 5] (a) We mozE THAT SINCE A amo E Me CONSTANT THE WICECTOM DE THE EMORMpE( PRRACLEL 7º TRA) 15 ConsSTAM7 AND THUT THE Ap MEMO IS ContameD tm ha franeo A SIMEE THE INTER SECOM OF A RÁNE Ano A CretnpeR ts Are EULIPsE, Farm 15 au Ersrose (Db) Baon sr Convarono mes Pesg dhiog For ne) fececenozas (EG ps) a n(Ê RE ep t(RE RE) ER =[ne tag «eu leç+l os20 etjane, - 06% “ER O = urh tema Vrtore* « HLlé! | Prab-et) =heE Werocrry (Ea 95) v= Rep + Rós, à U=A0 "er ame De, +actk lejymen so: Rest; Rea 88 velhos 4 Umert-cistero qzmhe, e LA «4 ne q0f0%r 6 cod aca A2 TEA smianE add azfitrendavar 4 ge =ant3) Zigfcos ame (nsinapaJam 47 Ra inicos int gumes É =c0; e te . Varia Tcoçtage Et Eca art &= -s9"4€, assra o! ContinveD) o ” s/ HLI6Z ULI6M | mem sorira or Pers 1.86 E RA O=zmt east Rizo á=27 DA sin vê cost . . DA sinznt R=0 80 =R2.3A cas art vELocTy (EG 4.49) Ve Rep trocar IR = LmAegrirA ch Amt ve rA/Ey sbiz7e « AcemetoTeN Legitso) ' | astE-Ra Dep +(RôseRó)e +84 =-sraLe 2a cosare R aspmalo + costeze « REZAR cost e 11.163 É Re-Zgamt é FP =-gRcost & ta) fara cr PáezeeE offer o 9 Cxmstose PRoJELTIAM 06 PH 9 Plane, K=Pos8 =2R code = 2A(iVircos 28) (u-R)=A cos ze g= FEMO = 2R205€ sin) g=2R(4) sino g=A snze WE MoTe quer (x Aegto Alca? tar sitrejo At THIS (8 EGUATIOS DE À CIRELE OF RAEUS À yisza Center 47 RR, jo sinee te vetoarr Cortona =p “5 COMSTINT, THE ERTH OS RB? ELI, VERTICAL ÁS 7 670 Rà « (é) vescerry ano aecestrnzrer La map! U= REptRÕE 1a VE( RA cintleg ticos rpR SImcE Ep Ea, R nam fr oRTHOSONAL TRAD reger = Ver sine) PER cost) p? «Vet tE + cost) rp? veto! WE NOTE THAT NT = CONSTANT, TEUS dufft =o EQuso: a-(E-réDE, «RÉ cera, + ER =(akcost het) ep + (0-4 Rsint)es + O = -(shcostle, - (eh sinto, as Wairar rap = 4Bflade + shPé aê 4 (c) apos ot corvarurE we tetra Tur ISO, LO has Te (Pecas pl ref -CRp sin pa) m=fpspoo fas pe) SMEE BOT VAMO O dE ly THE OZMATIM PLANE, dum NOZE THAT que furcerios 15 BRALLE! TO JAR, £ á ares lRprospt) epginçe) Apsiapt) o ARfcsspt) =-etptcospê é rep" singt det RS Hsipéreos pt) Aua= Rp-ecospti +Rpj re siopê 4] dez : = Que BETUEEY Cy ANO TRE 6 ASS cr, frxaeg rp lre] heat) Pp ie cos parcele sim pa)? cos = Fr Vigia? Ever Lincama! L OscuLATINE PANE, THE ANSLE BETWEEN OSculaTINS PANE Art) THE q As Ss E= 90% Lan =dan Go) cota fÊ qi dE q * q R aê bar ps É ? Basa +1).165 From soLorns or Prom UV: Er tTrstR REj Ae R SincE BOA SO AMO BLUE db LHE OSCULATINE PANE, WE MOTE THAT SINORMAL 15 BREBLLEA TE VAR, td 8 quad qe geo RE E 4 1208 a ur 4h Cade TST MEC” = 9208 EIMOEMAC IS unit vEtTOR Epi tra -B6c-8j +29.8] = i cos É » ops Se asa -« cos Te = -ooras ar 223º «ad a = sã = aanar S H.0 4 Mibá PANEL pro TIO (erica Bececem ATA) AT 70, WE SLhm/h= AS mts AT S=iom Ve Ando = PE nfs E abieza (5,5) (25)*=(15)"+ zas(iso ve) AT E=/oom: = Ut 2a (ss) ur = ces)" e 2(us28)(100 0) =4H67 ve APLD cale = 2047 mh* ap=hasz et * Ui zon? PO 2 TOPA! ACELERAR a =f0.223) Eizo)? ar a «fara acdvini: 4 169 CARLIA der be =+00005 vem ax (ga vt) ” e =9+R%) õ ! ; 1 Au A bold dlordiio 4 apt ade) DATAS iz cmol, US ds fe, Some ad? | fogoosk1ço)* PA E = 4448 he a dale EE =). E er mis (it trm68) b = (sos0Xo640) b=HBtm A LHS (E sa BEE -2cos ave) d (87 cos DEE H(47 sm ari) r Y a=Emeinnado «(Bfitos ané)5 E we Érrgs eriti 4 a =-femsignj +87 Eos em)4 e: o tem ar (EM) E) (4) farm or perco % ; s | de ns LeRriene, PIO TI Leds EEE ALELIRA DES ME MOTO qua MAO Merss = ÉE runas AT y= Yi, Vys so: Va “ta +2alg-u) os tuá 2227.2662) 404) 2 ; (orzsz ho (urigos td Since = 4OfE)s, We mera ia tiirto; a , (so Pu) = DZ + zera tdo Cedz seem Rg LET Eq TIME For WATER TO RENA For? A ATAS ao Cy ta)-3 Ém = 1605-322 é, Za O. s98p 5 For wazem vo Eepem E: Ez 2 099 S err Or MobDO mo fUNIF orar aa 700) DEL) ba (26.444 Novas) = 36,634 = 36.4 FE « SE NOVELGS ; xr deent ur-zesanê Been THAT cos Ze= sm Amp wire ya Ausarts (1 2swtnt) ty Bu, Simee, “= sm 123 OR smirt = %s EQ) Peer 3] I— = Ho sir ss Praga GEP, -« 33 vPWaRO memo |O (Posre Sense 1) Dobun vago Aga Tra Possrive Simoes +) emp rsss - Pb Peoerem HC! a= vas = -& + A a?) lg bela) = h n=2gd (2:22) A as Va = Carchr GR) [LO PRRRFRANAREEANIIEREOACLRIEALS ASAS RLENEARUAA PESAREANEIREAEERRARRAS: | CRIMAISREEARRIAS CREAM ARRAES SIR a Ea PROBLEM 12.C1 ajoe PROBLEM 11.01 . aa vs BASTE VERSION «jr FORTRAM VERSION - pree ce ermuteçe : e Definition of Symbols c Definition of Synbols c VO = initial velucity of rocket (m/s) c vo = initial yetac. -» of rocket (o/a! H = narinia height reached by rocket (mr c H = maximus height reached by rocket (n) À = Geceleration of rocket tn/8) e A = aeceleration ef rocket (nei) e Data REAL 4 40 Ka 0808 e Data 6 = e.6100m . En p.s005 6=9.8 PRINT vo H wo e PRINT" tuts) tm em cpRINT MRNE Co ttamado > vo H vei MRITE (eo COBGBNIS + (m/s) cm ca FORVO = TO 400 STEF ES WRITE des! C/s BRANDO e Hacioum height reached by rocket DO 19 1 = 6,400,25 H= CSPROLDEC Lea t90-2)/0) w=1 e velocsty of rocket as t hits ground c Haximia height reached by rockek VE = SQRIB/GÕA — EXPI=BEHMO 1) Ha 49.SAOSLOBC eita (UDEAE) 6) e PRINT USING" a savO E Weiocity of rocket as it hite around PRINT USING 0 Mad miau UE = SORTCA/KOLLO — ENPI-B MME) NEST e WRITE o28) vOsHyUF va w 28 FORMAT BKPS-l IBPT ERA? 18 CONTINUE qm/s3 em E 9 a.eo es as as so nsi9 a 252.86 o É Ba 128 ULI.96 eis o 125 ses: 1 tam anão asi 175 quan aé 199.3 : 119158 vaias eo nica 11.78 Ligas au 2es 178.46 19.0 acesa ds. 12 eso tása és s82/28 : e s85.79 gaia es 1878.96 12e.gi 7-$3.97 102.38 ae regia? 126.92 Bia 199.36 E 1858.78 1êg.e8 = asie sia mus.as t1u.7a 1275.44 nela as ergo.1s 131.22 ai ss ja 29 soe sara.as ts2.28 1579.98 124.81 ias Lu? 18s.se 1e53.78 189.49 980.12 raglga 2 132.20 36 00C0000000000000000000000000000000000000000000006 , LP eoerem Fhrw or Szpepro or Wehrer 5 mt É=o o | T 1 afio 42 do E enem E * TT gre gta sinto é - fgu* ReMCRANHEs AÍ Rs do!) fifa) » Z (EB cia dot +5 (a 12)-o cr deerrenta Mezom (DiiLorm) Us U casdo* VERTICRE POTRO (otros AECEERATES) 7 = UG sr bo! — 9 9 O z=B =À) coso)é = e. sf g +0:0? g =. SE nindo'» AE Ja 52 3 cá (9:42) £=05(-8 EEE ) S N, fr to, Pheqreee or hunzer Lemes A Wire velcery Vs AND Srs N Tht Eleoueas or Eur Ding N 87 TumE É, Wy ÀS MEME Of Vi INCRERSES WWE BAVES Esp JO Que cu CH, Ware EPuIMRES Glrua 67 4; dice - ATT X= Sm Qro GALÉ, mAze DTRMES tobil RT; Las E = é duo U<SIZM Wuarer ETwmis Poor Br; Foste 2— Fort GTO, WATER LTAKES Llovmi ATA, Fo FósÊ E=--VE os Cala: 2% use E) 70 cmo É mem use (PD) 30 Ermo Ne, Fropo o GETOSGÊ, Emo É= IA ve (2) ve cer gy, E=T mb) Crolêrata = 26/8 se (E) ve Eme E, rmem use Do FIND Xe; THEM BD 214% Be VT los Co luafa = Lp vsE (E) Te come É, Titio ese (D Demo 10 eresaas essi emmmaners | CS e er Bo PROBLEM 11.02 + | ce PROBLEM 11.02 x a a BASIC VERSION + Lo FORTRAM VERSIDA + “a va eres Re sRMASM Eat se seem | E: REMO NIE NI ASIA CTA NA LERALA MSM SAMA Latinas REM sa. e o XwY = coordinates: origin at the fest pf the firefignter | C Definition of Symbols 7” VD = initial velocity of the watar c À o: T > tive for mater to gove from the noztle to c XY = coorginates: origin at the feet of the fireman sa. where 1t strikes the ground or the building c VO e initial velocity of water 106 * c 7» time for water to aovo from the nozzla to c where 1t strikes the geaunt or the building 126 * y foras angle of 48 degress vit c B+C = coefficient in quadratic formula 136 + horizentatl therefore 1a + Vo Babe = Vileos 48 = vDxa,5 sem vo 1 VD sub VOssin de o 1=6 es — 1-2 176 rs 168 * 198 PRINT “Noter Strikes ground at à distance 4 from firefighter” MRITE (e ML/SBRBANICA “Water stei , ess * BRITE teccranro àrefigirter e Note = 1725 4,5 metera, mater mill ntrike buildirg. c COM TINurD Com tra S7 Potes tema LLC ConNTiseer Frosram Base Vizesso Continigts Fecigterra FORTRAM VERSO CONTINUE zes FOR VO = 4 TD dé DO IG M = 4,16 esa Bm SOR3I0/D: Ca -2.4/6 vD=N Bos T = &SWC-BASORID'O MUCH): Sm TAVD/E B = -SORTIa.Jav0/6 ese WHILE X <4.5 Er tis eo PRINT USINE "Vo = 88 0/5 “IVO Tm S.58-B+ERRTIDEE — 4.265) E PRINT USING "x = 88.888 ASIA * = TWaia, coa sora sig LE CI.LT.6.5) THEN Pa a VRITE Cu,86) VD, 288 fa let IFI>1 GOD 5 ea FORMAT E9ks Vo = PaF3.D afataOho! x = MoFbD 87) E PRINT nero eo ze PRINT "Witar atrikes tno wall at à distance y above ground"! E Note: x = 4.50 and y< é metara º BOTO MG e Ta ev ENDIF = Y = 1.2 + BAEEDAVDAT -.sage(TIE L E WILEY Cd ao 12141 E PRINT USING" Vo ké m/s 405 JE E.s7.1) BOTO Soo sa ORINTUGIHS = de déa WRITE to" (//4BKBANH? "Mater strikes the mail at”, 3a BOTO s18 + * distance y abeve ground” sea END WRITE tes t(/,AAM) “o ses. c Note nn 4Sa am yr da o DES MES cora asa as T=9.80 asa PRINT "Mater strikes the roof at a distance BD from B” e V= LB + O B6AOSVONT — O.5HGRITAER) Er Note: Height of roof ía ya & a; x < iB meters. se B =BOR(BHVO/6: E» 9,609081/5 e TE tao vos cao = 58 ox. x - fil QuE UPE IPDE ctmenha o nbavoa E) FORMAT CORO PO a ARS, mM SAPO! PD «so BD=x-4.5 BOTO ste “98 PRINT USING * Vo = Mk m/0 *9V9I e 309 PRINT USINE "BD = AMAS aº;8D Boto aa sa SOTO 18 ENDIF E e Co gn3+s “Me -3+ Ei Ena O Me TEMA ora ado TF (3.GT.L1 GUTO 450 ss PRINT MRITE (as! (Z7/BXSBAV) O Uator strikes tne roof at”, “Water strikes grouna at à aLetanca » from firafighter! . a distance BD from Rº ses + date: Water ques completely over Ouildino MRITE Gesrirrano 578 BD = SORIDAVD/O: Cm -2.4/6 c Note : Height of roof is ye bm. 1x (I2a ses T= SM-DASORIBPE ARCH): = TMMO/D 50 Dm cSURTIS.MAVOIG E PRINT USING" Vo = Mt m/s 140) S= abro 28 PRINT USING «yu 00.808 arsã Tm 0 5M-BASORTADeE — 4,40) s19 NET X= 6.5 + Der SF EXLT.12.> THEN RM E 24-45 Mater atrikes ground at a distance x from firerighter WRITE (+48) VD,BD Vo = 4m/s *= 1$8a “a FORMAT (Me Vo = sF3.6 m/s Mo BD = PoFhss! voa Sais xe 276 a GOTO &15 Voz 4 m/s += 27404 ELSE Goto 53 Slater strikes the mall at a distance y above ground ENDIF Vo = 7a/s y= 6.886 e c v- Ga/s v* 276 a se F=k+t vo e y= 689 a 1F (E.61.1) GOTO 578 vo = y= Dela WRITE (4,7 (//,BKBAU") "Water strikes ground at”, Vo ve Sta + * a distance x from firefighter” MRITE tea teZrANdoO Water atrikes the roof at a distance BD from E c Noto 1 Mater goes conplatsly over building vo = 12 8/6 BD= 4.138 a mo & = -EORI(INWOIG Vo = 19 0/5 Bb = 6.761 a € = BA 7» 0. 5eI-B+BRTIBRE — 4,60) Mater strikes ground at a distance x froa firefighter * = Tavares Vo = 16 m/s = 179% à MRITE (9,56) VOA Vo = 13 m/s y = 28.593 m SE FORMAT CS, Vo é EBD, M/S! = TaPR Dm) Vo = 16 m/s y- B3.07 e SB CONTINUE END Water strikes ground at a distante x fron firarignter vo o 4. ms +» toma ve = 5 mis “= Eb a vo = ma “= 37 Mater strikes the mall at distance y above ground vo + 7 ais * = Bs vo = 8. m/s “= Ema vo = 9. a/s * = 4a Vo = 18. m/s * = Sela Wo = 1 m/s + = Sim Mater strikes the roof at a distance BD fros E vo = 12, ais BD = 4.138 m vo = 13. m/s BN = 67% Water strikes ground at a Gistance x from firetigntar vo = 16. 0/5 * =17.96m vo = 15. m/s x = 26.88 m Vo = ió. m/s * = B32m ss Flrogenra Fi: Ensu Vesostar Projectile mith drag Constant ralates to gragr K = 2 7 v vr : as mts a "mes 771.35 5 mos st.as 19 91725 unas 15 9.85 2aB.35 28 19.5 127.35 es qi9.es 28.68 a vi9.05 196.48 35º q19.85 -S8s.65 “a ques 56.45 es o19.25 877.68 T= 08.00» NED= -s9.met Range = GE neters Tine at impact = 87.9) » Projectite with drag Constant related to drag: E = 7 T va vv s mis ts 2º mos ms 5 08% dee.34 18 eov.ê7 v35.81 15 esg.a 271.30 Bo sseso 128.49 es BB -Bp.8p zo gana -2t2,72 E BIS sm esta 587.86 os Buaze ABR és T=47.50s YE = -Seg.5s2 Range « 41328 sotera Tine at impact = 47.08 5 Projeckile nitn drag Constant related to drag: k = 7 7 va w s as ms 2 gas Ti 5 seg.m s7t.9%4 10 67.16 386.98 15 Tot e12.s 2» at? a E) 13.58 a Jo6.ze -BéB.1p as aaa 6,15 “8 ess 557.88 Tr a4S6s YNEG e 179.808 Range = 33466 meters Time at impact = 44,84 Projectite with dra Conatant related to rag: k = ? r we w s mis na D ge mas 5 Eeom sazize 10 251.56 235.45 15º 698.49 135.92 eo ag “11.46 as ago.46 17.2 3» 571.66 -B19,29 a Sauas aus. 05 “o 46.95 573.33 T=oLSOs YNEB= 13.669 Rargo a 27615 meters Tine at lepact = 61.48 4 8 as96 193 13789 15908 est EA 34770 1346 eos 2E-& x a usp E 12564 1796 eos oa Bose E E Pos 10E-6 x 2 usa E 12u8 16439 epaas Bessa e7se7 so7se vPOs = BOE-4, eua esses veos = Y a438 sgas 77nS sa asse seis E “sao mes seems asa,393 Fhossera MC 3 commp Prosena Rom: ferra Vezes Projectile uith drag Constant related to drag 1 K » & T vx w x Y - ms ais m a 2 9.25 8. 5. 5. ago asno. Basa, 1. mi9,as Pa, aos, 15 99.25 asg2s 19709. 798. 2. 9.25 eras 1aBAS. g987, Es. 9,05 965 apo. qem. 8 GI9.B5 9445 BOSTA. gêso, 3. MI9.25 asmos aos. 7e75- “5. 939.85 516,65 Sb77D. Saça. 450 OI9.25 MT7OS Giz6s. BI9B, T 00.995. NMES » 49,682 vPDS » 21,994 Range = 44BE. a Tine at impact = 47,9] 5, Projectile with drag Conatant related to drag : K =» EE-& T vx uy x Y 5 ais ns ” a e moas ras a 9. 5 S08.74 abas esta. quam, 1. So) assai 9009. assa. 15 OMbL BYLLDO 13664 7795. Bm GEBS4 IRIS 17h 83 es. 7.50 seg qegog, Bog, 3. G7.iS ceiaite Poma pela, 35 E59.Bb sra a1O4a. 4757. dB: GSI) -Sar.Bb anaas. asgo. 45. BUBLE3 ÓBD.As DOSTO. tune. T =47.50a. VNES = -spg.529 vos = 42,097 Range = 41329, a. Tine at impact = 47.98 6. erojactila with drag Constant related to day : K = 108-6 T va w x Y s ara aís n a e migas ras a. .. 5. eme sm auto. qast. 19. 87.16 386.99 esa. sms, mo 99% Bus ion. 228. Ba. MI? 95.08 L6b3g. 7082. Bs. 788.06 3,98 PASS. 270. Je. Josze aéBiP BaMhI. s75D. 55. 67H43 culhlS p74S7, 890. a8. 4B99 557.35 9754 cume. Tess, VHES = et9.897 vos = 163.289 Range = 32468. a. Tine at impact = 46.84 8. Projectile with drag Constant related to drag : K = Z0E-6 T ur w x r s aís as a a 9 om rms o 5. S. cam Sage ugo%, 3055, 19. mt.50 335.65 ses. só, 18. a9sat ISSO! 11H86. bhúbr BB. 497 ciiio6 IUBU6. TR69. ES. éMM4s 167.83 18903. asup. J S/l.ié -Bi325 BIB. 534. 35 BMth -MQ.05 eaIto. sa. vB. a96.95 333 eMBo, Ba Temos. vNES = 19.471 veos = ee3.a01 Range = E74]5, e. Tine at pat = 41,48 5. 41 Eecguem Wc aq td ô 2 alo? mx? q As ají+a, A E — 2 é? > p soco FE fes)t= afras ” A =Açtth a Vas) - a; <q Ave a, dé < For comam d Dorme (My TemvaL dt: - 4 2. Ax= VAL +7adÉ = = VAL + Elavjat ax = (rrgav)at <] -u + apa pt A <a 10» errenas Ereem 8 PROBLEM 11.04 [e PRIBLEM 11.C4 . ag e BASIC VERSION +| 6 FORTRON VERSION + “o cassa sentantass | O asscassrentao esses ese see e Definition of Syabols “o Definition of Symbols e % e ve velocity tfe/s) o» wa velocity trtzas c X = distance travaled by train (ft) > X + distanca travoles by train tft) e AA] = components of acceleration (ft/n452) 106 ANsAT = componente of accaleration (FtfarE) e Ta time since brakes var first applied (seconds) no X = tine since braes mero first applied (amconds) c DELT = tine interval aê uhich brakes are raset [1 second) 126 * DELT = time interval at which brakes are reset (4 second) e DT - final time interval until V becomes zero (seconds) ap + DT = final 2106 Anterval until V becones zero (seconds) | C ag * E initial velocity: 4 mi/h = de ste. 186 V = 685 PInitiul velocity: 46 aih» 8 Fte” v= 68. 149 DELT = 1: “Brake setting interval in seconds! c Brake sekting interval 170 » DELT = 1. 185 *Caleulate U and X at ore-gecond intervais until c sa - ng veineity becomes negativr 87.6.) THEN Boo WHILE V > & = VerE/ugaa. 21 A = Verao > -SORTCR. Sine - Axo) ses ar = -SoR(Z 502 = Amro) = ATSDELT 2a De = tATMDELT a tv + DUB. MADELT E DX É (44 DVJBIADELT =veD esa Va v+Dy K= 34 DX asa Xe xe Ta T+ DEI ao T=7+ DEI ELSE 28 vem soro 24 ema» Eni E “Asmune uniforaly aceelerated notion during ons-secord esta 18 310 "intarval uhen velocity b9coons zero. e 2 TaT- DEDO X=X-0R Va pr c brsune uniforaiy secelarated notion during the one-second 388 DT» V/MRSLA £ Anfervai Man VHOCIty Dacoda dera. do Ox = (rar 20 Te T- BAT 3a Tetebl Ka xD xa x>0x E v=v- De 378 PRINT "Total tame required to stop trains” DT = VIAS!) Das PRINT UBING * T = 84.44 seconds; TiPRINT DX = (vz. tupT É 4 PRINT “Total distance traveled by the train: TeT+BT noo PRINT USING * 1 OMG.MO Feet1z x= 45X e WRITE (6,38) T am à FORMAT (/sSks "Total tino raquiros ta StOp Erain 1ºs/420X6!T = MF Total time required to stop train: 45.85 seconda?) Tn 38.39 amcorea SRITE te,48) X “8 FORNAT (7,8X, "Total distance traveled by the train 25/520,X = Total distance traveled by the tram +P7.ês! fonte) X = [780.55 fast em Total tine required to stop train T = 29.39 seconds Total castanco traveled by the train = 788.55 feet +42 DO0000000000000900000000000000000000000090000000% [2.1 =32,09(1+ 0,0053 sin' de : [2.4] Mass of satellite 15 independent ot quavby: ê ( soa) Mb me S00 Rg = 32.09 fys* vy=26. X 102 Rem dh 5 g = Is sete” = (26.1 x10%m/hX is )= Z2Sx rim fo 22.26 Rfs L= mar= (500 RX 225 x 1025) Maçã: At al latifude m= od b L=3,63x0 “ego aé A 00 db cy ibe Cds Mm = CSSH bit ATE STE E Weight: Wo mg PO W=(bIssy bs/eifaz00m6h) = H9R/L A Pedi tl (050% JAP EA) = 500016 q] O Ny É ba: Wefoleserjersrampi= son al 2.2) 6) Hego wWomg = CANTE me) | W= 824 N e “o Mac * Sose os en modo: mo = 2h Dar? TZ frmg A=(sta "Wjel Drive sisp=o: E pl) = pd + 2homa H+k ma MW = Ma LR = pa EA = Pg =080(88)) [12.3 ja do 10 seole irdoeçier És | fre foue Fexertod | (b) Front -Whvel Drive or bhe sprink nc prq fo te w Fi ma cr to 4 0- bg frase: E a — LN DE z S : = So w) = F F =(IDRAnBImf) = 1962 N Fr ? ora = cuz Free Bed N-04W N=06W - as dé Fa g624N Tere Eedy: Boy Tom Lo Est . ma - 06 Rge oslosoyaB) = ma = AN mg a. poa ima (isto =1mAN-(58 Kim) . Ê os a = 308m- Bm! vie zen (Km con)= 565) mis? a =82mjs* + 4 W=l3IIMh =8Sbmlh E Lero Lente É) Reoc Wheel Drive Assu. unegue ) fever arms w F = ma ob op cos DEM=o: — me MD4W)= ma | =] bBh-ch=o0 o F a OEbW ora E MN - É É Epoca =06M > =orig- autos) Elevatr af rest = 3.139 m/s* ma NÉ, - W=0 Frhe AMA MM a= 1848 m/s” (a8t8 miss om )= 76 EE V=3069 mk =H05 Am/h «ad Substituto into CI): algra)= E (Saga) mt) O) Tomparme Edna (3), we havem =. Thus fhe Same [pass is oued fo balame bar. Mass indicated is still 15 figa am oh Substitae indo (1) | TS POR ( 16 be Om) = 171 ih mg = e (rtgs = m/s = 69.3 tm 6 amgt teia 5 2.6 |60) Umtemdy aereteroted motion to arte : derated m-€(i5tg) (2) “p= 198 fenloz 3 ms; V=0 va lar ULelera mia E Cio tigla v'ewtsrax: “5. o)! raa(rs) =Ha o a -- 600 mist a z G =6.00.D/8* e mt (tg)g - V=wtrat: 0=% -600t E=500s «df * Ficuf= ma Elisio - (sta lá, “ Eçre E = milgra) “E clstalgra) & = avo 7 a 2 FouN fu=W ma . bOQNh t M Gg O RBImp M= 0,612 -« H3
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