2 LEARNING GALS By sudying his chper, you will lern: How o describe srigh-line moion in erms of erge elociy, insnneous elociy, erge ccelerion, nd insnneous ccelerion. How o inerpre grphs of posiion ersus ime, elociy ersus ime, nd ccelerion ersus ime for srigh-line moion. How o sole problems inoling srigh-line moion wih consn ccelerion, including free-fll problems. How o nlyze srigh-line moion when he ccelerion is no consn. MTIN ALNG A STRAIGHT LINE? A ypicl spriner speeds up during he firs hird of rce nd slows grdully oer he res of he course. Is i ccure o sy h spriner is ccelering s he slows during he finl wo-hirds of he rce? Wh disnce mus n irliner rel down runwy before reching keoff speed? When you hrow bsebll srigh up in he ir, how high does i go? When glss slips from your hnd, how much ime do you he o cch i before i his he floor? These re he kinds of quesions you will lern o nswer in his chper. We re beginning our sudy of physics wih mechnics, he sudy of he relionships mong force, mer, nd moion. In his chper nd he ne we will sudy kinemics, he pr of mechnics h enbles us o describe moion. Ler we will sudy dynmics, which reles moion o is cuses. In his chper we concenre on he simples kind of moion: body moing long srigh line. To describe his moion, we inroduce he physicl quniies elociy nd ccelerion. These quniies he simple definiions in physics; howeer, hose definiions re more precise nd slighly differen hn he ones used in eerydy lnguge. An imporn pr of how physicis defines elociy nd ccelerion is h hese quniies re ecors. As you lerned in Chper 1, his mens h hey he boh mgniude nd direcion. ur concern in his chper is wih moion long srigh line only, so we won need he full mhemics of ecors jus ye. Bu using ecors will be essenil in Chper 3 when we consider moion in wo or hree dimensions. We ll deelop simple equions o describe srigh-line moion in he imporn specil cse when he ccelerion is consn. An emple is he moion of freely flling body. We ll lso consider siuions in which he ccelerion ries during he moion; in his cse, i s necessry o use inegrion o describe he moion. (If you hen sudied inegrion ye, Secion 2.6 is opionl.) 2.1 Displcemen, Time, nd Aerge Velociy Suppose drg rcer dries her AA-fuel drgser long srigh rck (Fig. 2.1). To sudy he drgser s moion, we need coordine sysem. We choose he -is o lie long he drgser s srigh-line ph, wih he origin he sring line. We lso choose poin on he drgser, such s is fron end, nd represen he enire drgser by h poin. Hence we re he drgser s pricle. A useful wy o describe he moion of he pricle h is, he poin h represens he drgser is in erms of he chnge in he pricle s coordine oer ime inerl. Suppose h 1. s fer he sr he fron of he drgser is poin P 1, 19 m from he origin, nd 4. s fer he sr i is poin P 2, 277 m from he origin. The displcemen of he pricle is ecor h poins from P 1 o P 2 (see Secion 1.7). Figure 2.1 shows h his ecor poins long he -is. The -componen of he displcemen is jus he chnge in he lue of, 1 277 m 2 19 m 2 5 258 m, h ook plce during he ime inerl of 1 4. s 2 1. s 2 5 3. s. We define he drgser s erge elociy during his ime inerl s ecor quniy whose -componen is he chnge in diided by he ime inerl: 1 258 m 2 / 1 3. s 2 5 86 m/s. In generl, he erge elociy depends on he priculr ime inerl chosen. For 3.-s ime inerl before he sr of he rce, he erge elociy would be zero becuse he drgser would be res he sring line nd would he zero displcemen. Le s generlize he concep of erge elociy. A ime 1 he drgser is poin P 1, wih coordine 1, nd ime 2 i is poin P 2, wih coordine 2. The displcemen of he drgser during he ime inerl from 1 o 2 is he ecor from P 1 o P 2. The -componen of he displcemen, denoed D, is jus he chnge in he coordine : D 5 2 2 1 The drgser moes long he -is only, so he y- nd z-componens of he displcemen re equl o zero. CAUTIN The mening of D Noe h D is no he produc of D nd ; i is single symbol h mens he chnge in he quniy. We lwys use he Greek cpil leer D (del) o represen chnge in quniy, equl o he finl lue of he quniy minus he iniil lue neer he reerse. Likewise, he ime inerl from 1 o 2 is D, he chnge in he quniy : D 5 2 2 1 (finl ime minus iniil ime). The -componen of erge elociy, or erge -elociy, is he -componen of displcemen, D, diided by he ime inerl D during which he displcemen occurs. We use he symbol for erge -elociy (he 2.1 Posiions of drgser wo imes during is run. - Posiion 1 5 1. s Posiion 2 5 4. s START FINISH P 1 P 2 1 5 19 m -coordine of drgser 1. s is posiie o he righ of he origin 12, negie o he lef of i. -is D 5 1 2 2 1 2 5 258 m Displcemen from 1 o 2 2 5 277 m 2.1 Displcemen, Time, nd Aerge Velociy 37 (2.1) When he drgser moes in he + direcion, he displcemen D is posiie nd so is he erge -elociy: - 5 D D 5 258 m 3. s 5 86 m /s -coordine of drgser 4. s 36
38 CHAPTER 2 Moion Along Srigh Line 2.2 Insnneous Velociy 39 subscrip signifies erge lue nd he subscrip indices h his is he -componen): - 5 2 2 1 2 2 1 5 D D (erge -elociy, srigh-line moion) (2.2) Drgser rck 1no o scle2 4 (m) For displcemen long he -is, n objec s erge -elociy - equls he slope of line connecing he corresponding poins on grph of posiion 12 ersus ime 12. 3 P 2 2 p 2 2.3 The posiion of drgser s funcion of ime. 2.2 Posiions of n officil s ruck wo imes during is moion. The poins P 1 nd P 2 now indice he posiions of he ruck, nd so re he reerse of Fig. 2.1. As n emple, for he drgser 1 5 19 m, 2 5 4. s, so Eq. (2.2) gies The erge -elociy of he drgser is posiie. This mens h during he ime inerl, he coordine incresed nd he drgser moed in he posiie -direcion (o he righ in Fig. 2.1). If pricle moes in he negie -direcion during ime inerl, is erge elociy for h ime inerl is negie. For emple, suppose n officil s ruck moes o he lef long he rck (Fig. 2.2). The ruck is 1 5 277 m 1 5 16. s nd is 2 5 19 m 2 5 25. s. Then D 5 1 19 m 2 277 m 2 5 2258 m nd D 5 1 25. s 2 16. s 2 5 9. s. The -componen of erge elociy is - 5D/D 5 1 2258 m 2 / 1 9. s 2 5229 m/s. Here re some simple rules for he erge -elociy. Wheneer is posiie nd incresing or is negie nd becoming less negie, he pricle is moing in he 1-direcion nd - is posiie (Fig. 2.1). Wheneer is posiie nd decresing or is negie nd becoming more negie, he pricle is moing in he 2-direcion nd is negie (Fig. 2.2). CAUTIN Choice of he posiie -direcion You migh be emped o conclude h posiie erge -elociy mus men moion o he righ, s in Fig. 2.1, nd h negie erge -elociy mus men moion o he lef, s in Fig. 2.2. Bu h s correc only if he posiie -direcion is o he righ, s we chose i o be in Figs. 2.1 nd 2.2. Hd we chosen he posiie -direcion o be o he lef, wih he origin he finish line, he drgser would he negie erge -elociy nd he officil s ruck would he posiie erge -elociy. In mos problems he direcion of he coordine is will be yours o choose. nce you e mde your choice, you mus ke i ino ccoun when inerpreing he signs of nd oher quniies h describe moion! - - 5 277 m 2 19 m 4. s 2 1. s 5 258 m 3. s 5 86 m /s - 2 5 277 m, 1 5 1. s, Wih srigh-line moion we someimes cll D simply he displcemen nd - simply he erge elociy. Bu be sure o remember h hese re relly he -componens of ecor quniies h, in his specil cse, he only -componens. In Chper 3, displcemen, elociy, nd ccelerion ecors will he wo or hree nonzero componens. Figure 2.3 is grph of he drgser s posiion s funcion of ime h is, n - grph. The cure in he figure does no represen he drgser s ph in spce; s Fig. 2.1 shows, he ph is srigh line. Rher, he grph is picoril wy o represen how he drgser s posiion chnges wih ime. The poins p 1 Posiion 2 5 25. s Posiion 1 5 16. s START FINISH P 2 P 1 2 5 19 m This posiion is now 2. D 5 1 2 2 1 2 5 2258 m D D Displcemen from 1 o 2 When he ruck moes in he 2-direcion, D is negie nd so is he erge -elociy: 2258 m - 5 5 5 229 m/s 9. s 1 5 277 m nd This posiion is now 1. 2 1 p 1 P 1 1 Slope 5 -elociy D 5 2 2 1 1 2 3 1 D 5 2 2 1 4 5 2 Slope 5 rise oer run 5 nd p 2 on he grph correspond o he poins P 1 nd P 2 long he drgser s ph. Line p 1 p 2 is he hypoenuse of righ ringle wih ericl side D 5 2 2 1 nd horizonl side D 5 2 2 1. The erge -elociy - 5D/D of he drgser equls he slope of he line p 1 p 2 h is, he rio of he ringle s ericl side D o is horizonl side D. The erge -elociy depends only on he ol displcemen D 5 2 2 1 h occurs during he ime inerl D 5 2 2 1, no on he deils of wh hppens during he ime inerl. A ime 1 moorcycle migh he rced ps he drgser poin P 1 in Fig. 2.1, hen blown is engine nd slowed down o pss hrough poin P 2 he sme ime 2 s he drgser. Boh ehicles he he sme displcemen during he sme ime inerl nd so he he sme erge -elociy. If disnce is gien in meers nd ime in seconds, erge elociy is mesured in meers per second 1 m/s 2. her common unis of elociy re kilomeers per hour 1 km/h 2, fee per second 1 f/s 2, miles per hour 1 mi/h 2, nd knos 1 1 kno 5 1 nuicl mile/h 5 68 f/h 2. Tble 2.1 liss some ypicl elociy mgniudes. Tes Your Undersnding of Secion 2.1 Ech of he following uomobile rips kes one hour. The posiie -direcion is o he es. (i) Auomobile A rels 5 km due es. (ii) Auomobile B rels 5 km due wes. (iii) Auomobile C rels 6 km due es, hen urns round nd rels 1 km due wes. (i) Auomobile D rels 7 km due es. () Auomobile E rels 2 km due wes, hen urns round nd rels 2 km due es. () Rnk he fie rips in order of erge -elociy from mos posiie o mos negie. (b) Which rips, if ny, he he sme erge -elociy? (c) For which rip, if ny, is he erge -elociy equl o zero? 2.2 Insnneous Velociy Someimes he erge elociy is ll you need o know bou pricle s moion. For emple, rce long srigh line is relly compeiion o see whose erge elociy, -, hs he grees mgniude. The prize goes o he compeior who cn rel he displcemen D from he sr o he finish line in he shores ime inerl, D (Fig. 2.4). Bu he erge elociy of pricle during ime inerl cn ell us how fs, or in wh direcion, he pricle ws moing ny gien ime during he inerl. To do his we need o know he elociy ny specific insn of ime or specific poin long he ph. This is clled insnneous elociy, nd i needs o be defined crefully. CAUTIN How long is n insn? Noe h he word insn hs somewh differen definiion in physics hn in eerydy lnguge. You migh use he phrse I lsed jus n insn o refer o somehing h lsed for ery shor ime inerl. Bu in physics n insn hs no durion ll; i refers o single lue of ime. (s) D D Tble 2.1 Typicl Velociy Mgniudes A snil s pce A brisk wlk Fses humn Running cheeh Fses cr Rndom moion of ir molecules Fses irplne rbiing communicions sellie Elecron orbiing in hydrogen om Ligh reling in cuum 1 23 m/s 2 m/s 11 m/s 35 m/s 341 m/s 5 m/s 1 m/s 3 m/s 2 3 1 6 m/s 3 3 1 8 m/s 2.4 The winner of 5-m swimming rce is he swimmer whose erge elociy hs he grees mgniude h is, he swimmer who rerses displcemen D of 5 m in he shores elpsed ime D.
4 CHAPTER 2 Moion Along Srigh Line 2.2 Insnneous Velociy 41 2.5 Een when he s moing forwrd, his cyclis s insnneous -elociy cn be negie if he s reling in he negie -direcion. In ny problem, he choice of which direcion is posiie nd which is negie is enirely up o you. To find he insnneous elociy of he drgser in Fig. 2.1 he poin P 1, we moe he second poin P 2 closer nd closer o he firs poin P 1 nd compue he erge elociy - 5D/D oer he eer-shorer displcemen nd ime inerl. Boh D nd D become ery smll, bu heir rio does no necessrily become smll. In he lnguge of clculus, he limi of D/D s D pproches zero is clled he deriie of wih respec o nd is wrien d/d. The insnneous elociy is he limi of he erge elociy s he ime inerl pproches zero; i equls he insnneous re of chnge of posiion wih ime. We use he symbol, wih no subscrip, for he insnneous elociy long he -is, or he insnneous -elociy: D 5 lim DS D 5 d d (insnneous -elociy, srigh-line moion) (2.3) The ime inerl D is lwys posiie, so hs he sme lgebric sign s D. A posiie lue of mens h is incresing nd he moion is in he posiie -direcion; negie lue of mens h is decresing nd he moion is in he negie -direcion. A body cn he posiie nd negie, or he reerse; ells us where he body is, while ells us how i s moing (Fig. 2.5). Insnneous elociy, like erge elociy, is ecor quniy; Eq. (2.3) defines is -componen. In srigh-line moion, ll oher componens of insnneous elociy re zero. In his cse we ofen cll simply he insnneous elociy. (In Chper 3 we ll del wih he generl cse in which he insnneous elociy cn he nonzero -, y-, nd z-componens.) When we use he erm elociy, we will lwys men insnneous rher hn erge elociy. The erms elociy nd speed re used inerchngebly in eerydy lnguge, bu hey he disinc definiions in physics. We use he erm speed o denoe disnce reled diided by ime, on eiher n erge or n insnneous bsis. We use he symbol wih no subscrips o denoe insnneous speed. Insnneous speed mesures how fs pricle is moing; insnneous elociy mesures how fs nd in wh direcion i s moing. For emple, pricle wih insnneous elociy 5 25 m/s nd second pricle wih 5225 m/s re moing in opposie direcions he sme insnneous speed 25 m/s. Insnneous speed is he mgniude of insnneous elociy, nd so insnneous speed cn neer be negie. CAUTIN Aerge speed nd erge elociy Aerge speed is no he mgniude of erge elociy. When Alender Popo se world record in 1994 by swimming 1. m in 46.74 s, his erge speed ws 1 1. m 2 / 1 46.74 s 2 5 2.139 m/s. Bu becuse he swm wo lenghs in 5-m pool, he sred nd ended he sme poin nd so hd zero ol displcemen nd zero erge elociy! Boh erge speed nd insnneous speed re sclrs, no ecors, becuse hese quniies conin no informion bou direcion. SLUTIN IDENTIFY: We use he definiions of displcemen, erge elociy, nd insnneous elociy. Using he firs wo of hese inoles lgebr; he ls one requires using clculus o ke deriie. SET UP: Figure 2.6b shows our skech of he cheeh s moion. To nlyze his problem we use Eq. (2.1) for displcemen, Eq. (2.2) for erge elociy, nd Eq. (2.3) for insnneous elociy. EXECUTE: () A ime 1 5 1. s he cheeh s posiion is A ime 2 5 2. s is posiion is The displcemen during his inerl is (b) The erge -elociy during his ime inerl is - 5 2 2 1 2 2 1 (c) Wih D 5.1 s, he ime inerl is from 1 5 1. s o 2 5 1.1 s. A ime 2, he posiion is 2.6 A cheeh cking n nelope from mbush. The nimls re no drwn o he sme scle s he is. () The siuion (b) ur skech 1 5 2 m 1 1 5. m/s 2 211. s 2 2 5 25 m 2 5 2 m 1 1 5. m/s 2 212. s 2 2 5 4 m D 5 2 2 1 5 4 m 2 25 m 5 15 m 2 5 2 m 1 1 5. m/s 2 211.1 s 2 2 5 26.5 m Vehicle 5 2 1 4 m 2 25 m 2. s 2 1. s 5 15 m 1. s 5 15 m /s Cheeh The erge -elociy during his inerl is - 5 26.5 m 2 25 m 1.1 s 2 1. s You should follow his sme pern o work ou he erge -elociies for he.1-s nd.1-s inerls. The resuls re 1.5 m/s nd 1.5 m/s. As D ges smller, he erge -elociy ges closer o 1. m/s, so we conclude h he insnneous -elociy ime 5 1. s is 1. m/s. (d) To find he insnneous -elociy s funcion of ime, ke he deriie of he epression for wih respec o. The deriie of consn is zero, nd for ny n he deriie of n is n n21, so he deriie of 2 is 2. Therefore 5 d d 5 1 5. m /s 2 212 2 5 1 1 m/s 2 2 A ime 5 1. s, 5 1 m/s s we found in pr (c). A ime 5 2. s, 5 2 m/s. EVALUATE: ur resuls show h he cheeh picked up speed from 5 (when i ws res) o 5 1. s 1 5 1 m/s 2 o 5 2. s 1 5 2 m/s 2. This mkes sense; he cheeh coered only 5 m during he inerl 5 o 5 1. s, bu coered 15 m during he inerl 5 1. s o 5 2. s. Anelope 5 1.5 m/s Emple 2.1 Aerge nd insnneous elociies A cheeh is crouched 2 m o he es of n obserer s ehicle elociy during he sme ime inerl. (c) Find he insnneous (Fig. 2.6). A ime 5 he cheeh chrges n nelope nd elociy ime 1 5 1. s by king D 5.1 s, hen D 5.1 s, begins o run long srigh line. During he firs 2. s of he hen D 5.1 s. (d) Derie generl epression for he insnneous elociy s funcion of ime, nd from i find 5 1. s ck, he cheeh s coordine ries wih ime ccording o he equion 5 2 m 1 1 5. m/s 2 2 2. () Find he displcemen of nd 5 2. s. he cheeh beween 1 5 1. s nd 2 5 2. s. (b) Find he erge 1 We drw n is. We poin i in he 2 We choose o plce he 3 We mrk he iniil posiions of he cheeh 4 We re ineresed in he cheeh s moion 5 We dd symbols for known nd unknown (c) ur hinking direcion he origin he nd he nelope. (We beween 1 s nd 2 s fer quniies. We use cheeh runs, so ehicle. won use he nelope s i begins running. We subscrips 1 nd 2 for h our lues will be posiie. posiion bu we don know h ye.) plce dos o represen hose poins. he poins 5 1 s nd 5 2 s.
42 CHAPTER 2 Moion Along Srigh Line 2.3 Aerge nd Insnneous Accelerion 43 2.7 Using n - grph o go from (), (b) erge -elociy o (c) insnneous -elociy. In (c) we find he slope of he ngen o he - cure by diiding ny ericl inerl (wih disnce unis) long he ngen by he corresponding horizonl inerl (wih ime unis). () (b) (c) (m) 4 D 5 2. s 3 D 5 15 m - 5 75 m/s 2 p 2 1 NLINE 1.1 Anlyzing Moion Using Digrms D p 1 D 1 2 3 4 5 (s) As he erge -elociy - is clculed oer shorer nd shorer ime inerls... Finding Velociy on n - Grph The -elociy of pricle cn lso be found from he grph of is posiion s funcion of ime. Suppose we wn o find he -elociy of he drgser in Fig. 2.1 poin P 1. As poin P 2 in Fig. 2.1 pproches poin P 1, poin p 2 in he - grphs of Figs. 2.7 nd 2.7b pproches poin p 1 nd he erge -elociy is clculed oer shorer ime inerls D. In he limi h D S, shown in Fig. 2.7c, he slope of he line p 1 p 2 equls he slope of he line ngen o he cure poin p 1. Thus, on grph of posiion s funcion of ime for srighline moion, he insnneous -elociy ny poin is equl o he slope of he ngen o he cure h poin. If he ngen o he - cure slopes upwrd o he righ, s in Fig. 2.7c, hen is slope is posiie, he -elociy is posiie, nd he moion is in he posiie -direcion. If he ngen slopes downwrd o he righ, he slope of he - grph nd he -elociy re negie, nd he moion is in he negie -direcion. When he ngen is horizonl, he slope nd he -elociy re zero. Figure 2.8 illusres hese hree possibiliies. Figure 2.8 cully depics he moion of pricle in wo wys: s () n - grph nd (b) moion digrm. A moion digrm shows he pricle s posi- (m) 4 3 2 1 D 5 1. s D 5 55 m - 5 55 m/s p 2 p 1 D D 1 2 3 4 5 (s)... is lue - 5 D/D pproches he insnneous -elociy. (m) 4 3 2 1 5 16 m 4. s 5 4 m/s Slope of ngen 5 insnneous -elociy p 1 4. s 16 m 1 2 3 4 5 (s) The insnneous -elociy ny gien poin equls he slope of he ngen o he - cure h poin. ion rious imes (like frmes from ideo of he pricle s moion) s well s rrows o represen he pricle s elociy ech insn. We will use boh - grphs nd moion digrms in his chper o help you undersnd moion. You will find i worh your while o drw boh n - grph nd moion digrm s pr of soling ny problem inoling moion. Tes Your Undersnding of Secion 2.2 Figure 2.9 is n - grph of he moion of pricle. () Rnk he lues of he pricle s -elociy he poins P, Q, R, nd S from mos posiie o mos negie. (b) A which poins is posiie? (c) A which poins is negie? (d) A which poins is zero? (e) Rnk he lues of he pricle s speed he poins P, Q, R, nd S from fses o slowes. 2.3 Aerge nd Insnneous Accelerion Jus s elociy describes he re of chnge of posiion wih ime, ccelerion describes he re of chnge of elociy wih ime. Like elociy, ccelerion is ecor quniy. When he moion is long srigh line, is only nonzero componen is long h line. As we ll see, ccelerion in srigh-line moion cn refer o eiher speeding up or slowing down. Aerge Accelerion Le s consider gin pricle moing long he -is. Suppose h ime 1 he pricle is poin P 1 nd hs -componen of (insnneous) elociy 1, nd ler ime 2 i is poin P 2 nd hs -componen of elociy 2. So he -componen of elociy chnges by n moun D 5 2 2 1 during he ime inerl D 5 2 2 1. We define he erge ccelerion of he pricle s i moes from P 1 o P 2 o be ecor quniy whose -componen - (clled he erge -ccelerion) equls D, he chnge in he -componen of elociy, diided by he ime inerl D: - 5 2 2 1 2 2 1 5 D D (erge -ccelerion, srigh-line moion) (2.4) 2.9 An - grph for pricle. Q P R S 2.8 () The - grph of he moion of priculr pricle. The slope of he ngen ny poin equls he elociy h poin. (b) A moion digrm showing he posiion nd elociy of he pricle ech of he imes lbeled on he - grph. () - grph A Slope zero: 5 C B Slope posiie:. D E Slope negie:, (b) Pricle s moion A 5 The seeper he slope (posiie or negie) of n objec s - grph, he greer is he objec s speed in he posiie or negie -direcion. B C D E 5 The pricle is, nd moing in he 1-direcion. From A o B i speeds up,...... nd from B o C i slows down, hen hls momenrily C. From C o D i speeds up in he 2-direcion,...... nd from D o E i slows down in he 2-direcion. For srigh-line moion long he -is we will ofen cll - simply he erge ccelerion. (We ll encouner he oher componens of he erge ccelerion ecor in Chper 3.) If we epress elociy in meers per second nd ime in seconds, hen erge ccelerion is in meers per second per second, or 1 m/s 2 /s. This is usully wrien s m/s 2 nd is red meers per second squred. CAUTIN Accelerion s. elociy Be ery creful no o confuse ccelerion wih elociy! Velociy describes how body s posiion chnges wih ime; i ells us how fs nd in wh direcion he body moes. Accelerion describes how he elociy chnges wih ime; i ells us how he speed nd direcion of moion re chnging. I my help o remember he phrse ccelerion is o elociy s elociy is o posiion. I cn lso help o imgine yourself riding long wih he moing body. If he body cceleres forwrd nd gins speed, you would feel pushed bckwrd in your se; if i cceleres bckwrd nd loses speed, you would feel pushed forwrd. If he elociy is consn nd here s no ccelerion, you would feel neiher sension. (We ll see he reson for hese sensions in Chper 4.)
44 CHAPTER 2 Moion Along Srigh Line 2.3 Aerge nd Insnneous Accelerion 45 Emple 2.2 Aerge ccelerion An sronu hs lef n orbiing spcecrf o es new personl mneuering uni. As she moes long srigh line, her prner on he spcecrf mesures her elociy eery 2. s, sring ime 5 1. s: 1. s 3. s 5. s 7. s Find he erge -ccelerion, nd describe wheher he speed of he sronu increses or decreses, for ech of hese ime inerls: () 1 5 1. s o 2 5 3. s; (b) 1 5 5. s o 2 5 7. s; (c) 1 5 9. s o 2 5 11. s; (d) 1 5 13. s o 2 5 15. s. SLUTIN IDENTIFY: We ll need he definiion of erge ccelerion -. To find he chnges in speed, we ll use he ide h speed is he mgniude of he insnneous elociy. SET UP: Figure 2.1 shows our grphs. We use Eq. (2.4) o find he lue of from he chnge in elociy for ech ime inerl. -.8 m/s 1.2 m/s 1.6 m/s 1.2 m/s EXECUTE: In he upper pr of Fig. 2.1, we grph he -elociy s funcion of ime. n his - grph, he slope of he line connecing he poins he beginning nd end of ech inerl equls he erge -ccelerion - 5D /D for h inerl. In he lower pr of Fig. 2.1, we grph he lues of -. We find: () - 5 1 1.2 m/s 2.8 m/s 2 / 1 3. s 2 1. s 2 5.2 m/s 2. The speed (mgniude of insnneous -elociy) increses from.8 m/s o 1.2 m/s. (b) - 5 1 1.2 m/s 2 1.6 m/s 2 / 1 7. s 2 5. s 2 5 2.2 m/s 2. The speed decreses from 1.6 m/s o 1.2 m/s. (c) - 5 321. m/s 2 1 2.4 m/s 24/ 1 11. s 2 9. s 2 5 9. s 11. s 13. s 15. s 2.11 A Grnd Pri cr wo poins on he srighwy. 2.4 m/s 21. m/s 21.6 m/s 2.8 m/s 2.1 ur grphs of -elociy ersus ime (op) nd erge -ccelerion ersus ime (boom) for he sronu. 2.3 m/s 2. The speed increses from.4 m/s o 1. m/s. (d).4 m/s 2. The speed decreses from 1.6 m/s o.8 m/s. EVALUATE: ur resuls show h when he erge -ccelerion hs he sme direcion (sme lgebric sign) s he iniil elociy, s in inerls () nd (c), he sronu goes fser; when i hs he opposie direcion (opposie lgebric sign), s in inerls (b) nd (d), she slows down. Thus posiie -ccelerion mens speeding up if he -elociy is posiie [inerl ()] bu slowing down if he -elociy is negie [inerl (d)]. Similrly, negie -ccelerion mens speeding up if he -elociy is negie [inerl (c)] bu slowing down if he -elociy is posiie [inerl (b)]. Insnneous Accelerion We cn now define insnneous ccelerion following he sme procedure h we used o define insnneous elociy. As n emple, suppose rce cr drier is driing long srighwy s shown in Fig. 2.11. To define he insnneous ccelerion poin P 1, we ke he second poin P 2 in Fig. 2.11 o be closer nd closer o P 1 so h he erge ccelerion is compued oer shorer nd shorer ime inerls. The insnneous ccelerion is he limi of he erge ccelerion s he ime inerl pproches zero. In he lnguge of clculus, insnneous ccelerion equls he insnneous re of chnge of elociy wih ime. Thus D 5 lim 5 d DS D d P 1 The slope of he line connecing ech pir of poins on he - grph...... equls he erge -ccelerion beween hose poins. - 5 32.8 m / s 2 1 21.6 m / s 24/ 1 15. s 2 13. s 2 5 Speed 1 -elociy 1 (insnneous -ccelerion, srigh-line moion) P 2 Speed 2 -elociy 2 (2.5) Noe h in Eq. (2.5) is relly he -componen of he ccelerion ecor, or he insnneous -ccelerion; in srigh-line moion, ll oher componens of his ecor re zero. From now on, when we use he erm ccelerion, we will lwys men insnneous ccelerion, no erge ccelerion. Emple 2.3 Suppose he -elociy gien by he equion Aerge nd insnneous ccelerions of he cr in Fig. 2.11 ny ime is () Find he chnge in -elociy of he cr in he ime inerl beween 1 5 1. s nd 2 5 3. s. (b) Find he erge -ccelerion in his ime inerl. (c) Find he insnneous -ccelerion ime 1 5 1. s by king D o be firs.1 s, hen.1 s, hen.1 s. (d) Derie n epression for he insnneous -ccelerion ny ime, nd use i o find he -ccelerion 5 1. s nd 5 3. s. SLUTIN IDENTIFY: This emple is nlogous o Emple 2.1 in Secion 2.2. (Now is good ime o reiew h emple.) There we found he erge -elociy oer shorer nd shorer ime inerls from he chnge in posiion, nd we deermined he insnneous -elociy by differeniing he posiion s funcion of ime. In his emple, we find he erge -ccelerion from he chnge in -elociy oer ime inerl. Likewise, we find he insnneous -ccelerion by differeniing he -elociy s funcion of ime. SET UP: We ll use Eq. (2.4) for erge -ccelerion nd Eq. (2.5) for insnneous -ccelerion. EXECUTE: () We firs find he -elociy ech ime by subsiuing ech lue of ino he equion. A ime 1 5 1. s, 1 5 6 m/s 1 1.5 m/s 3 211. s 2 2 5 6.5 m/s A ime 2 5 3. s, 2 5 6 m/s 1 1.5 m/s 3 213. s 2 2 5 64.5 m/s The chnge in -elociy D is 5 6 m/s 1 1.5 m/s 3 2 2 D 5 2 2 1 5 64.5 m/s 2 6.5 m/s 5 4. m/s The ime inerl is D 5 3. s 2 1. s 5 2. s. Finding Accelerion on - Grph or n - Grph In Secion 2.2 we inerpreed erge nd insnneous -elociy in erms of he slope of grph of posiion ersus ime. In he sme wy, we cn inerpre erge nd insnneous -ccelerion by using grph wih insnneous elociy on he ericl is nd ime on he horizonl is h is, - grph (Fig. 2.12). The poins on he grph lbeled p 1 nd p 2 correspond o poins P 1 nd P 2 in Fig. 2.11. The erge -ccelerion - 5D /D during his inerl is he slope of he line p 1 p 2. As poin P 2 in Fig. 2.11 pproches poin P 1, poin p 2 in he - grph of Fig. 2.12 pproches poin p 1, nd he slope of he line p 1 p 2 pproches he slope of he line ngen o he cure poin p 1. Thus, on grph of -elociy s funcion of ime, he insnneous -ccelerion ny poin is equl o he slope of he ngen o he cure h poin. Tngens drwn (b) The erge -ccelerion during his ime inerl is During he ime inerl from 1 5 1. s o 2 5 3. s, he -elociy nd erge -ccelerion he he sme lgebric sign (in his cse, posiie), nd he cr speeds up. (c) When D 5.1 s, 2 5 1.1 s nd we find - 5 D D You should do hese clculions for D 5.1 s nd D 5.1 s; he resuls re nd - 5 1.5 m/s 2-5 1.5 m/s 2, respeciely. As D ges smller, he erge -ccelerion ges closer o 1. m/s 2, so he insnneous -ccelerion 5 1. s is 1. m/s 2. (d) The insnneous -ccelerion is 5 d /d. The deriie of consn is zero nd he deriie of is 2, so When 5 1. s, When 5 3. s, - 5 2 2 1 5 4. m /s 5 2. m/s 2 2 2 1 2. s 2 5 6 m/s 1 1.5 m/s 3 211.1 s 2 2 5 6.65 m/s D 5.15 m/s 5.15 m /s.1 s 5 1.5 m/s 2 5 d 5 d d d 36 m /s 1 1.5 m/s 3 2 2 4 5 1.5 m/s 3 212 2 5 1 1. m/s 3 2 5 1 1. m/s 3 211. s 2 5 1. m/s 2 5 1 1. m/s 3 213. s 2 5 3. m/s 2 EVALUATE: Noe h neiher of he lues we found in pr (d) is equl o he erge -ccelerion found in pr (b). Th s becuse he cr s insnneous -ccelerion ries wih ime. The re of chnge of ccelerion wih ime is someimes clled he jerk. 2
46 CHAPTER 2 Moion Along Srigh Line 2.4 Moion wih Consn Accelerion 47 2.12 A - grph of he moion in Fig. 2.11. () - grph for n objec moing on he -is A B Slope zero: 5 C Slope posiie:. Slope negie:, The seeper he slope (posiie or negie) of n objec s - grph, he greer is he objec s ccelerion in he posiie or negie -direcion. D E 2 1 For displcemen long he -is, n objec s erge -ccelerion equls he slope of line connecing he corresponding poins on grph of -elociy ( ) ersus ime (). p 1 1 Slope 5 erge ccelerion D 5 2 2 1 differen poins long he cure in Fig. 2.12 he differen slopes, so he insnneous -ccelerion ries wih ime. CAUTIN The signs of -ccelerion nd -elociy By iself, he lgebric sign of he -ccelerion does no ell you wheher body is speeding up or slowing down. You mus compre he signs of he -elociy nd he -ccelerion. When nd he he sme sign, he body is speeding up. If boh re posiie, he body is moing in he posiie direcion wih incresing speed. If boh re negie, he body is moing in he negie direcion wih n -elociy h is becoming more nd more negie, nd gin he speed is incresing. When nd he opposie signs, he body is slowing down. If is posiie nd is negie, he body is moing in he posiie direcion wih decresing speed; if is negie nd is posiie, he body is moing in he negie direcion wih n -elociy h is becoming less negie, nd gin he body is slowing down. Figure 2.13 illusres some of hese possibiliies. The erm decelerion is someimes used for decrese in speed. Becuse i my men posiie or negie, depending on he sign of, we oid his erm. We cn lso lern bou he ccelerion of body from grph of is posiion ersus ime. Becuse 5 d /d nd 5 d/d, we cn wrie 5 d d p 2 2 D 5 2 2 1 Slope of ngen o - cure gien poin 5 insnneous -ccelerion h poin. 5 d d 1 d d 2 5 d 2 d 2 2.13 () A - grph of he moion of differen pricle hn h shown in Fig. 2.8. The slope of he ngen ny poin equls he -ccelerion h poin. (b) A moion digrm showing he posiion, elociy, nd ccelerion of he pricle ech of he imes lbeled on he - grph. The posiions re consisen wih he - grph; for insnce, from A o B he elociy is negie, so he pricle is more negie lue of hn A. B (b) bjec s posiion, elociy, nd ccelerion on he -is A 5 B C D E 5 5 5 (2.6) bjec is,, moing in he 2-direcion (, ), nd slowing down ( nd he opposie signs). bjec is,, insnneously res ( 5 ), nd bou o moe in he 1-direcion (. ). bjec is., moing in he 1-direcion (. ); is speed is insnneously no chnging ( 5 ). bjec is., insnneously res ( 5 ), nd bou o moe in he 2-direcion (, ). bjec is., moing in he 2-direcion (, ), nd speeding up ( nd he he sme sign).? 2.14 () The sme - grph s shown in Fig. 2.8. The -elociy is equl o he slope of he grph, nd he ccelerion is gien by he conciy or curure of he grph. (b) A moion digrm showing he posiion, elociy, nd ccelerion of he pricle ech of he imes lbeled on he - grph. () - grph A Slope zero: 5 Curure downwrd:, C E B Slope negie:, Curure zero: 5 Slope posiie:. Curure zero: 5 Slope posiie:. Curure upwrd:. Slope negie:, Curure upwrd:. The greer he curure (upwrd or downwrd) of n objec s - grph, he greer is he objec s ccelerion in he posiie or negie -direcion. D (b) bjec s moion A 5 Th is, is he second deriie of wih respec o. The second deriie of ny funcion is direcly reled o he conciy or curure of he grph of h funcion. A poin where he - grph is conce up (cured upwrd), he -ccelerion is posiie nd is incresing; poin where he - grph is conce down (cured downwrd), he -ccelerion is negie nd is decresing. A poin where he - grph hs no curure, such s n inflecion poin, he -ccelerion is zero nd he elociy is no chnging. Figure 2.14 shows ll hree of hese possibiliies. Emining he curure of n - grph is n esy wy o decide wh he sign of ccelerion is. This echnique is less helpful for deermining numericl lues of ccelerion becuse he curure of grph is hrd o mesure ccurely. B C D E 5 5 5 Tes Your Undersnding of Secion 2.3 Look gin he - grph in Fig. 2.9 he end of Secion 2.2. () A which of he poins P, Q, R, nd S is he -ccelerion posiie? (b) A which poins is he -ccelerion negie? (c) A which poins does he -ccelerion pper o be zero? (d) A ech poin se wheher he speed is incresing, decresing, or no chnging. 2.4 Moion wih Consn Accelerion The simples kind of ccelered moion is srigh-line moion wih consn ccelerion. In his cse he elociy chnges he sme re hroughou he moion. This is ery specil siuion, ye one h occurs ofen in nure. A flling body hs consn ccelerion if he effecs of he ir re no imporn. The sme is rue for body sliding on n incline or long rough horizonl surfce. Srigh-line moion wih nerly consn ccelerion lso occurs in echnology, such s n irplne being cpuled from he deck of n ircrf crrier. Figure 2.15 is moion digrm showing he posiion, elociy, nd ccelerion for pricle moing wih consn ccelerion. Figures 2.16 nd 2.17 depic his sme moion in he form of grphs. Since he -ccelerion is consn, he - grph (grph of -ccelerion ersus ime) in Fig. 2.16 is horizonl line. The grph of -elociy ersus ime, or - grph, hs consn slope becuse he ccelerion is consn, so his grph is srigh line (Fig. 2.17). bjec is,, moing in he 1-direcion (. ), nd speeding up ( nd he he sme sign). bjec is 5, moing in he 1-direcion (. ); speed is insnneously no chnging ( 5 ). bjec is., insnneously res ( 5 ), nd bou o moe in he 2-direcion (, ). bjec is., moing in he 2-direcion (, ); speed is insnneously no chnging ( 5 ). bjec is., moing in he 2-direcion (, ), nd slowing down ( nd he opposie signs). 2.15 A moion digrm for pricle moing in srigh line in he posiie -direcion wih consn posiie -ccelerion. The posiion, elociy, nd ccelerion re shown fie eqully spced imes. If pricle moes in srigh line wih consn -ccelerion...... he -elociy chnges by equl mouns in equl ime inerls. D 2D 3D 4D Howeer, he posiion chnges by differen mouns in equl ime inerls becuse he elociy is chnging.
48 CHAPTER 2 Moion Along Srigh Line 2.4 Moion wih Consn Accelerion 49 2.16 An ccelerion-ime 1-2 grph for srigh-line moion wih consn posiie -ccelerion. Consn -ccelerion: - grph is horizonl line (slope 5 ). Are under - grph 5 2 5 chnge in -elociy from ime o ime. 2.17 A elociy-ime 1-2 grph for srigh-line moion wih consn posiie -ccelerion. The iniil -elociy is lso posiie in his cse. Consn -ccelerion: - grph is srigh line. Slope 5 -ccelerion During ime inerl, he -elociy chnges by 2 5. Tol re under - grph 5 2 5 chnge in -coordine from ime o ime. NLINE 1.1 Anlyzing Moion Using Digrms 1.2 Anlyzing Moion Using Grphs 1.3 Predicing Moion from Grphs 1.4 Predicing Moion from Equions 1.5 Problem-Soling Sregies for Kinemics 1.6 Skier Rces Downhill When he -ccelerion is consn, he erge -ccelerion - for ny ime inerl is he sme s. This mkes i esy o derie equions for he posiion nd he -elociy s funcions of ime. To find n epression for, we firs replce - in Eq. (2.4) by : Now we le 1 5 nd le 2 be ny ler ime. We use he symbol for he -elociy he iniil ime 5 ; he -elociy he ler ime is. Then Eq. (2.7) becomes 5 1 (2.7) (consn -ccelerion only) (2.8) We cn inerpre his equion s follows. The -ccelerion is he consn re of chnge of -elociy h is, he chnge in -elociy per uni ime. The erm is he produc of he chnge in -elociy per uni ime,, nd he ime inerl. Therefore i equls he ol chnge in -elociy from he iniil ime 5 o he ler ime. The -elociy ny ime hen equls he iniil -elociy ( 5 ) plus he chnge in -elociy (see Fig. 2.17). Anoher inerpreion of Eq. (2.8) is h he chnge in -elociy 2 of he pricle beween 5 nd ny ler ime equls he re under he - grph beween hose wo imes. In Fig. 2.16, he re under he grph of -ccelerion ersus ime is recngle of ericl side nd horizonl side. The re of his recngle is, which from Eq. (2.8) is indeed equl o he chnge in elociy 2. In Secion 2.6 we ll show h een if he -ccelerion is no consn, he chnge in -elociy during ime inerl is sill equl o he re under he - cure, lhough in h cse Eq. (2.8) does no pply. Ne we ll derie n equion for he posiion s funcion of ime when he -ccelerion is consn. To do his, we use wo differen epressions for he erge -elociy - during he inerl from 5 o ny ler ime. The firs epression comes from he definiion of -, Eq. (2.2), which is rue wheher or no he ccelerion is consn. We cll he posiion ime 5 he iniil posiion, denoed by. The posiion he ler ime is simply. Thus for he ime inerl D 5 2 he displcemen is D 5 2, nd Eq. (2.2) gies We cn lso ge second epression for - h is lid only when he -ccelerion is consn, so h he - grph is srigh line (s in Fig. 2.17) nd he -elociy chnges consn re. In his cse he erge -elociy during ny ime inerl is simply he rihmeic erge of he -elociies he beginning nd end of he inerl. For he ime inerl o, - 5 1 2 (2.9) (consn -ccelerion only) (2.1) (This equion is no rue if he -ccelerion ries nd he - grph is cure, s in Fig. 2.13.) We lso know h wih consn -ccelerion, he -elociy ny ime is gien by Eq. (2.8). Subsiuing h epression for ino Eq. (2.1), we find - 5 1 2 1 1 1 2 5 1 1 2 5 2 2 1 2 2 1 5 2 2-5 2 or (consn -ccelerion only) (2.11) Finlly, we se Eqs. (2.9) nd (2.11) equl o ech oher nd simplify: 5 1 1 1 2 2 1 1 2 5 2 (consn -ccelerion only) (2.12) Here s wh Eq. (2.12) ells us: If ime 5 pricle is posiion nd hs -elociy, is new posiion ny ler ime is he sum of hree erms is iniil posiion, plus he disnce h i would moe if is -elociy were 1 consn, plus n ddiionl disnce 2 2 cused by he chnge in -elociy. A grph of Eq. (2.12) h is, n - grph for moion wih consn -ccelerion (Fig. 2.18) is lwys prbol. Figure 2.18b shows such grph. The cure inerceps he ericl is (-is), he posiion 5. The slope of he ngen 5 equls, he iniil -elociy, nd he slope of he ngen ny ime equls he -elociy h ime. The slope nd -elociy re coninuously incresing, so he -ccelerion is posiie; you cn lso see his becuse he grph in Fig. 2.18b is conce up (i cures upwrd). If is negie, he - grph is prbol h is conce down (hs downwrd curure). If here is zero -ccelerion, he - grph is srigh line; if here is consn -ccelerion, he ddiionl 2 2 erm in Eq. (2.12) for s funcion of 1 cures he grph ino prbol (Fig. 2.19). We cn nlyze he - grph in he sme wy. If here is zero -ccelerion his grph is horizonl line (he -elociy is consn); dding consn -ccelerion gies slope o he - grph (Fig. 2.19b). () A rce cr moes in he -direcion wih consn ccelerion. 5 1 During ime inerl, he -elociy chnges by 2 5. () An - grph for n objec moing wih posiie consn -ccelerion The grph wih consn -ccelerion: 1 5 1 1 2 2 The effec of -ccelerion: 1 2 2 The grph we would ge wih zero -ccelerion: 5 1 or (b) The - grph Slope 5 Consn -ccelerion: - grph is prbol. Slope 5 (b) The - grph for he sme objec The grph wih consn -ccelerion: 5 1 The dded elociy due o -ccelerion: The grph wih zero -ccelerion: 5 NLINE 1.8 Se Bels Se Lies 1.9 Screeching o Hl 1.1 Cr Srs, Then Sops 1.11 Soling Two-Vehicle Problems 1.12 Cr Cches Truck 1.13 Aoiding Rer-End Collision 2.18 () Srigh-line moion wih consn ccelerion. (b) A posiion-ime (-) grph for his moion (he sme moion s is shown in Figs. 2.15, 2.16, nd 2.17). For his moion he iniil posiion, he iniil elociy, nd he ccelerion re ll posiie. 2.19 () How consn -ccelerion ffecs body s () - grph nd (b) - grph.
5 CHAPTER 2 Moion Along Srigh Line 2.4 Moion wih Consn Accelerion 51 Jus s he chnge in -elociy of he pricle equls he re under he - grph, he displcemen h is, he chnge in posiion equls he re under he - grph. To be specific, he displcemen 2 of he pricle beween 5 nd ny ler ime equls he re under he - grph beween hose wo imes. In Fig. 2.17 he re under he grph is diided ino drkcolored recngle of ericl side nd horizonl side nd ligh-colored righ ringle of ericl side nd horizonl side. The re of he recngle 1 is nd he re of he ringle is 1 2 212 5 1 2 2, so he ol re under he - grph is in greemen wih Eq. (2.12). The displcemen during ime inerl cn lwys be found from he re under he - cure. This is rue een if he ccelerion is no consn, lhough in h cse Eq. (2.12) does no pply. (We ll show his in Secion 2.6.) We cn check wheher Eqs. (2.8) nd (2.12) re consisen wih he ssumpion of consn ccelerion by king he deriie of Eq. (2.12). We find which is Eq. (2.8). Differeniing gin, we find simply which grees wih he definiion of insnneous -ccelerion. I s ofen useful o he relionship beween posiion, -elociy, nd (consn) -ccelerion h does no inole he ime. To obin his, we firs sole Eq. (2.8) for, hen subsiue he resuling epression ino Eq. (2.12), nd simplify: 5 2 5 1 2 5 1 1 2 2 5 d d 5 1 d d 5 1 2 2 1 1 2 1 2 2 2 Equions (2.8), (2.12), (2.13), nd (2.14) re he equions of moion wih consn ccelerion. By using hese equions, we cn sole ny problem inoling srigh-line moion of pricle wih consn ccelerion. For he priculr cse of moion wih consn -ccelerion depiced in Fig. 2.15 nd grphed in Figs. 2.16, 2.17, nd 2.18, he lues of,, nd re ll posiie. We inie you o redrw hese figures for cses in which one, wo, or ll hree of hese quniies re negie. A specil cse of moion wih consn -ccelerion occurs when he -ccelerion is zero. The -elociy is hen consn, nd he equions of moion become simply Problem-Soling Sregy 2.1 5 5 consn 5 1 Moion wih Consn Accelerion IDENTIFY he relen conceps: In mos srigh-line moion problems, you cn use he consn-ccelerion equions. ccsionlly, howeer, you will encouner siuion in which he ccelerion isn consn. In such cse, you ll need differen pproch (see Secion 2.6). SET UP he problem using he following seps: 1. Firs decide where he origin of coordines is nd which is direcion is posiie. I is ofen esies o plce he pricle he origin ime 5 ; hen 5. I helps o mke moion digrm showing he coordines nd some ler posiions of he pricle. 2. Remember h your choice of he posiie is direcion uomiclly deermines he posiie direcions for -elociy nd -ccelerion. If is posiie o he righ of he origin, hen nd re lso posiie owrd he righ. 3. Rese he problem in words, nd hen rnsle i ino symbols nd equions. When does he pricle rrie cerin poin (h is, wh is he lue of )? Where is he pricle when is -elociy hs specified lue (h is, wh is he lue of when hs he specified lue)? Emple 2.4 sks, Where is he moorcyclis when his elociy is 25 m/s? In symbols, his sys Wh is he lue of when 5 25 m/s? 4. Mke lis of quniies such s,,,,, nd. In generl, some of hem will be known nd some will be unknown. Wrie down he lues of he known quniies, nd decide which of he unknowns re he rge ribles. Be on he lookou for implici informion. For emple, A cr sis sopligh usully mens 5. EXECUTE he soluion: Choose n equion from Eqs. (2.8), (2.12), (2.13), nd (2.14) h conins only one of he rge ribles. Sole his equion for he rge rible, using symbols only. Then subsiue he known lues nd compue he lue of he rge rible. Someimes you will he o sole wo simulneous equions for wo unknown quniies. EVALUATE your nswer: Tke hrd look your resuls o see wheher hey mke sense. Are hey wihin he generl rnge of lues you epeced? We rnsfer he erm Finlly, simplifying gies us 2 5 2 1 2 1 2 2 o he lef side nd muliply hrough by 2 : 2 1 2 2 5 2 2 2 2 1 2 2 2 2 1 (consn -ccelerion only) (2.13) We cn ge one more useful relionship by equing he wo epressions for -, Eqs. (2.9) nd (2.1), nd muliplying hrough by. Doing his, we obin 2 5 1 1 2 2 (consn -ccelerion only) (2.14) Noe h Eq. (2.14) does no conin he -ccelerion. This equion cn be hndy when is consn bu is lue is unknown. Emple 2.4 Consn-ccelerion clculions A moorcyclis heding es hrough smll Iow ciy cceleres fer he psses he signpos mrking he ciy limis (Fig. 2.2). His ccelerion is consn 4. m/s 2. A ime 5 he is 5. m es of he signpos, moing es 15 m/s. () Find his posiion nd elociy ime 5 2. s. (b) Where is he moorcyclis when his elociy is 25 m/s? 2.2 A moorcyclis reling wih consn ccelerion. SAGE 19651 AW 5 5. m 5 5 15 m/s 5 4. m/s 2 19651 AW 5? 5 2. s 5? (es) SLUTIN IDENTIFY: The problem semen ells us h he ccelerion is consn, so we cn use he consn-ccelerion equions. SET UP: We ke he signpos s he origin of coordines 1 5 2, nd choose he posiie -is o poin es (see Fig. 2.2, which lso seres s moion digrm). A he iniil ime 5, he iniil posiion is 5 5. m nd he iniil -elociy is 5 15 m/s. The consn -ccelerion is 5 4. m/s 2. The unknown rge ribles in pr () re he lues of he posiion nd he -elociy he ler ime 5 2. s; he rge rible in pr (b) is he lue of when 5 25 m/s. Coninued
52 CHAPTER 2 Moion Along Srigh Line 2.5 Freely Flling Bodies 53 EXECUTE: () We cn find he posiion 5 2. s by using Eq. (2.12), which gies s funcion of ime : 5 1 1 1 2 2 5 5. m 1 1 15 m/s 212. s 2 1 1 2 1 4. m /s 2 212. s 2 2 5 43 m We cn find he -elociy his sme ime by using Eq. (2.8), which gies s funcion of ime : 5 1 5 15 m/s 1 1 4. m/s 2 212. s 2 5 23 m/s (b) We wn o find he lue of when 5 25 m/s, bu we don know he ime when he moorcycle hs his -elociy. Hence we use Eq. (2.13), which inoles,, nd bu does no inole : 2 5 2 1 2 1 2 2 Soling for nd subsiuing in he known lues, we find 5 1 2 2 2 2 5 5. m 1 1 25 m /s 2 2 2 1 15 m/s 2 2 2 1 4. m/s 2 2 5 55 m An lernie bu longer roue o he sme nswer is o use Eq. (2.8) o firs find he ime when 5 25 m/s: Gien he ime, we cn find using Eq. (2.12): 5 1 1 1 2 2 5 5. m 1 1 15 m/s 212.5 s 2 1 1 2 1 4. m /s 2 212.5 s 2 2 5 55 m 5 2 5 1 so 5 25 m /s 2 15 m/s 4. m/s 2 5 2.5 s EVALUATE: Do hese resuls mke sense? According o our resuls in pr (), he moorcyclis cceleres from 15 m/s (bou 34 mi/h, or 54 km/h) o 23 m/s (bou 51 mi/h, or 83 km/h) in 2. s while reling disnce of 38 m (bou 125 f). This is prey brisk ccelerion, bu well wihin he cpbiliies of highperformnce bike. Compring our resuls in pr (b) wih hose in pr () ells us h he moorcycle ins n -elociy 5 25 m/s ler ime nd fer reling greer disnce hn when he moorcycle hd 5 23 m/s. This mkes sense, since he moorcycle hs posiie -ccelerion nd so is -elociy is incresing. prs () nd (c), nd Eq. (2.8) (which reles elociy nd ime) in pr (b). EXECUTE: () To find he lue of he ime when he mooris nd he police officer re he sme posiion, we pply Eq. (2.12), 5 1 1 1 2 2, o ech ehicle: M 5 1 M 1 1 2 1 2 2 5 M P 5 1 1 2 1 1 2 P 2 5 1 2 P 2 Since M 5 P ime, we se hese wo epressions equl o ech oher nd sole for : M 5 1 2 P 2 5 or 5 2 M P 5 2 1 15 m /s 2 3. m/s 2 5 1 s There re wo imes when boh he ehicles he he sme -coordine. The firs, 5, is he ime when he mooris psses he prked moorcycle he corner. The second, 5 1 s, is he ime when he officer cches up wih he mooris. Tes Your Undersnding of Secion 2.4 Four possible - grphs re shown for he wo ehicles in Emple 2.5. Which grph is correc? (b) We wn he mgniude of he officer s -elociy P he ime found in pr (). Her elociy ny ime is gien by Eq. (2.8): P 5 P 1 P 5 1 1 3. m/s 2 2 Using 5 1 s, we find P 5 3 m/s. When he officer oerkes he mooris, she is reling wice s fs s he mooris is. (c) In 1 s he disnce he mooris rels is M 5 M 5 1 15 m/s 211 s 2 5 15 m nd he disnce he officer rels is P 5 1 2 P 2 5 1 2 1 3. m /s 2 211 s 2 2 5 15 m This erifies h he ime he officer cches he mooris, hey he gone equl disnces. EVALUATE: Figure 2.21b shows grphs of ersus for ech ehicle. We see gin h here re wo imes when he wo posiions re he sme (where he wo grphs cross). A neiher of hese imes do he wo ehicles he he sme elociy (i.e., where he wo grphs cross, heir slopes re differen). A 5, he officer is res; 5 1 s, he officer hs wice he speed of he mooris. () (b) (c) (d) Emple 2.5 Two bodies wih differen ccelerions A mooris reling wih consn speed of 15 m/s (bou SET UP: We ke he origin he corner, so 5 for boh, 34 mi/h) psses school-crossing corner, where he speed limi is nd we ke he posiie direcion o he righ. Le P (for police) 1 m/s (bou 22 mi/h). Jus s he mooris psses, police officer be he officer s posiion nd M (for mooris) be he mooris s on moorcycle sopped he corner srs off in pursui wih consn ccelerion of 3. m/s 2 (Fig. 2.21). () How much ime officer nd M 5 15 m/s for he mooris; he consn -cceler- posiion ny ime. The iniil -elociies re P 5 for he elpses before he officer cches up wih he mooris? (b) Wh is ions re P 5 3. m/s 2 for he officer nd M 5 for he he officer s speed h poin? (c) Wh is he ol disnce ech mooris. ur rge rible in pr () is he ime when he officer ehicle hs reled h poin? cches he mooris h is, when he wo ehicles re he sme posiion. In pr (b) we re looking for he officer s speed SLUTIN (he mgniude of his elociy) he ime found in pr (). In IDENTIFY: The police officer nd he mooris boh moe wih pr (c) we wn o find he posiion of eiher ehicle his sme consn ccelerion (equl o zero for he mooris), so we cn ime. Hence we use Eq. (2.12) (which reles posiion nd ime) in use he formuls we he deeloped. 2.21 () Moion wih consn ccelerion oerking moion wih consn elociy. (b) A grph of ersus for ech ehicle. () SCHL CRSSING Police officer: iniilly res, consn -ccelerion PLICE P P 5 3. m/s 2 Mooris: consn -elociy M M 5 15 m/s (b) 16 12 (m) 8 The police officer nd mooris mee he ime where heir - grphs cross. Mooris 4 fficer (s) 2 4 6 8 1 12 Mooris fficer (s) 1 2.5 Freely Flling Bodies Mooris fficer (s) 1 The mos fmilir emple of moion wih (nerly) consn ccelerion is body flling under he influence of he erh s griionl rcion. Such moion hs held he enion of philosophers nd scieniss since ncien imes. In he fourh cenury B.C., Arisole hough (erroneously) h hey bodies fll fser hn ligh bodies, in proporion o heir weigh. Nineeen cenuries ler, Glileo (see Secion 1.1) rgued h body should fll wih downwrd ccelerion h is consn nd independen of is weigh. Eperimen shows h if he effecs of he ir cn be negleced, Glileo is righ; ll bodies priculr locion fll wih he sme downwrd ccelerion, regrdless of heir size or weigh. If in ddiion he disnce of he fll is smll compred wih he rdius of he erh, nd if we ignore smll effecs due o he erh s roion, he ccelerion is consn. The idelized moion h resuls under ll of hese ssumpions is clled free fll, lhough i includes rising s well s flling moion. (In Chper 3 we will eend he discussion of free fll o include he moion of projeciles, which moe boh ericlly nd horizonlly.) Figure 2.22 is phoogrph of flling bll mde wih sroboscopic ligh source h produces series of shor, inense flshes. As ech flsh occurs, n imge of he bll h insn is recorded on he phoogrph. There re equl Mooris fficer (s) 1 Mooris fficer (s) 1 2.22 Muliflsh phoo of freely flling bll.
54 CHAPTER 2 Moion Along Srigh Line 2.5 Freely Flling Bodies 55 NLINE 1.7 Blloonis Drops Lemonde 1.1 Pole-Vuler Lnds Emple 2.6 A freely-flling coin A one-euro coin is dropped from he Lening Tower of Pis. I srs from res nd flls freely. Compue is posiion nd elociy fer 1. s, 2. s, nd 3. s. SLUTIN IDENTIFY: Flls freely mens hs consn ccelerion due o griy, so we cn use he consn-ccelerion equions o deermine our rge ribles. 2.23 A coin freely flling from res. The Lening Tower ur skech for he problem ime inerls beween flshes, so he erge elociy of he bll beween successie flshes is proporionl o he disnce beween corresponding imges. The incresing disnces beween imges show h he elociy is coninuously chnging; he bll is ccelering downwrd. Creful mesuremen shows h he elociy chnge is he sme in ech ime inerl, so he ccelerion of he freely flling bll is consn. The consn ccelerion of freely flling body is clled he ccelerion due o griy, nd we denoe is mgniude wih he leer g. We will frequenly use he pproime lue of g or ner he erh s surfce: g 5 9.8 m/s 2 5 98 cm/s 2 5 32 f/s 2 (pproime lue ner he erh s surfce) The ec lue ries wih locion, so we will ofen gie he lue of g he erh s surfce o only wo significn figures. Becuse g is he mgniude of ecor quniy, i is lwys posiie number. n he surfce of he moon, he ccelerion due o griy is cused by he rcie force of he moon rher hn he erh, nd g 5 1.6 m/s 2. Ner he surfce of he sun, g 5 27 m/s 2. In he following emples we use he consn-ccelerion equions deeloped in Secion 2.4. You should reiew Problem-Soling Sregy 2.1 in h secion before you sudy he ne emples. SET UP: The righ side of Fig. 2.23 shows our moion digrm for he coin. The moion is ericl, so we use ericl coordine is nd cll he coordine y insed of. Then we replce ll he s in he consn-ccelerion equions by y s. We ke he origin he sring poin nd he upwrd direcion s posiie. The iniil coordine y nd he iniil y-elociy y re boh zero. The y-ccelerion is downwrd, in he negie y-direcion, so y 52g 529.8 m/s 2. (Remember h, by definiion, g iself is lwys posiie.) ur rge ribles re he lues of y nd y he hree gien imes. To find hese, we use Eqs. (2.12) nd (2.8) wih replced by y. EXECUTE: A ime fer he coin is dropped, is posiion nd y- elociy re y 5 y 1 y 1 1 2 y 2 5 1 1 1 2 1 2g 2 2 5 1 24.9 m/s 2 2 2 y 5 y 1 y 5 1 1 2g 2 5 1 29.8 m/s 2 2 When 5 1. s, y 5 1 24.9 m/s 2 211. s 2 2 524.9 m nd y 5 1 29.8 m/s 2 211. s 2 529.8 m/s; fer 1 s, he coin is 4.9 m below he origin (y is negie) nd hs downwrd elociy ( y is negie) wih mgniude 9.8 m/s. The posiion nd y-elociy 2. s nd 3. s re found in he sme wy. Cn you show h y 5219.6 m nd y 5219.6 m/s 5 2. s, nd h y 5244.1 m nd y 5229.4 m/s 5 3. s? EVALUATE: All our nswers for y re negie becuse we chose he posiie y-is o poin upwrd. Bu we could jus s well he chosen he posiie y-is o poin downwrd. In h cse he ccelerion would he been y 51g nd ll our nswers for y would he been posiie. Eiher choice of is is fine; jus mke sure h you se your choice eplicily in your soluion nd confirm h he ccelerion hs he correc sign. Emple 2.7 Up-nd-down moion in free fll You hrow bll ericlly upwrd from he roof of ll building. The bll lees your hnd poin een wih he roof riling wih n upwrd speed of 15. m/s; he bll is hen in free fll. n is wy bck down, i jus misses he riling. A he locion of he building, g 5 9.8 m/s 2. Find () he posiion nd elociy of he bll 1. s nd 4. s fer leing your hnd; (b) he elociy when he bll is 5. m boe he riling; (c) he mimum heigh reched nd he ime which i is reched; nd (d) he ccelerion of he bll when i is is mimum heigh. SLUTIN IDENTIFY: The words free fll in he semen of he problem men h he ccelerion is consn nd due o griy. ur rge ribles re posiion [in prs () nd (c)], elociy [in prs () nd (b)], nd ccelerion [in pr (d)]. SET UP: In Fig. 2.24 (which is lso moion digrm for he bll) he downwrd ph is displced lile o he righ of is cul posiion for clriy. Tke he origin he poin where he bll lees your hnd, nd ke he posiie direcion o be upwrd. The iniil posiion y is zero, he iniil y-elociy y is 115. m/s, nd he y-ccelerion is y 52g 529.8 m / s 2. We ll gin use Eqs. (2.12) nd (2.8) o find he posiion nd elociy s funcions of ime. In pr (b) we need o find he elociy cerin posiion rher hn cerin ime, so we ll use Eq. (2.13) for h pr. EXECUTE: () The posiion y nd y-elociy y ime fer he bll lees your hnd re gien by Eqs. (2.12) nd (2.8) wih s replced by y s: y 5 y 5 y 1 1 2 y 2 5 y 1 y 1 1 2 1 2g 2 2 5 1 2 1 1 15. m/s 2 1 1 2 1 29.8 m /s 2 2 2 y 5 y 1 y 5 y 1 1 2g 2 5 15. m/s 1 1 29.8 m/s 2 2 2.24 Posiion nd elociy of bll hrown ericlly upwrd. y The bll cully moes srigh up nd hen srigh down; we show y 5 U-shped ph for clriy. 5? 5 1. s, y 5? 5?, y 5? 5, y 5 15. m/s 5? y 5? 5 4. s y 5? y 5? y 5? y 5 5. m y 5 y 5? y 5 2g 5 29.8 m/s 2 When 5 1. s, hese equions gie The bll is 1.1 m boe he origin (y is posiie) nd moing upwrd ( y is posiie) wih speed of 5.2 m/s. This is less hn he iniil speed becuse he bll slows s i scends. When 5 4. s, he equions for y nd y s funcions of ime gie The bll hs pssed is highes poin nd is 18.4 m below he origin (y is negie). I hs downwrd elociy ( y is negie) wih mgniude 24.2 m/s. The bll loses speed s i scends, hen gins speed s i descends; i is moing he iniil 15.-m/s speed s i moes downwrd ps he bll s lunching poin (he origin), nd coninues o gin speed s i descends below his poin. (b) The y-elociy y ny posiion y is gien by Eq. (2.13) wih s replced by y s: y 2 5 y 2 1 2 y 1 y 2 y 2 5 y 2 1 2 1 2g 21y 2 2 5 1 15. m / s 2 2 1 2 1 29.8 m / s 2 2 y When he bll is 5. m boe he origin, y 515. m, so 2 y 5 1 15. m/s 2 2 1 2 1 29.8 m/s 2 215. m 2 5 127 m 2 /s 2 y 5611.3 m/s y 511.1 m y 515.2 m/s y 5218.4 m y 5224.2 m/s We ge wo lues of y becuse he bll psses hrough he poin y 515. m wice (see Fig. 2.24), once on he wy up so y is posiie nd once on he wy down so y is negie. (c) Jus he insn when he bll reches he highes poin, i is momenrily res nd y 5. The mimum heigh y 1 cn hen be found in wo wys. The firs wy is o use Eq. (2.13) nd subsiue y 5, y 5, nd y 52g: 5 y 2 1 2 1 2g 21y 1 2 2 y 1 5 2 y 2g 5 1 15. m /s 2 2 2 1 9.8 m/s 2 2 5111.5 m The second wy is find he ime which y 5 using Eq. (2.8), y 5 y 1 y, nd hen subsiue his lue of ino Eq. (2.12) o find he posiion his ime. From Eq. (2.8), he ime 1 when he bll reches he highes poin is gien by y 5 5 y 1 1 2g 2 1 1 5 y g 5 15. m /s 9.8 m/s 2 5 1.53 s Subsiuing his lue of ino Eq. (2.12), we find y 5 y 1 y 1 1 2 y 2 5 1 2 1 1 15 m/s 211.53 s 2 1 1 2 1 29.8 m /s 2 211.53 s 2 2 5111.5 m Noice h he firs wy of finding he mimum heigh is esier, since i s no necessry o find he ime firs. Coninued
56 CHAPTER 2 Moion Along Srigh Line 2.6 *Velociy nd Posiion by Inegrion 57 (d) CAUTIN A free-fll misconcepion I s common misconcepion h he highes poin of free-fll moion he elociy is zero nd he ccelerion is zero. If his were so, once he bll reched he highes poin i would hng here suspended in midir! Remember h ccelerion is he re of chnge of elociy. If he ccelerion were zero he highes poin, he bll s elociy would no longer chnge, nd once he bll ws insnneously res, i would remin res foreer. A he highes poin, he ccelerion is sill y 52g 5 29.8 m/s 2, he sme lue s when he bll is moing up nd when i s moing down. Th s becuse he bll s elociy is coninuously chnging, from posiie lues hrough zero o negie lues. EVALUATE: A useful wy o check ny moion problem is o drw he grphs of posiion nd elociy ersus ime. Figure 2.25 shows hese grphs for his problem. Since he y-ccelerion is consn nd negie, he y- grph is prbol wih downwrd curure nd he y - grph is srigh line wih negie slope. Emple 2.8 Two soluions or one? Find he ime when he bll in Emple 2.7 is 5. m below he roof riling. SLUTIN IDENTIFY: Agin his consn-ccelerion problem. The rge rible is he ime when he bll is cerin posiion. SET UP: We gin choose he y-is s in Fig. 2.24, so y, y, nd y 52g he he sme lues s in Emple 2.7. The posiion y s funcion of ime is gin gien by Eq. (2.12): y 5 y 1 y 1 1 2 y 2 5 y 1 y 1 1 2 1 2g 2 2 We wn o sole his for he lue of when y 525. m. Since his equion inoles 2, i is qudric equion for. EXECUTE: We firs rerrnge he equion ino he sndrd form of qudric equion for n unknown, A 2 1 B 1 C 5 : 1 1 2 2 g 2 1 1 2 y 2 1 1 y 2 y 2 5 A 2 1 B 1 C 5 so A 5 g/2, B 52 y, nd C 5 y 2 y. Using he qudric formul (see Appendi B), we find h his equion hs wo soluions: 5 2B6"B2 2 4AC 2A 5 2 1 2 y 2 6 " 1 2 y 2 2 2 4 1 g/2 21y 2 y 2 2 1 g/2 2 5 y 6 " 2 y 2 2g 1 y 2 y 2 g Subsiuing he lues y g 5 9.8 m/s 2 5, y 5115. m/s,, nd y 525. m, we find 5 1 15. m /s 2 6" 1 15. m/s 2 2 2 2 1 9.8 m/s 2 2125. m 2 2 9.8 m/s 2 513.36 s or 52.3 s 2.25 () Posiion nd (b) elociy s funcions of ime for bll hrown upwrd wih n iniil speed of 15 m/s. () y- grph (curure is downwrd becuse y 5 2g is negie) y (m) Before 5 1.53 s he 15 bll moes upwrd. 1 Afer 5 1.53 s he bll moes 5 downwrd. 25 21 215 22 1 2 3 4 (s) (b) y - grph (srigh line wih negie slope becuse y 5 2g is consn nd negie) y (m/s) 15 1 5 25 21 215 22 225 1 Before 5 1.53 s he y-elociy is posiie. 2 (s) 3 4 Afer 5 1.53 s he y-elociy is negie. To decide which of hese is he righ nswer, he key quesion o sk is, Are hese nswers resonble? The second nswer, 52.3 s, is simply no resonble; i refers o ime.3 s before he bll lef your hnd! The correc nswer is 513.36 s. The bll is 5. m below he riling 3.36 s fer i lees your hnd. EVALUATE: Where did he erroneous soluion 52.3 s come from? Remember h he equion y 5 y 1 y 1 1 1 2 2g 2 2 is bsed on he ssumpion h he ccelerion is consn for ll lues of, wheher posiie, negie, or zero. Tken fce lue, his equion ells us h he bll hs been moing upwrd in free fll eer since he dwn of ime; i eenully psses your hnd y 5 he specil insn we chose o cll 5, hen coninues in free fll. Bu nyhing h his equion describes hppening before 5 is pure ficion, since he bll wen ino free fll only fer leing your hnd 5 ; he soluion 52.3 s is pr of his ficion. You should repe hese clculions o find he imes when he bll is 5. m boe he origin 1 y 515. m 2. The wo nswers re 51.38 s nd 512.68 s. These re boh posiie lues of, nd boh refer o he rel moion of he bll fer leing your hnd. The erlier ime is when he bll psses hrough y 515. m moing upwrd; he ler ime is when i psses hrough his poin moing downwrd. [Compre his wih pr (b) of Emple 2.7.] You should lso sole for he imes which y 5115. m. In his cse, boh soluions inole he squre roo of negie number, so here re no rel soluions. This mkes sense; we found in pr (c) of Emple 2.7 h he bll s mimum heigh is only y 5111.5 m, so i neer reches y 5115. m. While qudric equion such s Eq. (2.12) lwys hs wo soluions, in some siuions one or boh of he soluions will no be physiclly resonble. Tes Your Undersnding of Secion 2.5 If you oss bll upwrd wih cerin iniil speed, i flls freely nd reches mimum heigh h ime fer i lees your hnd. () If you hrow he bll upwrd wih double he iniil speed, wh new mimum heigh does he bll rech? (i) h "2; (ii) 2h; (iii) 4h; (i) 8h; () 16h. (b) If you hrow he bll upwrd wih double he iniil speed, how long does i ke o rech is new mimum heigh? (i) /2; (ii) /"2 ; (iii) ; (i) "2; () 2. 2.6 *Velociy nd Posiion by Inegrion This opionl secion is inended for sudens who he lredy lerned lile inegrl clculus. In Secion 2.4 we nlyzed he specil cse of srigh-line moion wih consn ccelerion. When is no consn, s is frequenly he cse, he equions h we deried in h secion re no longer lid (Fig. 2.26). Bu een when ries wih ime, we cn sill use he relionship 5 d/d o find he -elociy s funcion of ime if he posiion is known funcion of ime. And we cn sill use 5 d /d o find he -ccelerion s funcion of ime if he -elociy is known funcion of ime. In mny siuions, howeer, posiion nd elociy re no known s funcions of ime, while ccelerion is. How cn we find he posiion nd elociy from he ccelerion funcion 1 2? This problem rises in niging n irliner beween Norh Americ nd Europe (Fig. 2.27). The pilos mus know heir posiion precisely ll imes, bu oer he ocen n irliner is usully ou of rnge of boh rdio nigion becons on lnd nd ir rffic conrollers rdr. To deermine heir posiion, irliners crry deice clled n ineril nigion sysem (INS), which mesures he irliner s ccelerion. This is done in much he sme wy h you cn sense chnges in he elociy of cr in which you re riding, een when your eyes re closed. (In Chper 4 we ll discuss how your body deecs ccelerion.) Gien his informion, long wih he irliner s iniil posiion (sy, priculr ge Mimi Inernionl Airpor) nd is iniil elociy (zero when prked he ge), he INS clcules he irliner s curren elociy nd posiion ll imes during he fligh. (Airliners lso use he Globl Posiioning Sysem, or GPS, for nigion, bu his supplemens INS rher hn replcing i.) ur gol in his secion is o see how hese clculions re done for he simpler cse of moion in srigh line wih ime-rying ccelerion. We firs consider grphicl pproch. Figure 2.28 is grph of -ccelerion ersus ime for body whose ccelerion is no consn. We cn diide he ime inerl beween imes 1 nd 2 ino mny smller inerls, clling ypicl one D. Le he erge -ccelerion during D be -. From Eq. (2.4) he chnge in -elociy D during D is Grphiclly, D equls he re of he shded srip wih heigh - nd widh D h is, he re under he cure beween he lef nd righ sides of D. The ol chnge in -elociy during ny inerl (sy, 1 o 2 ) is he sum of he -elociy chnges D in he smll subinerls. So he ol -elociy chnge is represened grphiclly by he ol re under he - cure beween he ericl lines 1 nd 2. (In Secion 2.4 we showed his for he specil cse in which he ccelerion is consn.) In he limi h ll he D s become ery smll nd heir number ery lrge, he lue of - for he inerl from ny ime o 1Dpproches he insnneous -ccelerion ime. In his limi, he re under he - cure is he inegrl of (which is in generl funcion of ) from 1 o 2. If 1 is he -elociy of he body ime nd is he elociy ime 2, hen 1 D 5 - D 2 2 2 2 1 5 3 d 5 3 d 1 1 The chnge in he -elociy is he ime inegrl of he -ccelerion. 2 (2.15) 2.26 When you push your cr s cceleror pedl o he floorbord, he resuling ccelerion is no consn: he greer he cr s speed, he more slowly i gins ddiionl speed. A ypicl cr kes wice s long o ccelere from 5 km/h o 1 km/h s i does o ccelere from o 5 km/h. 2.27 The posiion nd elociy of n irliner crossing he Alnic re found by inegring is ccelerion wih respec o ime. - To London Accelerion: Known Velociy: To be deermined Posiion: To be deermined From Mimi 2.28 An - grph for body whose -ccelerion is no consn. Tol re under he - grph from 1 o 2 5 ne chnge in -elociy from 1 o 2. W Are of his srip 5 D 5 chnge in -elociy during ime inerl D. 1 2 D N S E