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KINEMATICS USINGVECTOR ANALYSISStandard CompetencyAnalyzes the nature phenomenon and its regularity within thescope of particle’s MechanicsBase CompetencyAnalyzes linier, circular and parabolic motions using vectoranalysisLearning ObjectivesAfter completing this chapter, all students should be able to:1 Analyzes the quantity of displacement, velocity and acceleration on linier motion using vector analysis2 Applies the vector analysis of position, displacement, velocity and acceleration vectors on linier motion equations3 Calculates the velocity from position’s function4 Calculates the acceleration from velocity’s function5 Determines the position from the function of velocity and accelerationReferences[1] John D Cutnell dan Kenneth W. Johnson (2002). Physics 5th Ed with Compliments. John Wiley and Sons, Inc.[2] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MA untuk SMA/MA Kelas XI. CV Yrama Widya
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All object’s motions are described in terms of- position (x) and displacement (Δr),- velocity (v) and- acceleration (a)Where all the descriptions are considered occur in Cartesian orxy-coordinate. The discussion is categorize as follows- 1 Dimensional motion or LINIER MOTION: object moves on either x-axis or y-axis- 2 Dimensional motion or PLANE MOTION: object moves on xy-axis- 3 Dimensional motion or SPACE MOTION: object moves on xyz-axisIn describing motion, Physics is using Vector Analysis and somebasic Calculus (differential and integral’s concept)Motion Description: Position and DisplacementA particle position within Cartesian coordinate describes as r r = xi + y j or r = xi + yj rIn general term it writing as r = (± x i ± y j) y r vector position: vector that describes a r position of a particle in Cartesian coord ry i,j unit vector: vector that describes r unit scale of an axis r rx , ry vector component: projection of a j vector position on x and y-axis r x r = x i + y j description a vector position i rx on a plane (xy-plane)
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Vector position is a vector that describe the position of aparticle. It has vector components in x and y-direction andwritten as r = r x + ryThe value of vector position is written as r = x2 + y2This value is a magnitude of vector position r based on vectorcomponents Please note, the typing of vector position in bold, r is similar to the symbol of r r in vector notation, r .When a symbol is typed in italic format, such r, it indicates thatthe symbol is a scalar quantity. Otherwise, when it typed inbold format, such r, it indicates that the symbol is a vectorquantity.Vector Displacement Consider a particle that moves arbitrary randomly in arbitrary path on xy- y plane. The term of vector r r r Δr = r2 − r1 displacement is the difference of r vector positions where in unit r1 vector is r r2 Δr vector displacement is vectorj difference of r2 and r1 r r r x Δr = r2 − r1 i = ( x2 i + y2 j) − ( x1 i + y1 j) = ( x2 − x1 ) i + (y2 − y1 ) j = Δx i + Δy j
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Re-writing the equation Δr = Δxi + Δyjwhere Δr = rfinal − rinitial or Δr = r2 − r1 Δx = xfinal − xinitial or Δx = x2 − x1 Δy = yfinal − yinitial or Δr = y2 − y1The value of vector displacement is similar to the value ofvector position, i.e., Δr = Δx 2 + Δy 2Motion Description: VelocityVelocity is another motion description which indicate how fastor how slow is the object moves. It is very common in ourdaily life term, where usually describes how fast an objectmoves. v If an object moves in an arbitrary plane (xy- coordinate) then the direction is given as vy vy tan θ = vx vxVelocity VectorThe position vector r is a vector that goes from the origin ofthe coordinate system to a given point in the system. Thechange in position Δr (delta-r) is the difference between thestart point (r1) to end point (r2).We define the average velocity (vav) as: vav = (r2 - r1) / (t2 - t1) = Δr / Δt or r r Δr v = Δt
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Taking the limit as Δt approaches 0, we achieve the instanta-neous velocity v. In calculus terms, this is the derivativeof r with respect to t, or dr/dt which is written as r r dr v = dt r drIf r = at n then = a n t n −1 dtAs the difference in time reduces, the start and end pointsmove closer together. Since the direction of r is the samedirection as v, it becomes clear that the instantaneousvelocity vector at every point along the path is tangentto the path.Description of particle’s velocity onn Cartesian coordinate isdescribed using calculus r r r r r Δr r2 − r1 r dr d( x i + y j) v = = v = = Δt t2 − t1 dt dt dx dy = i+ j dt dt = v x i + vy jVelocity ComponentsThe useful trait of vector quantities is that they can be brokenup into their component vectors. The derivative of a vector isthe sum of its component derivatives, therefore: dx vx = dt dy vy = dtThe magnitude of the velocity vector is given by thePythagorean Theorem in the form: 2 2 v = v = vx + vy
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The direction of v is oriented theta degrees counter-clockwisefrom the x-component, and can be calculated from thefollowing equation: vy tan θ = vxMotion Description: AccelerationVarious changes in a particle’s motion may produce anacceleration. When an acceleration is build, it bringsconsequences, i.e.,* The magnitude of the velocity vector may change* The direction of the velocity vector may change (even if the magnitude remains constant)* Both may change simultaneously v=0 for an i t t
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Acceleration VectorAcceleration is the change of velocity over a given period oftime. Similar to the analysis above, we find that its Δv/Δt. Thelimit of this as Δt approaches 0 yields the derivative of v withrespect to t.In terms of components, the acceleration vector can be writtenas: dv x d2x ax = or ax = dt dt 2 dv y d 2y ay = ay = dt dt 2The magnitude and angle of the net acceleration vector arecalculated with components in a fashion similar to those forvelocity.Description of particle’s acceleration on Cartesian coordinateis described using calculus r r r r r Δv v2 − v1 r dv d(v x i + v y j) a= = a= = Δt t2 − t1 dt dt dv x dv y = i+ j dt dt = ax i + ay jDue to d ⎛ dx ⎞ d 2 x d ⎛ dy ⎞ d 2y ax = ⎜ ⎟= ay = ⎜ ⎟= dt ⎝ dt ⎠ dt 2 dt ⎝ dt ⎠ dt 2Then r d2x d 2y a= i+ j dt 2 dt 2
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If x and y component of a is perpendicular, then r 2 2 a= a = ax + ayIf their components made an angle of θ, then ay tan θ = axExample[1] Consider a particle in a Cartesian coordinate (xy- coordinate) where is initially positioned on P1 (4, −1) was then moved to P2 (8, 2) within 1 second. Finds (a) initial and final of vector position, (b) its vector displacement and its value (c) its average velocity and its valueKnown variables: initial position: P1 (4, −1) final position: P2 (8, 2) time elapsed: Δt = 1 sAsked: (a) r1 and r2 (b) Δr and Δr r (c) v and vAnswer (a) r1 = 4i − j (b) Δr = 4i + 3j r2 = 8i + 2j Δr = 42 + 32 = 5 units r r Δr (c) v = Δt = 4i + 3j v = 5 units
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Example[2] A particle moves on a circular track r = 2t + t3 with r in meter and t in second. Calculate the velocity of particle when (a) t = 0, (b) t = 2sKnown variables: r = 2t + t3 Δt = 0 s Δt = 2 sAsked: (a) vo (b) v2Answer r dr 2t + t 3 v = = dt dt (a) vo = 2 + 3t2 = 2 m/s = 2 + 3t 2 (b) v2 = 2 + 3t2 = 14 m/sExample[3] Given velocity components at time t, i.e., vx = 2t and vy = (t2 + 4) where t is in second and v is in m/s. Determine its average acceleration between t = 1 s and t = 2 s along with its direction.Answer v x = 2t v x = 2(1) v x = 2(2) v y = (t 2 + 4) v y = (1)2 + 4 v y = (2)2 + 4 Δv x v x 2 − v x 1 4 − 2 ax = = = = 2 m/s2 Δt t2 − t1 1 Δv y v y2 − v y 1 8−5 ay = = = = 3 m/s2 Δt t2 − t1 1
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average vector acceleration : r a = ax i + ax j = (2 m/s2 ) i + (3 m/s2 ) j magnitude : 2 2 a= ax + ay → a = 22 + 32 = 3,6 m/s2 a 3 direction : tan θ = y = → θ = 56,3o ax 2Exercises[1] A particle moves from point A (1,0) to point B (5,4) on xy- plane. Write down the displacement vector from A to B and determine its value[2] A tennis ball moves on xy-plane. The coordinat position of point X and Y of the ball is describe by an equation such that x = 18t and y = 4t − 5t2 and a relevant constant. Write down an equation for vector position r with respect to time t using unit vector i and j.[3] Position of a particle due to time change on xy-plane was described by vector position r(t) = (at2 + bt)i + (ct + d)j with a, b, c, and d are constants of similar units. Determine its displacement vector between t = 1 second and t = 2 seconds, and define the value of its displacement.[4] Position of a particle describes by an equation such as r = 2t − 5t2 with r in meter and t in second. Determine (a) the initial velocity of particle (b) the velocity of particle at t = 2 seconds (c) its maximum distance that can be reached by the particle in positive direction[5] An object moves with velocity of 20 m/s by the direction of 210o counter clockwise related to x-axis. Determine the-x and y components of such velocity.
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Motion Description Using Differential ConceptConsider a motion of two men (blue and red clothes as shownon the picture). They both on xy-plane, where the blue clotheman is walking and the red clothe man is running.The dotted line showed their real motion and short bold lineshowed instant and short range motion.The relevance of motion on xy-plane or Cartesian coordinatewith linier line as follows- in textbook format, distance denote as d or s- in Cartesian coordinate, distance denote as x if the motion lies on horizontal line or x-axis and denotes as y if it lies on vertical line or y-axisIn general, any linier motion is described by equation distance = velocity x time s = v.t or x = v. t (1)
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For motion with small path or short distance and refer asinstant distance, the equation modify as instant (distance) = instant (velocity x time) Δx = Δ(v. t ) (1a)instant velocity means that it has constant velocity Δx = v. Δt (1b)Consider if the distance of motion become shorter and shorterand even in very tiny distance where it can be realized. Thiscondition refers as infinity small path. dx dx = v. dt or v = (1c) dtTerm dx → refer as the rate of change of position dt dxv = → (read as) velocity is the differential of position over dt the time (mean) rate of change of position is equal to velocityThe differential concept indeed is very accurate way to describea very tiny or even infinity path.Similar to equation (1c), we will define differential term ofacceleration with respect to velocity. Recall that v = vo + at.If vo = 0, then v = a.t → Δv = a.Δt → dv = a dt dvForm a = → is similar to term velocity in differential way. dtIt reads as acceleration is the differential of velocity over thetime and means that rate of change of velocity is equal toacceleration.
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In term of double rate of change, the differential of velocityover the time can be re-write as ⎛ dx ⎞ d⎜ ⎟ ⎝ dt ⎠ d2x a= → dt dt 2Acceleration is double differential of position.In other way around, it could be conclude that Differentiating position Differentiating velocity could yielding velocity could yielding accelerationMost Important Differential Rules y = xn x = a.t n ; a : constant dy dx = n.x n −1 = a.n.t n −1 dx dtExample[1] A vector position of particle describes as r r = (2t 2 + 4) i + (2t 3 + 4t ) j where t is in second and r in meter. Determine (a) its instantaneous velocity at t = 2 s, and (b) the magnitude and direction of (a)Answer r r = x i+y j x = 2t 2 + 4 r y = 2t 3 + 5t r = (2t 2 + 4) i + (2t 3 + 4t ) j
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dx d(a) v x = = (2t 2 + 4) = 4t m/s dt dt dy d vy = = (2t 3 + 5t ) = (6t + 5) m/s dt dt r v = v x i + v y j = (4t ) i + (6t 2 + 5) j r if t = 2 s → v = 8 i + 29 j(b) magnitude : 2 2 v = v x + vy → v = 82 + 292 = 30,08 m/sIntegral: Opposed to Differential ConceptWhen both sides of differential form of position dx = v dt beingintegralled, it can be found that ∫ dx = ∫ v dt → x = ∫ v dtThis is the integral form of position. Related to the concept ofdifferential of motion, position of an object could be tracedback by perform integral operation on velocity.As for reminder, integral is the way to turned back a differentialequation into “its original” equationSimillary, dv = a dt when integralling on both sides gives ∫ dv = ∫ a dt → v = ∫ a dt
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In a short diagram, Integralling velocity yields position “original” equation differentialled x = v.t dx = v.dt similar to its original integralled equation x = v.t + C x = ∫ v.dt yielded Integralling acceleration yields velocity “original” equation differentialled v = a. t dv = a. dt similar to integralled its original equation v = a. t + C v = ∫ a. dt yieldedMost Important Integral Rules x2 ∫x n y = dx ∫ax n Eq. Function: y = x1 a x2 Solution: y = x n +1 + C 1 n +1 y = x n −1 n +1 x1
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Example[1] A particle moves on xy-plane at initial position (2,4) m. Their velocity components are v x = 5t and v y = (4 + 3t 2 ) Determine (a) the equation of its position and (b) its position at t = 3 sAnswer(a) Initial position (2,4) means xo = 2m and yo = 4 m t t x = xo + ∫ v x dt y = y o + ∫ v y dt 0 0 t t = 2 + ∫ 5t dt ∫ (4 + 3t ) dt 2 =4+ 0 0 = 2 + 5 12 t ( 2 ) ( = 4 + 4t + 3 13 t 3 ) ( = 2 + 2,5 t 2 )m ( = 4 + 4t + t 3 m ) r Vector position of the particle is r = x i + y j r = (2 + 2,5t 2 ) i + (4 + 4t + t 3 ) j r(b) Particle’s position at t = 3 s r ( ) ( r = 2 + 2,5t 2 i + 4 + 4t + t 3 j ) = (24,5) i + (43) jExample[2] An object moves from rest with acceleration of r a = (6t 2 − 4) i + 6 j Determine object’s velocity at t = 4 sAnswer r r r r v = v o + ∫ a dt → vo = 0 =0+ ∫ [(6t − 4) i + 6 j] dt = (6 2 t 2 − 4t ) i + 6t j →t =4s r v = (6 2 42 − 4.4) i + 6.4 j = 32 i + 24 j value of its velocity is (object s vel) : 2 2 v = v x + vy → v = 322 + 242 = 40 m/s
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Integral as an area beneath the curvev (t) x(t) Δv t to Δt tThe curve path which is known as displacement is defined as x(t ) = ∑ Δv ΔtFrom the figure, displacement sample Δt is not appropriate forthe curve path, i.e, it is not depicted the real path of the curvetherefore we could pick an infinity small displacement.Mathematics provide the infinity situation by adding limit. x(t ) = ∑ lim Δv Δt Δt → 0v (t) x(t) Δv t to Δt tIn limit form, Δv tends to be constant where the value is v.Form Σ lim is simplify by the symbol of integral (means:summation of infinity small parts). The equation is then end upas t x(t ) = ∫ v dt to
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It can then be concluded that the integration of area under thecurve (path) which is likely the integration of velocity willprovide the whole object’s distance which represented by thefunction of x(t).Exercises[1] An object is throwing out to xy-plane such as its vector velocity was described as v = 50 m/s i + (100 m/s − 10 m/s2 t)j, The positive direction of y-axis to be vertical direction. At t = 0, the object is on its origin (0, 0). (a) determine the vector position of object as a function of time (b) its position when t = 2 s (c) the maximum height that can be reach by the object[2] Determine the particle’s position as a function of time if the particle velocity is (a) v = 2t + 6t2 1 3 (b) v x = 3t 2 + 5t 2 v y = sin 5t (c) v = 4ti + 3j The particle is initially at its origin (0, 0)[3] An object is moving on xy-plane with vector velocity of v = { (3 − 3t2)i + 2tj } m/s Determine the value of its diplacement of the object between t = 1 s and t = 2 s.
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