Kinematics Vocabulary. Kinematics and One Dimensional Motion. Position. Coordinate System in One Dimension. Kinema means movement 8.

Transcription

1 Kinemaics Vocabulary Kinemaics and One Dimensional Moion 8.1 WD1 Kinema means movemen Mahemaical descripion of moion Posiion Time Inerval Displacemen Velociy; absolue value: speed Acceleraion Averages of he laer wo quaniies. Coordinae Sysem in One Dimension Used o describe he posiion of a poin in space A coordinae sysem consiss of: 1. An origin a a paricular poin in space. A se of coordinae aes wih scales and labels 3. Choice of posiive direcion for each ais: uni vecors 4. Choice of ype: Caresian or Polar or Spherical Posiion A vecor ha poins from origin o body. Posiion is a funcion of ime In one dimension: () = ()ˆ i Eample: Caresian One-Dimensional Coordinae Sysem 1

2 Displacemen Vecor Change in posiion vecor of he objec during he ime inerval Δ = 1 Concep Quesion: Displacemen An objec goes from one poin in space o anoher. Afer he objec arrives a is desinaion, he magniude of is displacemen is: 1) eiher greaer han or equal o ) always greaer han 3) always equal o Δr ( ) ( ) ˆi Δ( ) ˆi ( ) 1 4) eiher smaller han or equal o 5) always smaller han 6) eiher smaller or larger han he disance i raveled. Average Velociy The average velociy, v (), is he displacemen Δr divided by he ime inerval Δ Δr Δ v() = ˆi = v () ˆ i Δ Δ The -componen of he average velociy is given by v () = Δ Δ Insananeous Velociy For each ime inerval Δ, we calculae he - componen of he average velociy. As Δ, we generae a sequence of he -componen of he average velociies. The limiing value of his sequence is defined o be he -componen of he insananeous velociy a he ime. Δ v () lim v Δ = lim Δ Δ = lim +Δ ( ) Δ Δ () d d

3 Insananeous Velociy -componen of he velociy is equal o he slope of he angen line of he graph of -componen of posiion vs. ime a ime Table Problem: Hedge Fund Ride Home A hedge fund manager usually akes he rain home and is me a he rain saion eacly a 6:3 by his chauffeur who drives him 6 miles o his esae. One day he leaves work early, arriving a he rain saion a 6:. Finding his cell phone discharged, he jogs owards his home a 6 mph. Afer 4 minues, he mees his chauffeur who urns around and akes him home. a How much earlier han usual does he arrive home? b Wha else can you deermine from his informaion? Suck? Represen he Problem in New Way Graphical Picures wih descripions Pure verbal Equaions Could You Solve i If? The problem were simplified? You knew some oher fac/relaionship Solve any par of problem, even simple one posiion Esae Graphical Represenaion Limousine Manager Today min Usually Saion 4 3 ime 3

4 CQ: Represening Moion - 1d Which of hese is a possible graph of he posiion, ()? CQ: Represening Moion - 1d Which of hese is a possible graph of he - componen of velociy? Average Acceleraion Change in insananeous velociy divided by he ime inerval Δ = 1 ( v, v,1 ) Δv Δv ˆ ˆ Δv a = i = i = ˆi = a ˆ i Δ Δ Δ Δ The -componen of he average acceleraion a = Δv Δ Insananeous Acceleraion For each ime inerval Δ, we calculae he - componen of he average acceleraion. As Δ, we generae a sequence of -componen of average acceleraions. The limiing value of his sequence is defined o be he -componen of he insananeous acceleraion a he ime. ( +Δ ) ( ) ˆ ˆ Δv ˆ v v ( ) ( ) lim lim lim ˆ dv a = a ˆ i ai = i = i i Δ Δ Δ Δ Δ d a () = dv d 4

5 Insananeous Acceleraion The -componen of acceleraion is equal o he slope of he angen line of he graph of he -componen of he velociy vs. ime a ime Concep Quesion: One- Dimensional Kinemaics The graph shows he posiion as a funcion of ime for wo rains running on parallel racks. For imes greaer han =, which is rue: 1. A ime B, boh rains have he same velociy.. Boh rains speed up all he ime. 3. Boh rains have he same velociy a some ime before B,. 4. Somewhere on he graph, boh rains have he same acceleraion. Summary: Consan Acceleraion Acceleraion a = consan Eample: Free Fall Choose coordinae sysem wih -ais verical, origin a ground, and posiive uni vecor poining upward. Velociy Posiion v () = v, + a 1 () = + v,+ a Acceleraion: Velociy Posiion: a = g = (9.8 m s - ) v () = v, g () = + v, 1 g 5

6 Concep Quesion: One- Dimensional Kinemaics You are hrowing a ball sraigh up in he air. A he highes poin, he ball s 1) velociy and acceleraion are zero ) velociy is nonzero bu is acceleraion is zero 3) acceleraion is nonzero, bu is velociy is zero 4) velociy and acceleraion are boh nonzero. Concep Quesion: One Dimensional Kinemaics A person sanding a he edge of a cliff hrows one ball sraigh up and anoher ball sraigh down, each a he same iniial speed. Neglecing air resisance, which ball his he ground below he cliff wih he greaer speed: 1. ball iniially hrown upward;. ball iniially hrown downward; 3. neiher; hey boh hi a he same speed. Worked Eample: Runner Problem Solving Sraegies: One-Dimensional Kinemaics A runner acceleraes from res wih a consan -componen of acceleraion. m s - for. s and hen ravels a a consan velociy for an addiional 6. s. How far did he runner ravel? 6

7 I. Undersand ge a concepual grasp of he problem Quesion 1: How many objecs are involved in he problem? Quesion : How many differen sages of moion occur? Quesion 3: For each objec, how many independen direcions are needed o describe he moion of ha objec? Quesion 4: Wha choice of coordinae sysem bes suis he problem? I. Undersand ge a concepual grasp of he problem sage 1: consan acceleraion sage : consan velociy. Tools: Coordinae sysem Kinemaic equaions Quesion 5: Wha informaion can you infer from he problem? Skech he problem Choose a coordinae sysem II. Devise a Plan Wrie down he complee se of equaions for he posiion and velociy funcions; idenify any specified quaniies; clean up he equaions. Finding he missing links : coun he number of independen equaions and he number of unknowns. You can solve a sysem of n independen equaions if you have eacly n unknowns. II. Devise a Plan Sage 1: consan acceleraion Iniial condiions: = v, = 1 Kinemaic Equaions: () = a v () = a Final Condiions: end acceleraion a posiion: velociy 1 a ( = a) = aa v,a v ( = a ) = a a = a Look for consrain condiions 7

8 III. Devise a Plan Sage : consan velociy, ime inerval runs a a consan velociy for he ime final posiion [, ] a b b ( ) ( ) = = + v b b a, a b a a III. Solve Design a sraegy for solving a sysem of equaions. Check your algebra and dimensions. Subsiue in numbers. Check your resuls and unis. III. Solve hree independen equaions 1 a = a a v = a, a a ( ) = + v b a, a b a Si symbols: b, a, v,a, a, a, b III. Solve Solve for disance he runner has raveled 1 1 ( = ) = a + a ( ) = a a b b a a b a a b a Three given quaniies specified in problem a, a, b 8

9 IV. Review Check your resuls, do hey make sense (hink!) Check limis of an algebraic epression (be creaive) Think abou how o eend model o cover more general cases (hinking ouside he bo) Solved problems ac as models for hinking abou new problems. (Mechanics provides a foundaion of of solved problems.) Numerical resuls: Runner acceleraed for Iniial acceleraion: Runs a a consan velociy for IV. Review Toal ime of running b = a + 6.s =.s + 6.s = 8.s Toal disance running 1-1 oal = aab aa = (. m s )(.s)( 8.s) (. m s )( ) 1.s =.8 1 m - Final velociy =.s a a =.m s b a - = 6.s ( )( ) va, = aa =.m s.s = 4.m s - -1 Table Problem: One Dimensional Kinemaics Bus and car A he insan a raffic ligh urns green, a car sars from res wih a given consan acceleraion, m s -. Jus as he ligh urns green, a bus, raveling wih a given consan speed, m s -1, passes he car. The car speeds up and passes he bus some ime laer. How far down he road has he car raveled, when he car passes he bus? Summary: Time Dependen Acceleraion Acceleraion is a nonconsan funcion of ime a () Change in velociy v () v = a ( ) d, Change in posiion () = v ( ) d 9

10 Eample: Non-consan acceleraion Consider an objec released a ime = wih an iniial - componen of velociy v, locaed a posiion,, and acceleraing according o a () = b + b+ b 1 Find he velociy and posiion as a funcion of ime. Velociy: Eample: Non-consan acceleraion Posiion: () =, + ( ) 1 = 1 3 = v, ( b b 1 b ) d v, b b 1 b = = 3 = v, b b 1 b v v a d = = = () = + v( ) d = (, 1 ), = + v + b + b + b d = v + b + b + b 1