Motion Along a Straight Line

Transcription

1 PH 1-3A Fall 010 Min Alng a Sraigh Line Lecure Chaper (Halliday/Resnick/Walker, Fundamenals f Physics 8 h ediin)

2 Min alng a sraigh line Sudies he min f bdies Deals wih frce as he cause f changes in min Deals wih he mahemaical descripin f min in erms f psiin, velciy and accelerain Key Cnceps 1. Ideal paricle a pin-like mass f infiniesimal size. Recangular crdinaes and rigin pin 3. Reference frame crdinae grid wih suiably adjused clcks

3 Min Alng a Sraigh Line In his chaper we will sudy kinemaics i.e. hw bjecs mve alng a sraigh line. The fllwing parameers will be defined: Displacemen Average velciy Average Speed Insananeus velciy Average and insananeus accelerain Fr cnsan accelerain we will develp he equains ha give us he velciy and psiin a any ime. In paricular we will sudy he min under he influence f graviy clse he surface f he earh. Finally we will sudy a graphical inegrain mehd ha can be used analyze he min when he accelerain is n cnsan 3

4 Kinemaics is he par f mechanics ha describes he min f physical bjecs. We say ha an bjec mves when is psiin as deermined by an bserver changes wih ime. In his chaper we will sudy a resriced class f kinemaics prblems Min will be alng a sraigh line We will assume ha he mving bjecs are paricles i.e. we resric ur discussin he min f bjecs fr which all he pins mve in he same way. The causes f he min will n be invesigaed. This will be dne laer in he curse. Cnsider an bjec mving alng a sraigh line aken be he x-axis. The bjec s psiin a any ime is described by is crdinae x() defined wih respec he rigin O. The crdinae x can be psiive r negaive depending wheher he bjec is lcaed n he psiive r he negaive par f he x-axis 4

5 Displacemen. If an bjec mves frm psiin x 1 in psiin is described by he displacemen Δ x= x x 1 psiin x, he change O. x. x 1. x x-axis min Fr example if x 1 = 5 m and x = 1 m hen x = 1 5 = 7 m. The psiive sign f x indicaes ha he min is alng he psiive x-direcin If insead he bjec mves frm x 1 = 5 m and x = 1 m hen x = 1 5 = -4 m. The negaive sign f x indicaes ha he min is alng he negaive x- direcin Displacemen is a vecr quaniy ha has bh magniude and direcin. In his resriced ne-dimensinal min he direcin is described by he algebraic sign f x Ne: The acual disance fr a rip is irrelevan as far as he displacemen is cncerned Cnsider as an example he min f an bjec frm an iniial psiin x 1 = 5 m x = 00 m and hen back x = 5 m. Even hugh he al disance cvered is 390 m he displacemen hen x = 0 5

6 Disance and Displacemen One afernn, a cuple walks hree-furhs f he way arund a circular lake, he radius f which is 1.50 km. They sar a he wes side f he lake and head due suh a he beginning f heir walk. a) Wha is he disance hey ravel? b) Wha are he magniude and direcin (relaive due eas) f he cuple s displacemen? 6

7 Average Velciy One mehd f describing he min f an bjec is pl is psiin x() as funcin f ime. In he lef picure we pl x versus fr an bjec ha is sainary wih respec he chsen rigin O. Nice ha x is cnsan. In he picure he righ we pl x versus fr a mving armadill. We can ge an idea f hw fas he armadill mves frm ne psiin x 1 a ime 1 a new psiin x a ime by deermining he average velciy beween 1 and. v avg x x Δx Δ 1 = = 1 Here x and x 1 are he psiins x( ) and x( 1 ), respecively. The ime inerval is defined as: = 1 The unis f v avg are: m/s Ne: Fr he calculain f v avg bh 1 and mus be given. 7

8 Graphical deerminain f v avg On an x versus pl we can deermine v avg frm he slpe f he sraigh line ha cnnecs pin ( 1, x 1 ) wih pin (, x ). In he pl belw 1 =1 s, and = 4 s. The crrespnding psiins are: x 1 = - 4 m and x = m v avg x x 1 = = = = 1 ( 4) 6 m m/s s Average Speed s avg The average speed is defined in erms f he al disance raveled in a ime inerval (and n he displacemen x as in he case f v avg ) s avg = al disance Ne: The average velciy and he average speed fr he same ime inerval can be quie differen Δ 8

9 A persn wh walks fr exercise prduces he psiin-ime graph a) Wihu ding any calculains, decide which segmens f he graph (A, B, C r D) indicae psiive, negaive and zer average velciies. b) Calculae he average velciy fr each segmen verify yur answers par A a) The sign f he average velciy during a segmen crrespnds he sign f he slpe f he segmen. The slpe, and hence he average velciy, is psiive fr segmens A and D, and negaive fr C, zer fr segmen B b) 9

10 A runner runs 100m in 10s, ress 60s and reurns a a walk in 80s. Wha is he average speed fr he cmplee min? Wha is he average velciy? -The runner mved a al disance d = = 00 m -The rund rip k = = 150 s V av = d/ = 00m/150s = 1.3 m/s -Afer he min, he runner is precisely lcaed a he saring pin His psiin didn change V av = x/ = 0/ = 0 10

11 Prblem 5. The psiin f an bjec mving alng an x axis is given by x= , where x is in meers and is in secnds. Find he psiin f he bjec a he fllwing values f : (a) 1 s, (b) s, (c) 3 s, (d) 4 s. (e) Wha is he bjec s displacemen beween =0 and =4 s? (f) Wha is he average velciy fr he ime inerval frm = s and = 4 s? (g) Graph x versus fr 0 4 s and indicae hw he answer fr (f) can be fund n he graph. Using x = wih SI unis undersd is efficien (and is he apprach we will use), bu if we wished make he unis explici we wuld wrie x = (3 m/s) (4 m/s ) + (1 m/s 3 ) 3. We will que ur answers ne r w significan figures, and n ry fllw he significan figure rules rigrusly. (a) Plugging in = 1 s yields x = = 0. (b) Wih = s we ge x = 3() 4() +() 3 = m. (c) Wih = 3 s we have x = 0 m. (d) Plugging in = 4 s gives x = 1 m. Fr laer reference, we als ne ha he psiin a = 0 is x = 0. (e) The psiin a = 0 is subraced frm he psiin a = 4 s find he displacemen Δx = 1 m. (f) The psiin a = s is subraced frm he psiin a = 4 s give he displacemen Δx = 14 m. Eq. -, hen, leads v avg Δ x 14 m = = = Δ s 7 m / s. (g) The hriznal axis is 0 4 wih SI unis undersd. A sraigh line drawn frm he pin a (, x) = (, ) he highes pin shwn (a = 4 s) wuld represen he answer fr par (f). 11

12 Insananeus Velciy The average velciy v avg deermined beween imes 1 and prvide a useful descripin n hw fas an bjec is mving beween hese w imes. I is in realiy a summary f is min. In rder describe hw fas an bjec mves a any ime we inrduce he nin f insananeus velciy v (r simply velciy). Insananeus velciy is defined as he limi f he average velciy deermined fr a ime inerval as we le 0. Δx v = lim = Δ Δ 0 dx d Frm is definiin insananeus velciy is he firs derivaive f he psiin crdinae x wih respec ime. Is is hus equal he slpe f he x versus pl. Speed We define speed as he magniude f an bjec s velciy vecr 1

13 13

14 Prblem 15. (a) If a paricle s psiin is given by x= (where is in secnds and x is in meers), wha is he velciy a = 1 s? (b) is i mving in he psiive r negaive direcin f x jus hen? (c) Wha is he speed jus hen? (d) Is he speed increasing r decreasing jus hen? (Try answering he nex w quesins wihu furher calculain) (e) is here ever an insan when he velciy is zer? If s, give he ime ; if n, answer n. (f) is here a ime afer 3 s when he paricle is mving in he negaive direcin f x? If s, give he ime ; if n, answer n. (a) The velciy f he paricle is iii. v dx d = = ( ) = d d Thus, a = 1 s, he velciy is v = ( 1 + (6)(1)) = 6 m/s. (b) Since v < 0, i is mving in he negaive x direcin a = 1 s. (c) A = 1 s, he speed is v = 6 m/s. (d) Fr 0 < < s, v decreases unil i vanishes. Fr < < 3 s, v increases frm zer he value i had in par (c). Then, v is larger han ha value fr > 3 s. (e) Yes, since v smhly changes frm negaive values (cnsider he = 1 resul) psiive (ne ha as +, we have v + ). One can check ha v = 0 when = s. 14

15 Average Accelerain We define as he average accelerain a avg beween 1 and as: a avg v v Δv Δ 1 = = 1 Insananeus Accelerain Unis: m/s If we ake he limi f a avg as 0 we ge he insananeus accelerain a which describes hw fas he velciy is changing a any ime Δv dv dv d dx d x a = lim =, a = = = Δ d d d d d Δ 0 The accelerain is he slpe f he v versus pl Ne: The human bdy des n reac velciy bu i des reac accelerain 15

16 Accelerain The average accelerain is defined prvide a measure f hw much he velciy changes per uni f elapsed ime. a av is a vecr ha pins in he same direcin as V + and indicae w pssible direcins fr he accelerain vecr a av direced he lef i s cmpnen n he psiive x direcin is negaive Insananeus accelerain 16

17 Equains f kinemaics fr cnsan accelerain Assume x 0 = 0 when 0 = 0 => x = x x 0 = x Since min is alng a sraigh line all he vecrs f displacemen, velciy and accelerain are alng his line and we will subsiue hem wih heir magniudes having plus r minus signs cnveying he direcin f hese vecrs Assume v = v 0 a 0 = 0 and v a Assume a = cnsan => a av = a = v-v 0 / r v = v 0 + a Frm he definiin f he v av v av = x-x 0 /- 0 = x/ r x = v av Because a = cnsan, v increases a a cnsan rae. => v av is midway beween v 0 and v final v av = ½(v 0 +v) => x = ½(v 0 +v) => x = v av = ½(v 0 +v) = ½(v 0 +v 0 +a) = v 0 + a / x = v 0 + a / x = ½(v 0 +v) = ½(v 0 +v)(v-v 0 /a) = (v -v 0 )/a => v = v 0 + ax x = ½(v 0 +v) = ½(v-a+v) = v - a / => x = v - a / 17

18 Equains f kinemaics fr cnsan accelerain Equain Number Equain v = v0 + a x x0 1 x x0 = ( v0 + v ) 1 x x0 = v0+ a a v v = v0 + a( x x0) 1 x x v a v 0 = 0 Missing Quaniy 18

19 Anher lk a Cnsan Accelerain dv a = dv= ad d dv = ad = a d v = a + C If we inergrae bh sides f he equain we ge: Here C is he inegrain cnsan C can be deermined if we knw he velciy v = v( 0 ) a = 0 v(0) = v = ( a)(0) + C C = v v = v + a 0 (eqs.1) dx v = dx = vd = ( v0 + a) d = v0d + ad If we inegrae bh sides we ge: d a dx = vd 0 + a d x = v + + C Here C is he inegrain cnsan C can be deermined if we knw he psiin x = x( 0 ) a = 0 a x(0) = x = ( v)(0) + (0) + C C = x a x () = x + v + (eqs.) 19

20 a v= v0 + a (eqs.1) ; x= x + v+ (eqs.) If we eliminae he ime beween equain 1 and equain we ge: ( ) v v = a x x (eqs.3) Belw we pl he psiin x(), he velciy v() and he accelerain a versus ime a x= x + v+ The x() versus pl is a parabla ha inerceps he verical axis a x = x v= v0 + a The v() versus pl is a sraigh line wih Slpe = a and Inercep = v The accelerain a is a cnsan 0

21 a) Wha is he magniude f he average accelerain f a skier wh, saring frm res, reaches a speed f 8.0 m/s when ging dwn a slpe fr 5.0s? b) Hw far des he skier ravel in his ime? 5. Slve he equain wih respec he unknwn kinemaic variable 0 V =0 1. Make a drawing a 3. Fill up he able f given kinemaic variables wih apprpriae + r signs. 4. Verify ha given infrmain cnains a leas hree f he five kinemaics variables and chse ne kinemaic equain relaing hree given kinemaic variables wih he unknwn ne. V 1 X (+). Shw psiive direcin f min and chse a cnvenien rigin pin

22 Equains f Kinemaics The driver f an aumbile raveling a 95 km/h perceives an bsacle n he rad and slams n he brakes. Calculae he al spping disance in meers. Assume ha he reacin ime f he driver is 0.75s (s here is a ime inerval f 0.75s during which he aumbile cninues a cnsan speed while he driver ges ready apply he brakes and ha he decelerain f he aumbile is 7.8 m/s when he brakes are applied. Ofen he min f an bjec is divided in segmens, each wih a differen accelerain. When slving each prblem, i is impran realize ha he final velciy fr ne segmen is he iniial velciy fr he nex segmen.

23 Equains f Kinemaics The perain manual f a passenger aumbile saes ha he spping disance is 50m when he brakes are fully applied a 96 km/h. Wha is he decelerain? 3

24 Equains f Kinemaics A speedba sars frm res and acceleraes a +.01m/s fr 7.00s. A he end f his ime he ba cninues fr an addiinal 6.00s wih an accelerain f m/s. Fllwing his, he ba acceleraes a -1.49m/s fr 8.00s. a) Wha is he velciy f he ba a = 1.0s? b) Find he al displacemen f he ba. a) The velciy a he end f he firs (7.00s) perid is v 1 = v 0 + a 1 1 = (.01m/s )(7.00s) - a he end f he secnd perid: v = v 1 + a = v 1 + (0.518m/s )(6.00s) - a he end f he hird perid: v 3 = v + a 3 3 = v + (-1.49m/s )(8.00s) = 5.6m/s b) The displacemen fr he firs ime perid is fund frm x 1 = v /a 1 1 x 1 = ½(.01m/s )(7.00s) = 49.m Similarly, x = 93.7m; x 3 = 89.7m Tal displacemen: x = x 1 + x + x 3 = 33m!!!. When he min f an bjec is divided in segmens, remember ha he final velciy f ne segmen is he iniial velciy fr he nex segmen 4

25 !!! The min f bjecs may be inerrelaed s hey share a cmmn variable. The fac ha he mins are inerrelaed is an impran piece f infrmain. In such cases, daa fr nly w variables need be specified fr each bjec. Tw sccer players sar frm res 48m apar. They run direcly ward each her, bh players acceleraing. The firs player has an accelerain whse magniude is 0.50m/s. The secnd player s accelerain has a magniude f 0.30m/s. a) Hw much ime passes befre hey cllide? b) A he insan hey cllide, hw far has he firs player run? 5

26 !!! Smeimes here are pssible answers a kinemaics prblem, each answers crrespnding a differen siuain. Example The spacecraf is raveling wih a velciy f +350m/s. Suddenly he rerrckes are fired and he spacecraf begins slw dwn wih an accelerain whse magniude is 10.0m/s. Wha is he velciy f he spacecraf when he displacemen f he craf is +15km relaive he pin where he rerrckes began firing? Reasning: Since he spacecraf is slwing dwn, a has a direcin ppsie he velciy => a = -10.0m/s Sluin: v = v 0 + ax = (350m/s) + (-10.0m/s )(15000m) = 6.3x10 6 m /s v = ± 6.3x10 6 m /s = ±500 m/s Bh f hese answers crrespnd he same displacemen x = +15km, bu each arises in a differen par f he min. 6

27 Freely Falling Bdies Galile Galilei Galile Galilei was an Ialian scienis wh frmulaed he basic law f falling bdies, which he verified by careful measuremens. In he absence f air resisance, he fund ha all bdies a he same lcain abve he earh fall verically wih he same accelerain The Leaning Twer f Pisa buil in

28 Freely Falling Bdies The effec f graviy causes bjecs fall dwnward In he absence f air resisance, i is fund ha all bdies a he same lcain abve he earh fall verically wih he same accelerain If he disance f he fall is small cmpared he radius f he earh, he accelerain remains cnsan hrughu he fall Idealized min, in which air resisance is negleced is knwn as free fall Since he accelerain is nearly cnsan in free fall, he equains f kinemaics can be used Since he min ccurs in verical r y direcin we simply replace x wih y in kinemaics equains v = v + a v = v g 1 1 y = ( v + v) y = ( v + v) y = v 1 v + = v + ay a y = v 1 v = v gy g 8

29 A Falling Sne. The Velciy f a Falling Sne A sne is drpped frm res frm he p f a all building. Afer 3.00s f free fall a) Wha is he displacemen f he sne? b) Wha is he velciy f he sne? Because f he accelerain due graviy, he magniude f he sne s dwnward velciy increases by 9.80 m/s during each secnd f free fall v = v 0 + a Since he sne is mving dwnward in he negaive direcin he value fr v shuld be negaive v = v 0 + a = (-9.80m/s )(3.00s) = -9.4 m/s 9

30 Hw High Des I G? A glf ball rebunds frm he flr and ravels sraigh upward wih a speed f 5.0 m/s. T wha max heigh des he ball rise? 30

31 Hw Lng Is I in he Air? An arrw is fired frm grund level sraigh upward wih an iniial speed f 15 m/s. Hw lng is he arrw in he air befre i srikes he grund? During he ime he arrw ravels upward, graviy causes is speed decrease 0 On he way dwn, graviy causes he arrw regain he ls speed => he ime fr he arrw g up is equal he ime g dwn 31

32 Free Fall Clse he surface f he earh all bjecs mve wards he cener f he earh wih an accelerain whse magniude is cnsan and equal 9.8 m/s We use he symbl g indicae he accelerain f an bjec in free fall If we ake he y-axis pin upwards hen he accelerain f an bjec in free fall a = -g and he equains fr free fall ake he frm: a y v = v0 g (eqs.1) ; B g x = x + v (eqs.3) v v = g x x (eqs.4) ( ) A Ne: Even hugh wih his chice f axes a < 0, he velciy can be psiive ( upward min frm pin A pin B). I is mmenarily zer a pin B. The velciy becmes negaive n he dwnward min frm pin B pin A Hin: In a kinemaics prblem always indicae he axis as well as he accelerain vecr. This simple 3 precauin helps avid algebraic sign errrs.

33 Prblem 55. A ball f mis clay falls 15.0 m he grund. I is in cnac wih he grund fr 0.0 ms befre spping. (a) wha is he magniude f he average accelerain f he ball during he ime i is in cnac wih he grund? (Trea he ball as a paricle.) (b) Is he average accelerain up r dwn? (a) We firs find he velciy f he ball jus befre i his he grund. During cnac wih he grund is average accelerain is given by v a = Δ avg Δ where Δv is he change in is velciy during cnac wih he grund and 3 Δ= s is he durain f cnac. Nw, find he velciy jus befre cnac, we pu he rigin a he pin where he ball is drpped (and ake +y upward) and ake = 0 be when i is drpped. The ball srikes he grund a y = 15.0 m. Is velciy here is fund frm v = gy. Therefre, v= = = gy (9.8 m/s )( 15.0 m) 17.1 m/s where he negaive sign is chsen since he ball is raveling dwnward a he mmen f cnac. Cnsequenly, he average accelerain during cnac wih he grund is a 0 ( 17.1m/s) = = s avg m/s. (b) The fac ha he resul is psiive indicaes ha his accelerain vecr pins upward. In a laer chaper, his will be direcly relaed he magniude and direcin f he frce exered by he grund n he ball during he cllisin. 33

34 Graphical Inegrain in Min Analysis (nn-cnsan accelerain) When he accelerain f a mving bjec is n cnsan we mus use inegrain deermine he velciy v() and he psiin x( ) f he bjec. The inegain can be dne eiher using he analyic r he graphical apprach dv a = dv = ad dv = ad v v = ad v = v + ad d ad = 1 1 [ Area under he a versus curve beween and ] vd = dx v= dx= vd dx= vd d x x = vd x = x + vd [ Area under he v versus curve beween and ] 1 34

35 Prblem 70. w paricles mve alng an x axis. The psiin f paricle 1 is given by x = (in meers and secnds); he accelerain f paricle is given by a = ( in meers per secnds squared and secnds) and, a = 0, is velciy is 0 m/s. When he velciies f he paricles mach, wha is heir velciy? T slve his prblem, we ne ha velciy is equal he ime derivaive f a psiin funcin, as well as he ime inegral f an accelerain funcin, wih he inegrain cnsan being he iniial velciy. Thus, he velciy f paricle 1 can be wrien as dx1 d v = 1 ( ) d = d + + = +. Similarly, he velciy f paricle is v v a d d = 0 + = ( 8.00 ) = The cndiin ha v1 = v implies iv = = 0 which can be slved give (aking psiive r) = ( 3+ 6)/= 1.05s. Thus, he velciy a his ime is v1 = v = 1.0(1.05) = 15.6 m/s. 35