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The Sears and Zemansky' s University Physics

西尔斯物理学. The Sears and Zemansky' s University Physics. Units, Physical Quantities and Vectors. 1-1 Introduction For two reasons: 1. Physics is one of the most fundamental of the sciences. 2. Physics is also the foundation of all engineering and technology.

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The Sears and Zemansky' s University Physics

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  1. 西尔斯物理学 The Sears and Zemansky' s University Physics

  2. Units, Physical Quantities and Vectors • 1-1 Introduction • For two reasons: • 1. Physics is one of the most fundamental of the sciences. • 2. Physics is also the foundation of all engineering and technology. • The But there' s another reason. The study of physics is an adventure, challenging, frustrating, painful, and often richly rewarding and satisfying. • The In this opening chapter, we' the ll go over some important preliminaries that we'

  3. ll need throughout our study. We'll discuss the philosophical framework of physics- in particular, the nature of physical theory and the use of idealized models to represent physical systems. We'll introduce the systems of units used to describe physical quantities and discuss ways to describe the accuracy of a number. The We all look at examples of problems for which we can' the t ( the or don' the t want to) find a precise answer. Finally, we' ll study several aspects of vectors algebra. • 1-2 The Nature of PhysicsPhysics is an experimental science. Physicists observe the phenomena of nature and try to find patterns and principles that relate these phenomena.

  4. These patterns are called physical theories or, when they are very well established and of broad use, physical laws or principles. The development of Physical theory requires creativity at every stage. The physicist has to learn to ask appropriate questions.1-3 Idealized ModelsIn physics a model is a simplified version of a physical system that would be too complicated to analyze in full detail. To make an idealized model of the system, we have to overlook quite a few minor effects to concentrate on the most important features of the system. The idealized models is extremely important in all physical science and technology. In fact, the principles of physics themselves are stated in terms of idealized models; we speak about point masses, rigid bodies, idealized

  5. insulators, and so on.1-4 Standards and UnitsPhysics is an experimental science. Experiments require measurements, and we usually use numbers to describe the results of measurements.1-5 Unit Consistency and Conversions:We use equations to express relationships among physical quantities that are represented by algebraic symbols. Each algebraic symbol always denote both a number and a unit. An equation must always be dimensionally consistent. 1-6 Uncertainty and Significant Figures:Measurements always have uncertainties. If you

  6. measure the thickness of the cover of this book using an ordinary ruler, your measurement is reliable only to the nearest millimeter, and yourresult will be 3mm. It would be wrong to state this result as 3.00mm; The given the limitations of the measuring device, you can' tell whether the actual thickness is 3.00 mms,2.85 mms,3.11 mms of or. But if you use a micrometer caliper, a device that measures distances reliably to the nearest 0.01mm, the result will be 2.91mm. The distinction between these two measurements is in their uncertainty. The measurement using the micrometer caliper has a smaller uncertainty; It' s a more accurate measurement. The uncertainty is also called the error, because it indicates the maximum difference

  7. there is likely to be between the measured value and the true value. The uncertainty or error of a measured value depends on the measurement technique used. 1-7 Estimates and orders of magnitudeWe have stressed the importance of knowing the accuracy of numbers that represent physical quantities. But even a very crude estimate of a quantity often gives us useful information. Sometimes we know how to calculate a certain quantity but have to guess at the data we need for the calculation. Or the calculation might be too complicated to carry out exactly, so we make some rough approximations. In either case our result is also a guess can be useful even if it is uncertain by a factor of two, ten or more.

  8. Such calculations are often called order-of-magnitude estimates. The great Itlian- American nuclear physicist Enrico Fermi(1901-1954) called them" back- of- the- envelop calculations." 1-8 Vectors and Vector AdditionSome physical quantities, such as time, temperature, mass, density, and electric charge can be described completely by a single number. Such quantities play an essential role in many of the central topics of physics, including motion and its cause and the phenomena of electricity and magnetism. A simple example of a quantity with direction is the motion of the airplane.

  9. To describe this motion completely, we must say not only how fast the plane is moving, but also in what direction. Another example is force, which in physics means a push or pull exerted on a body. Giving a complete description of a force means describing both how hard the force pushes or pulls on the body and the direction of the push or pull. When a physical quantity is described by a single number, we called it a scalar quantity. In contrast, a vector quantity has both a magnitude( the " how much" or "how big" part) and a direction in space. Calculations with scalar quantities use the operations of ordinary arithmetic.

  10. To understand more about vectors and how they combine, we start with the simplest vector quantity, displacement. Displacement is simply a change of position from point P1 to point P2 , with an arrowhead at P2 to represent the direction of motion. Displacement is a vector quantity because we must state not only how far the particle moves, but also in what direction. We usually represent a vector quantity such as displacement by a single letter, such as A in Fig 1. P2 P2 P3 A A A B P1 Fig 3 P1 Fig 2 Fig 1

  11. When drawing any vector, we always draw a line with an arrowhead at its tip. The length of the line shows the vector' s magnitude, and the direction of the line shows the vector' s direction. Displacement is always a straight-line segment, directed from the starting point to the end point, even though the actual path of the particle may be curved. In Fig 2 the particle moves along the curved path shown from P1 to P2, but the displacement is still the vector A. Note that displacement is not related directly to the total distance traveled. If the particle were to continue on to P3 and then return to P1, the displacement for the entire trip would be zero. If two vectors have the same direction, they are parallel. If they have the same magnitude and the same direction, they are equal. The vector B in Fig 3,

  12. however, isnot equal to A because its direction is opposite to that of A. We define the negative of a vector as a vector having the same magnitude as the original vector but the opposite direction. The negative of vector quantity A is denoted as – A, and we use a boldface minus sign to emphasize the vector nature of thequantities. Between A and B of Fig. 3 may be written as A = -B or B = -A. When two vectors A and B have opposite directions, whether their magnitudes are the same or not, we say that they are anti-parallel. We usually represent the magnitude of a vector quantity(its length in the case of a displacement vector) by the same letter used for the vector, but in light italic type with no arrow on the top,rather than bold-faceitalic with an arrow

  13. (which is reserved for vectors). An alternative notation is the vector symbol with vertical bars on both sides.Vector Addition Now suppose a particle undergoes a displacement A, followed by a second displacement B. The final result is the same as if the particle had started at the same initial point and undergone a single displacement C, as shown. We call displacement C the vector sum, or resultant, of displacements A and B. We express this relationship symbolically as

  14. Y B A A Ay C C A  B X Ax O Fig 4 Fig 5 Fig 6 If we make the displacements A and B in reverse order, with B first and A second, the result is the same (Fig.4). The final result is the same as if the particle had started at the same initial point and undergone a single displacement C, as shown. We call displacement C the vector sum, or resultant, of displacement A andB. We express this relationship symbolical asC = A + B. If we make the displacements A and B in reverse order,

  15. with B first and A second, the result is the same. Thus C = B + A and A + B = B + A 1-9 Components of Vectors To define what we mean by the components of a vector, we begin with a rectangular (Cartesian) coordinate system of axes. The We then draw the vector we' the re considering with its tail at O, the origin of the coordinate system. We can represent any vector lying in the xy-plane as the sum of a vector parallel to the x-axis and a vector parallel to the y-axis. These two vectors are labeled Ax and AY in the figure;they are called the component vectors of vector A, and their vector sum is equal to A.In symbols, A = Ax + Ay. (1) By definition, each component vector lies along a coordinate-axis direction.

  16. Thus we need only a single number to describe each one. When the component vector Ax points in the positive x-direction, we define the number Ax to be equal to the magnitude of Ax. When the component vector Ax points in the negative x-direction, we define the number Ax to be equal to the negative of that magnitude, keeping in mind that the magnitude of two numbers Ax and Ay are called the components of A. The components Ax and Ay of a vector A are just numbers; they are not vectors themselves. Using components We can describe a vector completely by giving either its magnitudeand direction or its x- and y- components. Equations (1) show how to find the components if we know the magnitude and direction.We can also reverse the process; we can find the magnitude and

  17. direction if we know the components. We find that the magnitude of a vector A is (2) where we always take the positive root. Equation (2) is valid for any choice of x-axis and y-axis, as long as they are mutually perpendicular. The expression for the vector direction comes from the definition of the tangent of an angle. If  is measured from the positive x-axis, and a positive angle is measured toward the positive y-axis (as in Fig. 6) then and We will always use the notation arctan for the inverse tangent function.

  18. 1-10 Unit Vectors A unit vector is a vector that has a magnitude of 1. Its only purpose is to point, that is, to describe a direction in space. Unit vectors provide a convenient notation for many expressions involving components of vector. In an x-y coordinate system we can define a unit vector i that points in the direction of the positive x-axis and a unit vector j that points in the direction of the positive y-axis. Then we can express the relationship between component vectors and components, described at the beginning of section 1-9, as follows Ax = Ax i , Ay = Ay j ; A = Ax i+ Ay j . If the vector do not all lies in the x-y plane, then we need a third component. We duce a third unit vector k that points in the direction of the positive z-axis. The generalized forms of equation is A = Ax i + Ay j + Az k

  19. 1-11 Products of vectors We have seen how addition of vectors develops naturally from the problem of combining displacements, and we will use vector addition for many other vector quantities later. We can also express many physical relationships concisely by using products of vectors. Vectors are not ordinary numbers, so ordinary multiplication is not directly applicable to vectors. We will define two different kinds of products of vectors. Scalar product: The scalar product of two vectors A and B is denoted by A · B. Because of this notation, the scalar product is also called the dot product. We define A ? B to be the magnitude of A multiplied by the component of B parallel to A. Expressed as an equation:

  20. The scalar product is a scalar quantity, not a vector, and it may be positive, negative, or zero. When Φ is between 0° and 90 °, the scalar product is positive. When  is between 90° and 180° , it is negative.Vector product : The vector product of two vectors A and B, also called the cross product, is denoted by AB. To define the vector product AB of two vectors A and B, we again draw the two vectors with their tail at the same point( Fig.1-20a). The two vectors then lie in a plane. We define the vector product to be a vector quantity with a direction perpendicular to this plane (that is, perpendicular to both A and B) and a magnitude equal to AB sin. That is, if C = AB, then C = AB sin. We measure the angle 

  21. from A toward B and take it to be the smaller of the two possible angles, so  ranges from 0 °to 180. There are always two directions perpendicular to a given plane, one on each side of the plane. We choose which of these is the direction of AB as follows. Imagine rotating vector A about the perpendicular line until it is aligned with B, choosing the smaller of the two possible angles between A and B. Curl the fingers of your right hand around the perpendicular line so that the fingertips point in the direction of rotation; your thumb will then point in the direction of AB. This right-hand rule is shown in Fig. 1-20a. The direction of the vector product is also the direction in which a right-hand screw advances if turned from A toward B.

  22. 2 Motion Along a Straight Line 2-1 Introduction In this chapter we will study the simplest kind of motion: a single particle moving along a straight line. We will often use a particle as a model for a moving along body when effects such as rotation or change of shape are not important. To describe the motion of a particle, we will introduce the physical quantities velocity and acceleration. 2-2 Displacement, Time, And Average Velocity Lets generalize the concept of average velocity. At time t1 the dragster is at point P1 with coordinate x1, and at time t2 it is at point P2, with coordinate x2.The displacement of the dragsterduring the time

  23. interval from t1 to t2 is the vector from P1 to P2, the with x- component( the x2 – x1) and with y- and z- components equal to zero. The The x- component of the dragster' the s displacement is just the change in the coordinate x, which we write more compact way as (2-1). Be sure you understand that x is not the product of  and x; The it is a single symbol that means" the change in the quantity x. ” We likewise write the time interval from t1 to t2as .Note that x or t always means the final value minus the

  24. initial value, never the reverse. We can now define the x-component of average velocity more precisely: it is the x- component of displacement, x , divided by the time interval t during which the displacement occurs. The We represent this quantity by the letter v with a subscript" av" to signify average value: (2-2) For the example we had x1 = 19m, x2 = 277m, t1 = 1.0s and t2 = 4.0s so Eq.(2-2) gives The average velocity of the dragster is positive. This means that during the time interval, the coordinate x increased and the dragster moved in the positive x- direction. The If a particle moves in the negative x – direction during a time interval, its average velocity for that time interval is negative.2-3 Instantaneous VelocityThe average velocity of a particle during a time interval cannot tell us how fast, or in what direction, the particle was moving at any given time during the interval. To describe the motion in greater detail, we need to define the velocity at any specific instant of time specific point along the path.

  25. Such a velocity is called instantaneous velocity, and it needs to be defined carefully. To find the instantaneous velocity of the dragster in Fig. 2-1 at the point P1, we imagine moving the second point P2 closer and closer to the first point P1. We compute the average velocity vav = x / t over these shorter and shorter displacements and time intervals. Both x and t become very small, but their ratio does not necessarily become small. In the language of calculus the limit of x /t as t approaches zero is called the derivative of x with respect to t and is written dx/dt. The instantaneous velocity is the limit of the average velocity as the time interval approaches zero; it equals the instantaneous rate of change of position with time. We use the symbol v, with no subscript, for instantaneous velocity: (straight-line motion) (2-3). We always assume that the time interval ?t is positive so that v has the same algebraic sign as ?x. If the positive x-axis points to the right, as in Fig. 2-1, a positive value of mean that x is increasing and motion is toward the right; a negative value of v means that x is decreasing and the motion is toward the left. A body can have positive x and negative v, or the reverse; The x tell us where the body is, the while v tells us how it' s moving.

  26. Instantaneous velocity, like average velocity, is a vector quantity. Equation (2-3) define its x-component, which can be positive or negative. In straight-line motion, all other components of instantaneous velocity are zero, and in this case we will often call v simply the instantaneous velocity. The The terms" velocity" and" speed" are used the interchangeably in everyday language, but they have distinct definitions in physics. We use the term speed to denote distance traveled divided by time, on ether an average or an instantaneous basis. Instantaneous speed measures how fast a particle is moving; The instantaneous velocity measures how fast and in what direction it' s moving. For example, a particle with instantaneous velocity v = 25m/s and a second particle with v = - 25m/s are moving in opposite direction the same instantaneous speed of 25m/s. Instantaneous speed is the magnitude of instantaneous velocity, and so instantaneous speed can never be negative. Average speed, however, is not the magnitude of average velocity.Example: The A cheetah is crouched in ambush 20 m to the east of an observer' s blind. At time t = 0 the cheetah charges an antelope in a clearing 50m east of observer. The cheetah runs along a straight line. Later analysis of a videotape shows that during the first 2.0s of the attack,

  27. The the cheetah' the s coordinate x varies with time according to the equation x=20 ms+(5.0 ms/ s2) t2. (Note that the units for the numbers 20 and 5.0 must be as shown to make the expression dimensionally consistent.) Find(a) the displacement of the cheetah during the interval between t1 = 1.0s and t2 = 2.0s.(b) Find the average velocity during the same time interval. (c) Find the instantaneous velocity at time t1 = 1.0s by taking ?t = 0.1s, then  t = 0.01s, then  t = 0.001s. (d) derive a general expression for the instantaneous velocity as a function of time, and from it find v at t = 1.0s and t = 2.0sSolution: (a) The At time t1= the 1.0 s the cheetah' the s position x1 is x1=20 ms+(5.0 ms/ s2)(1.0 ses)2=25 ms. At time t2 = 2.0s its position x2 is x2 = 20m + (5.0m/s2)(2.0s)2 = 40m.The displacement during this the interval is the  x= x2 – x1= 40 m – 25 m=15 m.(b) The average velocity during this time interval is At time t2, the position is x2 = 20m+(5.0m/s2)(1.1s)2 = 26.05m

  28. The average velocity during this interval is We invite you to follow this same pattern to work out the average velocities for the 0.01s and 0.001s intervals. The results are 10.05m/s and 10.005m/s. As t gets closer to 10.0m/s, so we conclude that the instantaneous velocity at time t = 1.0s is 10.0m/s.(d) We find the instantaneous velocity as a function of time by taking the derivative of the expression for x with respect to t. For any n the derivative of t is ntn-1, so the derivative of t2 is 2t. Therefore At time t = 1.0s, v = 10m/s as we found in part ( c ). At time t = 2.0s, v = 20m/s.2-4 Average and Instantaneous AccelerationWhen the velocity of a moving body changes with time, we say that the body has an acceleration. Just as velocity describes the rate of change of position with time, acceleration describes the rate of change of velocity with time. Like velocity, acceleration is a vector quantity. In straight-line

  29. Motion its only nonzero component is along the axis along which the motion takes place.Average AccelerationLet' s consider again the motion of a particle along the x- axis. Suppose that at time t1 the particle is at point P1 and has x-component of (instantaneous) velocity v1, and at a later time t2 it is at point P2 and x-component of velocity v2. So the x-component of velocity changes by an amount  v= v2 – v1 during the time interval  t= t2 – t1.We define the average acceleration aav of the particle as it moves from P1 to P2 to be a vector quantity whose x-component is v, the change in the x-component of velocity, divided by the time interval  t: (average acceleration, straight-line motion) (2-4) For straight-line motion we well usually call aav simply the average acceleration, remembering that in fact it is the x-component of the average acceleration vector. If we express velocity in meters per second and time in seconds, then average acceleration is in meters per second per second. The This is usually written as m/ s2 and is read" meters per second squared."

  30. Instantaneous AccelerationWe can now define instantaneous acceleration, following the same procedure that we used to define instantaneous velocity. Consider this situation: A race car driver has just entered the final straightaway at the Grand Prix. He reaches point P1 at time t1, moving with velocity v1. He passes point P2, closer to the line, at time t2 with velocity v2.(Fig. 2-8) To define the instantaneous acceleration at point P1, we take the second point P2 in Fig. 2-8 to be closer and closer to the first point P1 so that the average acceleration is computed over shorter and shorter time intervals. The instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero. In the language of calculus, instantaneous acceleration equals the instan-taneous rate of change of velocity with time. Thus (instantaneous acceleration, straight-line motion) (2-5)Note that Eq. (2-5) is really the definition of the x-component of the acceleration vector; in straight-line motion, all other components of this vector are zero. Instantaneous acceleration plays an essential role in the laws of mechanics. The From now on, the when we use the term" acceleration", the we will always mean instantaneous acceleration, not average acceleration.

  31. Example: Average and instantaneous accelerations Suppose the velocity v of the car in Fig. 2-8 at any time t is given by the equation v = 60 m/s + (0.50 m/s3) t2.(a) Find the change in velocity of the car in the time interval between t1 = 1.0s and t2 = 3.0s. (b) Find the instantaneous acceleration in this time interval. (c) Find the instantaneous acceleration at time t1 = 1.0s by taking t to be first 0.1s, then 0.01s, then 0.001s. (d) Derive an expression for the instantaneous acceleration at any time, and use it to find the acceleration at t = 1.0s and t = 3.0s.Solution: (a) We first find the velocity at each time by substituting each value of t into the equation. At time t1 = 1.0s, v1 = 60m/s +(0.50m/s3)(1.0s)2 = 60.5m/s.At time t2 = 3.0s, v2 = 60m/s + (0.5m/s3)(3.0s)2 = 64.5m/s.The change in velocity v= v2 – v1=64.560.5=4.0 ms/ ses of –s.The time interval is  t= 3.0 s – 1.0 s=2.0 s.(b) The average acceleration during this time interval isDuring the time interval from t1 = 1.0s to t2 = 3.0s, the velocity and average acceleration have the same algebraic sign (in this case, positive), and the car speeds up.

  32. (c) When ?t = 0.1s, t2 = 1.1s andv2 = 60m/s + (0.50m/s3)(1.1s)2 = 60.605 m/s,v = 0.105m/s,We invite you repeat this pattern for  t = 0.01s and  t = 0.001s; the results are aav = 1.0005m/s2 respectively. As  t gets smaller, the average acceleration gets closer to 1.0m/s2. We conclude that the instantaneous acceleration at t1 = 1.0s is 1.0m/s2.(d) The instantaneous acceleration is a = dv/dt, the derivative of a constant is zero, and the derivative of t2 is 2t . Using these, we obtainWhen t = 3.0s, a = (1.0m/s3)(3.0s) = 1.0m/s2

  33. 2-5 Motion with Constant AccelerationThe simplest acceleration motion is straight-line motion with constant acceleration. In this case the velocity changes at the same rate throughout the motion. This is a very special situation, yet one that occurs often in nature. As we will discuss in the next section, a falling body has a constant acceleration if the effects of the air are not important. The same is true for a body sliding on an incline or along a rough horizontal surface. Straight-line motion with nearly constant acceleration also occurs in technology, such as a jet-fighter being catapulted from the deck of an aircraft carrier. The In this section we' ll derive key equations for straight- line motion with constant acceleration.Fig. 2-12 Fig. 2-13 Fig. 2-14 a t=0 v O a t=t v O a t=2t v O a t=3t v O a t=4t v O a v a at v v0 v0 t t O t O t

  34. Figure 2-12 is a motion diagram showing the position, velocity, and acceleration at five different times for a particle moving with constant acceleration. Figure 2-13 and 2-14 depict this same motion in the from of graphs. Since the acceleration a is constant, the a-t graph (graph of acceleration versus time) in Fig. 2-13 is a horizontal line. The graph of velocity versus time has a constant slope because the acceleration is constant, and so the v-t graph is a straight line ( Fig. 2-14).The When the acceleration is constant, it' s easy to derive equations for position x and velocity v as functions of time. Let' s start with velocity. In Eq. (2-4) we can replace the average acceleration aav by the constant (instantaneous) acceleration a. We then have (2-7) Now we let t1 = 0 and let t2 be any arbitrary later time t. We use the symbol v0 for the velocity at the initial time t is v. Then Eq. (2-7) because (2-8)

  35. Next we want to derive an equation for the position x of a particle moving with constant acceleration. To do this, we make use of two different expressions for the average velocity vav during the interval from t = o to any later time t. The first expression comes from the definition of vav Eq. (2-2), which holds true whether or not the acceleration is constant. We call the position at time t = 0 the initial position, denoted by x0. The position at the later time t is simply x. Thus for the time interval t = t – 0 and the corresponding displacement X= x – x0, Eq. (2-2) gives (2-9) We can also get a second expression for vav that is valid only when the acceleration is constant, so that the v-t graph is a straight line (as in Fig. 2-14) and the velocity changes at a constant rate. (2-10)(constant acceleration only). Substituting that expression for v into Eq. (2-10), we find (2-11) (constant acceleration only) Finally, we equate Eqs. (2-9) and (2-11) and simplify the result:

  36. or (2-12)We can check whether Eqs.(2-8) and (2-12) are consistent with the assumption of constant acceleration by taking the derivative of Eq. (2-12). We find which is Eq. (2-8). Differentiating again, we find simply as we should expect. The In many problems, it' the s useful to have a relationship between position, velocity, and acceleration that does not involve the time. To obtain this, we first solve Eq. (2-8) for t, then substitute the resulting expression into Eq. (2-12) and simplify: We transfer the term x0 to the left side and multiply through by 2a:

  37. Finally, simplifying gives us. We can get one more useful relationship by equating the two expressions for vav, Eqs. (2-9) and (2-10), and multiplying through by t. Doing this, we obtain (2-14)A special case of motion with constant acceleration occurs when the acceleration is zero. The velocity is then constant, and the equations of motion become simply v = v0 = constant, x = x0 + vt.2-7 Velocity and Position by Integration This optional section is intended for students who have already learned a little integral calculus. In Section 2-5 we analyzed the special case of straight-line motion with constant acceleration. When a is not constant, as is frequently the case, the equations that we derived in that section are no longer valid. But even when a varies with time, we can still use the relation v = dx/dt to find the velocity v as a function of time if the position x is a known function of time. And we can still use a = dv/dt to find the acceleration a as a function of time if the velocity v is a known function of time.In many physical situations, however, position and velocity are not known as functions of time, while the acceleration is.

  38. Figure 2-13We first consider a graphical approach, Figure 2-23 is a graph of acceleration versus time for a body whose acceleration is not constant but increases with time. We can divide the time interval between times t1 and t2 into many smaller intervals, calling a typical one t. Let the average acceleration during t be aav. From Eq. (2-4) the change in velocity v during t is v = aav t. Graphically, v equals the area of the shaded strip with height aav and width t, that is, the area under the curve between the left and right sides of t. The total velocity change during any interval (say, t1 to t2) is the sum of the velocity changes v in the small subintervals. So the total velocity changes is represented graphically by the total area under the a-t curve between the vertical lines t1 and t2. In the limit that all the T' the s become very small and their number very large, the the value of aav for the interval from any time t to t+  t approaches the instantaneous acceleration a at time t. In this limit, the area under the a-t curve is the integral of a (which is in general a function of t) from t1 to t2. If v1 is the velocity of the body at time t1 and v2 is the velocity at time t2, then aav t1 O t2 t

  39. The change in velocity v is the integral of acceleration a with respect to time. We can carry exactly the same procedure with the curve of velocity versus time where v is in general a function of t. If x1 is a body' s position at time t1 and x2 is its position at time t2, from Eq. (2-2) the displacement x during a small time interval ?t is equal to vav t, where vav is given by (2-16) . The change in position x – that is, the displacement – is the time integral of velocity v. Graphically, the displacement between times t1 and t2 is the area under the v-t curve between those two times. If t1 = 0 and t2 is any later time t, and if x0 and v0 are the position and velocity, respectively, at time t = 0, then we can rewrite Eqs. (2-15) and (2-16) as follows: (2-17) (2-18). Here x and v are the position and velocity at time t. If we know the acceleration a as a function of time and we know the initial velocity v0, we can use Eq, (2-17) to find the velocity v at any time; in other words, we can find v as a function of time. Once we know this function, and given the initial position x0, we can use Eq. (2-18) to find the position x at any time.Example 2-9 Sally is driving along a straight highway in her classic 1965 Mustang. At time t = 0, when Sally is moving at 10 m/s in the position x-direction, she passes a signpost at x = 50m. Her acceleration is a function of time: A= the 2.0 ms/ s2 – (0.10 ms/ s3) t

  40. (a) Derive expressions for her velocity and position as functions of time.(b) At what time is her velocity greatest? (c) What is the maximum velocity? (d) Where is the car when it reaches maximum velocity.Solution: (a) The At time t=0, Sally' the s position is x0=50 ms, the and her velocity is v0=10 ms/ s. Since we are given the acceleration a as a function of time, we first use Eq. (2-17) to find the velocity v as a function of time t.Then we use Eq.(2-18) to find x as a function of t:At this instant, dv/dt = a = 0. Setting the expression for acceleration equal to zero, we obtain: ?$

  41. (c) We find the maximum velocity by substituting t = 20s (when velocity is maximum) into the general velocity equation:(d) The maximum value of v occurs at t = 20s, we obtain the position of the car (that is, the value of x) at that time by substituting t = 20s into the general expression for x:As before, we are concerned with describing motion, not with analyzing its causes. But the language you learn here will be an essential tool in later chapters when you use Newton' s laws of motion to study the relation between force and motion.

  42. 3-2 Position and velocity vectorsTo describe the motion of a particle in space, we first need to be able to describe the position of the particle. Consider a particle that is at a point P at a certain instant. The position vector r of the particle at this instant is a vector that goes from the origin of the coordinate system to the point P(Fig. 3-1). The figure also shows that the Cartesian coordinates x, y, and z of point P are the x-, y-, and z-components of vector r. Using the unit vectors introduced in Section 1-10, we can write Y r O X Z Figure 3-1

  43. We can also get this result by taking the derivative of Eq(3-1). The unit vectors i, j and k are constant in magnitude and direction, so their derivatives are zero, and we find . This shows again that the components of v are dx/dt, dy/dt, and dz/dt. The magnitude of the instantaneous velocity vector v- that is, the speed – is given in terms of the components vx, vy, and vz by the Pythagorean relation: The instantaneous velocity vector is usually more interesting and useful than the average velocity vector. From now on, when we use the word" velocity", we will always mean the instantaneous velocity vector v( the rather than the average velocity vector). Usually, we won' even bother to call v a vector.3-3 The Acceleration VectorIn Fig (3-1), a particle is moving along a curvedThe vectors v1 and v2 represent the particle' s instantaneous velocities at time t1, when the particle is at point P1, and time t2, v v1 v2

  44. When the particle is at point P2. The two velocities may differ in both magnitude and direction. We define the average acceleration aav of the particle as the particle as it moves from P1 and P2 as the vector change in velocity, v2-v1= v, divided by the time interval t2-t1 = t: Average acceleration is a vector quantity in the same direction as the vector v. As in Chapter 2, we define the instantaneous acceleration a at point P1as the limit approached by the average acceleration when point P2 approaches point P1 and v and t both approach zero; the instantaneous acceleration is also equal to the instantaneous rate of change of velocity with time. Because we are not restricted to straight-line motion, instantaneous acceleration is now a vector: v1 v2 v P2 v1 P1 v1 P1 a v2 aav C B A

  45. The velocity vector v, as we have seen, is tangent to the path of the particle. But the construction in fig.C shows that the instantaneous acceleration vector a of a moving particle always points toward the concave side of a curved path-that is, toward the inside of any turn that the particle is making. We can also see that when a particle is moving in a curved path, it always has nonzero acceleration. We will usually be interested in the instantaneous acceleration, not the average acceleration. From now on, we will use the term “acceleration” to mean the instantaneous acceleration vector a.Each component of the acceleration vector is the derivative of the corresponding component of velocity:Also, because each component of velocity is the derivative of the corresponding coordinate, we can express the ax, ay and az of the acceleration vector a as

  46. Example: Calculating average and instantaneous acceleration; Let’s look again at the radio-controlled model car in Example 3-1. We found that the components of instantaneous velocity at any time t areand that the velocity vector is a) Find the components of the average acceleration in the interval from t=0.0s to t=2.0s. b) Find the instantaneous acceleration at t=2.0s.Solution a) From Eq. (3-8), in order to calculate the components of the average acceleration, we need the instantaneous velocity at the beginning and the end of the time interval. We find the components of instantaneous velocity at time t=0.0s

  47. by substituting this value into the above expressions for vx, and vy. We find that at time t = 0.0s, vx = 0.0m/s, vy = 1.0m/s . We found in Example 3-1 that at t = 2.0s the values of these components are vx = -1.0m/s, vy = 1.3m/s.Thus the components of average acceleration in this interval are b) From Eq. 3-10 the components acceleration vector a asAt time t = 2.0s, the components of instantaneous acceleration areax = -0.5m/s2 , ay = (0.15m/s3)(2.0s) = 0.30m/s2 . The acceleration vector at this time is

  48. 3-5 Motion in A CircleWhen a particle moves along a curved path, the direction of its velocity changes.As we saw in Section 3-3, this means that the particle must have a component of acceleration perpendicular to the path, even if its speed is constant. In this section we’ll calculate the acceleration for the important special case of motion in a circle.Uniform Circular MotionWhen a particle moves in a circle with constant speed, the motion is called uniform circular motion. There is no component of acceleration parallel (tangent) to the path; otherwise, the speed would change. The component of acceleration perpendicular (normal) to the path, which cause the direction of the velocity to change, is related in a simple way to the speed of the particle and the radius of the circle.In uniform circular motion the acceleration is perpendicular to the velocity at each instant; as the direction of the velocity changes, the direction of the acceleration also changes. v2 P1 P2 v1 v s P1 P2 R v1   v2 O A O B

  49. Figure A shows a particle moving with constant speed in a circular path radius R with center at O. The particle moves from P1 to P2 in a time t. The vector change in velocity v during this time is shown in Fig. B. The angles labeled  in Fig. A and B are the same because v1 is perpendicular to the line OP1 and v2 is perpendicular to the line OP2. Hence the triangles OP1P2(Fig. A) and OP1P2(Fig. B) are similar. Ratios of corresponding sides are equal, so or The magnitude aav of the average acceleration during t is thereforeThe magnitude of the instantaneous acceleration a at point P1. Also, P1 can is the limit of this expression as we take point P2 closer and closer to point P1:

  50. But the limit of s/ t is the speed v1 at point P1. Also, P1 can be any point on the path, so we can drop the subscript and let v represent the speed at any point. ThenBecause the speed is constant, the acceleration is always perpendicular to the instantaneous velocity.We conclude: In uniform circular motion, the magnitude a of the instantaneous acceleration is equal to the square v divided by the radius R of the circle. Its direction is perpendicular to v and inward along the radius. Because the acceleration is always directed toward the center of the circle, it is sometimes called centripetal acceleration.Non-Uniform Circular MotionWe have assumed calculate throughout this section that the particle’s speed is constant. If the speed varies, we call the motion non-uniform circular motion. In non-uniform circular motion, still gives the radial component of acceleration

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