2-1 MOTION ALONG A STRAIGHT LINE. 2-2 Motion. 2-3 Position and Displacement WHAT IS PHYSICS? CHAPTER

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1 hall-isv_c_3-37hr.qxd 6--9 :7 Page 3 - WHAT IS PHYSICS? move, for example, and how far hey move in a given amoun of ime. NASCAR engineers are fanaical abou his aspec of physics as hey deermine he performance of heir cars before and during a race. Geologiss use his physics o measure econic-plae moion as hey aemp o predic earhquakes. Medical researchers need his physics o map he blood flow hrough a paien when diagnosing a parially closed arery, and mooriss use i o deermine how hey migh slow sufficienly when heir radar deecor sounds a warning. There are counless oher examples. In his chaper, we sudy he basic physics of moion where he objec (race car, econic plae, blood cell, or any oher objec) moves along a single axis. Such moion is called one-dimensional moion. - Moion MOTION ALONG A STRAIGHT LINE One purpose of physics is o sudy he moion of objecs how fas hey The world, and everyhing in i, moves. Even seemingly saionary hings, such as a roadway, move wih Earh s roaion, Earh s orbi around he Sun, he Sun s orbi around he cener of he Milky Way galaxy, and ha galaxy s migraion relaive o oher galaxies. The classificaion and comparison of moions (called kinemaics) is ofen challenging.wha exacly do you measure, and how do you compare? Before we aemp an answer, we shall examine some general properies of moion ha is resriced in hree ways.. The moion is along a sraigh line only. The line may be verical, horizonal, or slaned, bu i mus be sraigh.. Forces (pushes and pulls) cause moion bu will no be discussed unil Chaper 5. In his chaper we discuss only he moion iself and changes in he moion. Does he moving objec speed up, slow down, sop, or reverse direcion? If he moion does change, how is ime involved in he change? 3. The moving objec is eiher a paricle (by which we mean a poin-like objec such as an elecron) or an objec ha moves like a paricle (such ha every porion moves in he same direcion and a he same rae). A siff pig slipping down a sraigh playground slide migh be considered o be moving like a paricle; however, a umbling umbleweed would no. -3 Posiion and Displacemen To locae an objec means o find is posiion relaive o some reference poin, ofen he origin (or zero poin) of an axis such as he x axis in Fig. -. The posiive direcion of he axis is in he direcion of increasing numbers (coordinaes), which is o he righ in Fig. -.The opposie is he negaive direcion. CHAPTER Posiive direcion Negaive direcion x (m) 3 3 Origin Fig. - Posiion is deermined on an axis ha is marked in unis of lengh (here meers) and ha exends indefiniely in opposie direcions. The axis name, here x, is always on he posiive side of he origin. 3

2 hall-isv_c_3-37hr.qxd 6--9 :7 Page 4 4 CHAPTER MOTION ALONG A STRAIGHT LINE For example, a paricle migh be locaed a x 5 m, which means i is 5 m in he posiive direcion from he origin. If i were a x 5 m, i would be jus as far from he origin bu in he opposie direcion. On he axis, a coordinae of 5 m is less han a coordinae of m, and boh coordinaes are less han a coordinae of 5 m. A plus sign for a coordinae need no be shown, bu a minus sign mus always be shown. A change from posiion x o posiion x is called a displacemen x, where x x x. (-) (The symbol, he Greek uppercase dela, represens a change in a quaniy, and i means he final value of ha quaniy minus he iniial value.) When numbers are insered for he posiion values x and x in Eq. -, a displacemen in he posiive direcion (o he righ in Fig. -) always comes ou posiive, and a displacemen in he opposie direcion (lef in he figure) always comes ou negaive. For example, if he paricle moves from x 5 m o x m, hen he displacemen is x ( m) (5 m) 7 m. The posiive resul indicaes ha he moion is in he posiive direcion. If, insead, he paricle moves from x 5 m o x m, hen x ( m) (5 m) 4 m. The negaive resul indicaes ha he moion is in he negaive direcion. The acual number of meers covered for a rip is irrelevan; displacemen involves only he original and final posiions. For example, if he paricle moves from x 5 m ou o x m and hen back o x 5 m, he displacemen from sar o finish is x (5 m) (5 m). A plus sign for a displacemen need no be shown, bu a minus sign mus always be shown. If we ignore he sign (and hus he direcion) of a displacemen, we are lef wih he magniude (or absolue value) of he displacemen. For example, a displacemen of x 4 m has a magniude of 4 m. Displacemen is an example of a vecor quaniy, which is a quaniy ha has boh a direcion and a magniude. We explore vecors more fully in Chaper 3 (in fac, some of you may have already read ha chaper), bu here all we need is he idea ha displacemen has wo feaures: () Is magniude is he disance (such as he number of meers) beween he original and final posiions. () Is direcion, from an original posiion o a final posiion, can be represened by a plus sign or a minus sign if he moion is along a single axis. Wha follows is he firs of many checkpoins you will see in his book. Each consiss of one or more quesions whose answers require some reasoning or a menal calculaion, and each gives you a quick check of your undersanding of a poin jus discussed.the answers are lised in he back of he book. CHECKPOINT Here are hree pairs of iniial and final posiions, respecively, along an x axis. Which pairs give a negaive displacemen: (a) 3 m, 5 m; (b) 3 m, 7 m; (c) 7 m, 3 m? -4 Average Velociy and Average Speed A compac way o describe posiion is wih a graph of posiion x ploed as a funcion of ime a graph of x(). (The noaion x() represens a funcion x of, no he produc x imes.) As a simple example, Fig. - shows he posiion funcion x() for a saionary armadillo (which we rea as a paricle) over a 7 s ime inerval.the animal s posiion says a x m. Figure -3 is more ineresing, because i involves moion. The armadillo is apparenly firs noiced a when i is a he posiion x 5 m. I moves

3 hall-isv_c_3-37hr.qxd 6--9 :7 Page 5-4 AVERAGE VELOCITY AND AVERAGE SPEED 5 PART Fig. - The graph of x() for an armadillo ha is saionary a x m. The value of x is m for all imes. This is a graph of posiion x versus ime for a saionary objec. Same posiion for any ime. x (m) x() oward x, passes hrough ha poin a 3 s, and hen moves on o increasingly larger posiive values of x. Figure -3 also depics he sraigh-line moion of he armadillo (a hree imes) and is somehing like wha you would see. The graph in Fig. -3 is more absrac and quie unlike wha you would see, bu i is richer in informaion. I also reveals how fas he armadillo moves. Acually, several quaniies are associaed wih he phrase how fas. One of hem is he average velociy v avg, which is he raio of he displacemen x ha occurs during a paricular ime inerval o ha inerval: v avg x x x. (-) The noaion means ha he posiion is x a ime and hen x a ime. A common uni for v avg is he meer per second (m/s). You may see oher unis in he problems, bu hey are always in he form of lengh/ime. On a graph of x versus, v avg is he slope of he sraigh line ha connecs wo paricular poins on he x() curve: one is he poin ha corresponds o x and, and he oher is he poin ha corresponds o x and. Like displacemen, v avg has boh magniude and direcion (i is anoher vecor quaniy). Is magniude is he magniude of he line s slope. A posiive v avg (and slope) ells us ha he line slans upward o he righ; a negaive v avg (and slope) ells us ha he line slans downward o he righ. The average velociy v avg always has he same sign as he displacemen x because in Eq. - is always posiive. This is a graph of posiion x versus ime for a moving objec. I is a posiion x = 5 m when ime = s. Tha daa is ploed here. x (m) A x = m when = 4 s. Ploed here. x() 5 x (m) A x = m when = 3 s. Ploed here. 4 s A 5 x (m) s Fig. -3 The graph of x() for a moving armadillo.the pah associaed wih he graph is also shown, a hree imes. 5 x (m) 3 s

4 hall-isv_c_3-37hr.qxd 6--9 :7 Page 6 6 CHAPTER MOTION ALONG A STRAIGHT LINE A Fig. -4 Calculaion of he average velociy beween s and 4 s as he slope of he line ha connecs he poins on he x() curve represening hose imes. This is a graph of posiion x versus ime. To find average velociy, firs draw a sraigh line, sar o end, and hen find he slope of he line. Sar of inerval x (m) x() 4 5 v avg = slope of his line rise x = = run 3 4 End of inerval This verical disance is how far i moved, sar o end: x = m ( 4 m) = 6 m This horizonal disance is how long i ook, sar o end: = 4 s s = 3 s Figure -4 shows how o find v avg in Fig. -3 for he ime inerval s o 4s. We draw he sraigh line ha connecs he poin on he posiion curve a he beginning of he inerval and he poin on he curve a he end of he inerval. Then we find he slope x/ of he sraigh line. For he given ime inerval, he average velociy is v avg 6 m m/s. 3 s Average speed s avg is a differen way of describing how fas a paricle moves. Whereas he average velociy involves he paricle s displacemen x, he average speed involves he oal disance covered (for example, he number of meers moved), independen of direcion; ha is, s avg oal disance. (-3) Because average speed does no include direcion, i lacks any algebraic sign. Someimes s avg is he same (excep for he absence of a sign) as v avg. However, he wo can be quie differen. Sample Problem Average velociy, bea-up pickup ruck You drive a bea-up pickup ruck along a sraigh road for 8.4 km a 7 km/h, a which poin he ruck runs ou of gasoline and sops. Over he nex 3 min, you walk anoher. km farher along he road o a gasoline saion. (a) Wha is your overall displacemen from he beginning of your drive o your arrival a he saion? KEY IDEA Assume, for convenience, ha you move in he posiive direcion of an x axis, from a firs posiion of x o a second posiion of x a he saion. Tha second posiion mus be a x 8.4 km. km.4 km.then your displacemen x along he x axis is he second posiion minus he firs posiion. Calculaion: From Eq. -, we have x x x.4 km.4 km. (Answer) Thus, your overall displacemen is.4 km in he posiive direcion of he x axis. (b) Wha is he ime inerval from he beginning of your drive o your arrival a he saion? KEY IDEA We already know he walking ime inerval wlk (.5 h), bu we lack he driving ime inerval dr. However, we know ha for he drive he displacemen x dr is 8.4 km and he average velociy v avg,dr is 7 km/h. Thus, his average

5 hall-isv_c_3-37hr.qxd 6--9 :7 Page 7-5 INSTANTANEOUS VELOCITY AND SPEED 7 PART velociy is he raio of he displacemen for he drive o he ime inerval for he drive. Calculaions: We firs wrie Rearranging and subsiuing daa hen give us dr x dr v avg,dr So, dr wlk. h.5 h.6 h. (Answer) (c) Wha is your average velociy v avg from he beginning of your drive o your arrival a he saion? Find i boh numerically and graphically. KEY IDEA From Eq. - we know ha v avg for he enire rip is he raio of he displacemen of.4 km for he enire rip o he ime inerval of.6 h for he enire rip. Calculaion: Here we find v avg x v avg,dr x dr dr. 8.4 km 7 km/h.4 km.6 h 6.8 km/h 7 km/h.. h. (Answer) To find v avg graphically, firs we graph he funcion x() as shown in Fig. -5, where he beginning and arrival poins on he graph are he origin and he poin labeled as Saion. Your average velociy is he slope of he sraigh line connecing hose poins; ha is, v avg is he raio of he rise ( x.4 km) o he run (.6 h), which gives us v avg 6.8 km/h. (d) Suppose ha o pump he gasoline, pay for i, and walk back o he ruck akes you anoher 45 min. Wha is your Posiion (km) average speed from he beginning of your drive o your reurn o he ruck wih he gasoline? x Driving Driving ends, walking sars. Walking KEY IDEA Your average speed is he raio of he oal disance you move o he oal ime inerval you ake o make ha move. Calculaion: The oal disance is 8.4 km. km. km.4 km. The oal ime inerval is. h.5 h.75 h.37 h.thus, Eq. -3 gives us s avg.4 km.37 h..4.6 Time (h) 9. km/h. Saion Fig. -5 The lines marked Driving and Walking are he posiion ime plos for he driving and walking sages. (The plo for he walking sage assumes a consan rae of walking.) The slope of he sraigh line joining he origin and he poin labeled Saion is he average velociy for he rip, from he beginning o he saion. (Answer) Slope of his line gives average velociy. How far: x =.4 km How long: =.6 h Addiional examples, video, and pracice available a WileyPLUS -5 Insananeous Velociy and Speed You have now seen wo ways o describe how fas somehing moves: average velociy and average speed, boh of which are measured over a ime inerval. However, he phrase how fas more commonly refers o how fas a paricle is moving a a given insan is insananeous velociy (or simply velociy) v. The velociy a any insan is obained from he average velociy by shrinking he ime inerval closer and closer o.as dwindles, he average velociy approaches a limiing value, which is he velociy a ha insan: v lim : x dx d. (-4)

6 hall-isv_c_3-37hr.qxd 6--9 :7 Page 8 8 CHAPTER MOTION ALONG A STRAIGHT LINE Noe ha v is he rae a which posiion x is changing wih ime a a given insan; ha is, v is he derivaive of x wih respec o. Also noe ha v a any insan is he slope of he posiion ime curve a he poin represening ha insan. Velociy is anoher vecor quaniy and hus has an associaed direcion. Speed is he magniude of velociy; ha is, speed is velociy ha has been sripped of any indicaion of direcion, eiher in words or via an algebraic sign. (Cauion: Speed and average speed can be quie differen.) A velociy of 5 m/s and one of 5 m/s boh have an associaed speed of 5 m/s. The speedomeer in a car measures speed, no velociy (i canno deermine he direcion). CHECKPOINT The following equaions give he posiion x() of a paricle in four siuaions (in each equaion, x is in meers, is in seconds, and ): () x 3 ; () x 4 ; (3) x / ; and (4) x. (a) In which siuaion is he velociy v of he paricle consan? (b) In which is v in he negaive x direcion? Sample Problem Velociy and slope of x versus, elevaor cab Figure -6a is an x() plo for an elevaor cab ha is iniially saionary, hen moves upward (which we ake o be he posiive direcion of x), and hen sops. Plo v(). KEY IDEA We can find he velociy a any ime from he slope of he x() curve a ha ime. Calculaions: The slope of x(), and so also he velociy, is zero in he inervals from o s and from 9 s on, so hen he cab is saionary. During he inerval bc, he slope is consan and nonzero, so hen he cab moves wih consan velociy.we calculae he slope of x() hen as x v 4 m 4. m 8. s 3. s 4. m/s. (-5) The plus sign indicaes ha he cab is moving in he posiive x direcion. These inervals (where v and v 4 m/s) are ploed in Fig. -6b. In addiion, as he cab iniially begins o move and hen laer slows o a sop, v varies as indicaed in he inervals s o 3 s and 8 s o 9 s. Thus, Fig. -6b is he required plo. (Figure -6c is considered in Secion -6.) Given a v() graph such as Fig. -6b, we could work backward o produce he shape of he associaed x() graph (Fig. -6a). However, we would no know he acual values for x a various imes, because he v() graph indicaes only changes in x. To find such a change in x during any inerval, we mus, in he language of calculus, calculae he area under he curve on he v() graph for ha inerval. For example, during he inerval 3 s o 8 s in which he cab has a velociy of 4. m/s, he change in x is x (4. m/s)(8. s 3. s) m. (-6) (This area is posiive because he v() curve is above he axis.) Figure -6a shows ha x does indeed increase by m in ha inerval. However, Fig. -6b does no ell us he values of x a he beginning and end of he inerval. For ha, we need addiional informaion, such as he value of x a some insan. -6 Acceleraion When a paricle s velociy changes, he paricle is said o undergo acceleraion (or o accelerae). For moion along an axis, he average acceleraion a avg over a ime inerval is a avg v v v, (-7) where he paricle has velociy v a ime and hen velociy v a ime.the insananeous acceleraion (or simply acceleraion) is a dv d. (-8)

7 hall-isv_c_3-37hr.qxd 6--9 :7 Page 9-6 ACCELERATION 9 PART x Posiion (m) 5 5 x() 5 b a Time (s) (a) Slope of x() v x = 4. m a = 3. s x = 4 m a = 8. s c x d Slopes on he x versus graph are he values on he v versus graph. 4 b v() c Velociy (m/s) 3 Fig. -6 (a) The x() curve for an elevaor cab ha moves upward along an x axis. (b)thev() curve for he cab. Noe ha i is he derivaive of he x() curve (v dx/d). (c) The a() curve for he cab. I is he derivaive of he v() curve (a dv/d).the sick figures along he boom sugges how a passenger s body migh feel during he acceleraions. Acceleraion (m/s ) a a Acceleraion 3 a b Time (s) (b) a() 4 5 (c) c 8 Deceleraion d Slopes on he v versus graph are he values on he a versus graph. d Wha you would feel. Addiional examples, video, and pracice available a WileyPLUS In words, he acceleraion of a paricle a any insan is he rae a which is velociy is changing a ha insan. Graphically, he acceleraion a any poin is he slope of he curve of v() a ha poin.we can combine Eq. -8 wih Eq. -4 o wrie a dv d d d dx d d x d. (-9) In words, he acceleraion of a paricle a any insan is he second derivaive of is posiion x() wih respec o ime. A common uni of acceleraion is he meer per second per second: m/(s s) or m/s. Oher unis are in he form of lengh/(ime ime) or lengh/ime. Acceleraion has boh magniude and direcion (i is ye anoher vecor quaniy). Is algebraic sign represens is direcion on an axis jus as for displacemen and velociy; ha is, acceleraion wih a posiive value is in he posiive direcion of an axis, and acceleraion wih a negaive value is in he negaive direcion.

8 hall-isv_c_3-37hr.qxd 6--9 :7 Page CHAPTER MOTION ALONG A STRAIGHT LINE Fig. -7 Colonel J. P. Sapp in a rocke sled as i is brough up o high speed (acceleraion ou of he page) and hen very rapidly braked (acceleraion ino he page). (Couresy U.S. Air Force) Figure -6 gives plos of he posiion, velociy, and acceleraion of an elevaor moving up a shaf. Compare he a() curve wih he v() curve each poin on he a() curve shows he derivaive (slope) of he v() curve a he corresponding ime. When v is consan (a eiher or 4 m/s), he derivaive is zero and so also is he acceleraion. When he cab firs begins o move, he v() curve has a posiive derivaive (he slope is posiive), which means ha a() is posiive.when he cab slows o a sop, he derivaive and slope of he v() curve are negaive; ha is, a() is negaive. Nex compare he slopes of he v() curve during he wo acceleraion periods. The slope associaed wih he cab s slowing down (commonly called deceleraion ) is seeper because he cab sops in half he ime i ook o ge up o speed. The seeper slope means ha he magniude of he deceleraion is larger han ha of he acceleraion, as indicaed in Fig. -6c. The sensaions you would feel while riding in he cab of Fig. -6 are indicaed by he skeched figures a he boom. When he cab firs acceleraes, you feel as hough you are pressed downward; when laer he cab is braked o a sop, you seem o be sreched upward. In beween, you feel nohing special. In oher words, your body reacs o acceleraions (i is an acceleromeer) bu no o velociies (i is no a speedomeer). When you are in a car raveling a 9 km/h or an airplane raveling a 9 km/h, you have no bodily awareness of he moion. However, if he car or plane quickly changes velociy, you may become keenly aware of he change, perhaps even frighened by i. Par of he hrill of an amusemen park ride is due o he quick changes of velociy ha you undergo (you pay for he acceleraions, no for he speed).a more exreme example is shown in he phoographs of Fig. -7, which were aken while a rocke sled was rapidly acceleraed along a rack and hen rapidly braked o a sop. Large acceleraions are someimes expressed in erms of g unis, wih g 9.8 m/s (g uni). (-) (As we shall discuss in Secion -9, g is he magniude of he acceleraion of a falling objec near Earh s surface.) On a roller coaser, you may experience brief acceleraions up o 3g, which is (3)(9.8 m/s ), or abou 9 m/s, more han enough o jusify he cos of he ride. In common language, he sign of an acceleraion has a nonscienific meaning: posiive acceleraion means ha he speed of an objec is increasing, and negaive acceleraion means ha he speed is decreasing (he objec is deceleraing). In his book, however, he sign of an acceleraion indicaes a direcion, no wheher an objec s speed is increasing or decreasing. For example, if a car wih an iniial velociy v 5 m/s is braked o a sop in 5. s, hen a avg 5. m/s. The acceleraion is posiive, bu he car s speed has decreased. The reason is he difference in signs: he direcion of he acceleraion is opposie ha of he velociy. Here hen is he proper way o inerpre he signs: If he signs of he velociy and acceleraion of a paricle are he same, he speed of he paricle increases. If he signs are opposie, he speed decreases.

9 hall-isv_c_3-37hr.qxd 6--9 :7 Page -6 ACCELERATION PART CHECKPOINT 3 A womba moves along an x axis. Wha is he sign of is acceleraion if i is moving (a) in he posiive direcion wih increasing speed, (b) in he posiive direcion wih decreasing speed, (c) in he negaive direcion wih increasing speed, and (d) in he negaive direcion wih decreasing speed? Sample Problem Acceleraion and dv/d A paricle s posiion on he x axis of Fig. - is given by which has he soluion x 4 7 3, 3 s. (Answer) wih x in meers and in seconds. (a) Because posiion x depends on ime, he paricle mus be moving. Find he paricle s velociy funcion v() and acceleraion funcion a(). KEY IDEAS () To ge he velociy funcion v(), we differeniae he posiion funcion x() wih respec o ime. () To ge he acceleraion funcion a(), we differeniae he velociy funcion v() wih respec o ime. Calculaions: Differeniaing he posiion funcion, we find v 7 3, (Answer) wih v in meers per second. Differeniaing he velociy funcion hen gives us a 6, (Answer) wih a in meers per second squared. (b) Is here ever a ime when v? Calculaion: Seing v() yields 7 3, Thus, he velociy is zero boh 3 s before and 3 s afer he clock reads. (c) Describe he paricle s moion for. Reasoning: We need o examine he expressions for x(), v(), and a(). A, he paricle is a x() 4 m and is moving wih a velociy of v() 7 m/s ha is, in he negaive direcion of he x axis. Is acceleraion is a() because jus hen he paricle s velociy is no changing. For 3 s, he paricle sill has a negaive velociy, so i coninues o move in he negaive direcion. However, is acceleraion is no longer bu is increasing and posiive. Because he signs of he velociy and he acceleraion are opposie, he paricle mus be slowing. Indeed, we already know ha i sops momenarily a 3 s. Jus hen he paricle is as far o he lef of he origin in Fig. - as i will ever ge. Subsiuing 3 s ino he expression for x(), we find ha he paricle s posiion jus hen is x 5 m. Is acceleraion is sill posiive. For 3 s, he paricle moves o he righ on he axis. Is acceleraion remains posiive and grows progressively larger in magniude. The velociy is now posiive, and i oo grows progressively larger in magniude. Addiional examples, video, and pracice available a WileyPLUS

10 hall-isv_c_3-37hr.qxd 6--9 :7 Page CHAPTER MOTION ALONG A STRAIGHT LINE -7 Consan Acceleraion: A Special Case In many ypes of moion, he acceleraion is eiher consan or approximaely so. For example, you migh accelerae a car a an approximaely consan rae when a raffic ligh urns from red o green. Then graphs of your posiion, velociy, and acceleraion would resemble hose in Fig. -8. (Noe ha a() in Fig. -8c is consan, which requires ha v() in Fig. -8b have a consan slope.) Laer when you brake he car o a sop, he acceleraion (or deceleraion in common language) migh also be approximaely consan. Such cases are so common ha a special se of equaions has been derived for dealing wih hem. One approach o he derivaion of hese equaions is given in his secion. A second approach is given in he nex secion. Throughou boh secions and laer when you work on he homework problems, keep in mind ha hese equaions are valid only for consan acceleraion (or siuaions in which you can approximae he acceleraion as being consan). When he acceleraion is consan, he average acceleraion and insananeous acceleraion are equal and we can wrie Eq.-7,wih some changes in noaion,as a a avg v v. Here v is he velociy a ime and v is he velociy a any laer ime.we can recas his equaion as v v a. (-) As a check, noe ha his equaion reduces o v v for, as i mus.as a furher check, ake he derivaive of Eq. -. Doing so yields dv/d a, which is he definiion of a. Figure -8b shows a plo of Eq. -, he v() funcion; he funcion is linear and hus he plo is a sraigh line. In a similar manner, we can rewrie Eq. - (wih a few changes in noaion) as v avg x x x x() Posiion Slope varies x v (a) v() Slopes of he posiion graph are ploed on he velociy graph. Velociy Slope = a Fig. -8 (a) The posiion x() of a paricle moving wih consan acceleraion. (b) Is velociy v(), given a each poin by he slope of he curve of x(). (c) Is (consan) acceleraion, equal o he (consan) slope of he curve of v(). Acceleraion v a (b) a() Slope = (c) Slope of he velociy graph is ploed on he acceleraion graph.

11 hall-isv_c_3-37hr.qxd 6--9 :7 Page 3-7 CONSTANT ACCELERATION: A SPECIAL CASE 3 PART and hen as x x v avg, (-) in which x is he posiion of he paricle a and v avg is he average velociy beween and a laer ime. For he linear velociy funcion in Eq. -, he average velociy over any ime inerval (say, from o a laer ime ) is he average of he velociy a he beginning of he inerval ( v ) and he velociy a he end of he inerval ( v). For he inerval from o he laer ime hen, he average velociy is v avg (v v). (-3) Subsiuing he righ side of Eq. - for v yields, afer a lile rearrangemen, v avg v a. (-4) Finally, subsiuing Eq. -4 ino Eq. - yields x x v a. (-5) As a check, noe ha puing yields x x, as i mus. As a furher check, aking he derivaive of Eq. -5 yields Eq. -, again as i mus. Figure -8a shows a plo of Eq. -5; he funcion is quadraic and hus he plo is curved. Equaions - and -5 are he basic equaions for consan acceleraion; hey can be used o solve any consan acceleraion problem in his book. However, we can derive oher equaions ha migh prove useful in cerain specific siuaions. Firs, noe ha as many as five quaniies can possibly be involved in any problem abou consan acceleraion namely, x x, v,, a, and v. Usually, one of hese quaniies is no involved in he problem, eiher as a given or as an unknown. We are hen presened wih hree of he remaining quaniies and asked o find he fourh. Equaions - and -5 each conain four of hese quaniies, bu no he same four. In Eq. -, he missing ingredien is he displacemen x x. In Eq. -5, i is he velociy v. These wo equaions can also be combined in hree ways o yield hree addiional equaions, each of which involves a differen missing variable. Firs, we can eliminae o obain v v a(x x ). (-6) This equaion is useful if we do no know and are no required o find i. Second, we can eliminae he acceleraion a beween Eqs. - and -5 o produce an equaion in which a does no appear: x x (v v). (-7) Finally, we can eliminae v, obaining x x v a. (-8) Noe he suble difference beween his equaion and Eq. -5. One involves he iniial velociy v ; he oher involves he velociy v a ime. Table - liss he basic consan acceleraion equaions (Eqs. - and -5) as well as he specialized equaions ha we have derived. To solve a simple consan acceleraion problem, you can usually use an equaion from his lis (if you have he lis wih you). Choose an equaion for which he only unknown variable is he variable requesed in he problem. A simpler plan is o remember only Eqs. - and -5, and hen solve hem as simulaneous equaions whenever needed. CHECKPOINT 4 The following equaions give he posiion x() of a paricle in four siuaions: () x 3 4; () x ; (3) x / 4/; (4) x 5 3. To which of hese siuaions do he equaions of Table - apply? Table - Equaions for Moion wih Consan Acceleraion a Equaion Missing Number Equaion Quaniy - v v a x x -5 x x v a v -6 v v a(x x ) -7 x x (v v) a -8 x x v a v a Make sure ha he acceleraion is indeed consan before using he equaions in his able.

12 hall-isv_c_3-37hr.qxd 6--9 :7 Page 4 4 CHAPTER MOTION ALONG A STRAIGHT LINE Sample Problem Consan acceleraion, graph of v versus x Figure -9 gives a paricle s velociy v versus is posiion as i moves along an x axis wih consan acceleraion. Wha is is velociy a posiion x? KEY IDEA We can use he consan-acceleraion equaions; in paricular, we can use Eq. -6 (v v a(x x )), which relaes velociy and posiion. Firs ry: Normally we wan o use an equaion ha includes he requesed variable. In Eq. -6, we can idenify x as and v as being he requesed variable. Then we can idenify a second pair of values as being v and x. From he graph, we have wo such pairs: () v 8 m/s and x m, and () v and x 7 m. For example, we can wrie Eq. -6 as (8 m/s) v a( m ). (-9) However, we know neiher v nor a. Second ry: Insead of direcly involving he requesed variable, le s use Eq. -6 wih he wo pairs of known daa, idenifying v 8 m/s and x m as he firs pair and v m/s and x 7 m as he second pair. Then we can wrie ( m/s) (8 m/s) a(7 m m), which gives us a.64 m/s. Subsiuing his value ino Eq. -9 and solving for v (he velociy associaed wih he posiion of x ), we find v (m/s) 8 x (m) Fig. -9 The velociy is 8 m/s when he posiion is m. 7 The velociy is when he posiion is 7 m. Velociy versus posiion. v 9.5 m/s. (Answer) Commen: Some problems involve an equaion ha includes he requesed variable. A more challenging problem requires you o firs use an equaion ha does no include he requesed variable bu ha gives you a value needed o find i. Someimes ha procedure akes physics courage because i is so indirec. However, if you build your solving skills by solving los of problems, he procedure gradually requires less courage and may even become obvious. Solving problems of any kind, wheher physics or social, requires pracice. Addiional examples, video, and pracice available a WileyPLUS -8 Anoher Look a Consan Acceleraion* The firs wo equaions in Table - are he basic equaions from which he ohers are derived. Those wo can be obained by inegraion of he acceleraion wih he condiion ha a is consan. To find Eq. -, we rewrie he definiion of acceleraion (Eq. -8) as dv a d. We nex wrie he indefinie inegral (or aniderivaive) of boh sides: dv a d. Since acceleraion a is a consan, i can be aken ouside he inegraion. We obain dv a d or v a C. (-) To evaluae he consan of inegraion C, we le, a which ime v v. Subsiuing hese values ino Eq. - (which mus hold for all values of, *This secion is inended for sudens who have had inegral calculus.

13 hall-isv_c_3-37hr.qxd 6--9 :7 Page 5 PA R T -9 FREE-FALL ACCELERATION including ) yields 5 v (a)() C C. Subsiuing his ino Eq. - gives us Eq. -. To derive Eq. -5, we rewrie he definiion of velociy (Eq. -4) as dx v d and hen ake he indefinie inegral of boh sides o obain 冕 冕 dx v d. Nex, we subsiue for v wih Eq. -: 冕 冕 dx (v a) d. Since v is a consan, as is he acceleraion a, his can be rewrien as 冕 冕 冕 dx v d a d. Inegraion now yields x v a C, (-) where C is anoher consan of inegraion. A ime, we have x x. Subsiuing hese values in Eq. - yields x C. Replacing C wih x in Eq. - gives us Eq Free-Fall Acceleraion If you ossed an objec eiher up or down and could somehow eliminae he effecs of air on is fligh, you would find ha he objec acceleraes downward a a cerain consan rae. Tha rae is called he free-fall acceleraion, and is magniude is represened by g. The acceleraion is independen of he objec s characerisics, such as mass, densiy, or shape; i is he same for all objecs. Two examples of free-fall acceleraion are shown in Fig. -, which is a series of sroboscopic phoos of a feaher and an apple. As hese objecs fall, hey accelerae downward boh a he same rae g. Thus, heir speeds increase a he same rae, and hey fall ogeher. The value of g varies slighly wih laiude and wih elevaion. A sea level in Earh s midlaiudes he value is 9.8 m/s (or 3 f/s), which is wha you should use as an exac number for he problems in his book unless oherwise noed. The equaions of moion in Table - for consan acceleraion also apply o free fall near Earh s surface; ha is, hey apply o an objec in verical fligh, eiher up or down, when he effecs of he air can be negleced. However, noe ha for free fall: () The direcions of moion are now along a verical y axis insead of he x axis, wih he posiive direcion of y upward. (This is imporan for laer chapers when combined horizonal and verical moions are examined.) () The free-fall acceleraion is negaive ha is, downward on he y axis, oward Earh s cener and so i has he value g in he equaions. The free-fall acceleraion near Earh s surface is a g 9.8 m/s, and he magniude of he acceleraion is g 9.8 m/s. Do no subsiue 9.8 m/s for g. Suppose you oss a omao direcly upward wih an iniial (posiive) velociy v and hen cach i when i reurns o he release level. During is free-fall fligh (from jus afer is release o jus before i is caugh), he equaions of Table - apply o is Fig. - A feaher and an apple free fall in vacuum a he same magniude of acceleraion g. The acceleraion increases he disance beween successive images. In he absence of air, he feaher and apple fall ogeher. (Jim Sugar/Corbis Images)

14 hall-isv_c_3-37hr.qxd 6--9 :7 Page 6 6 CHAPTER MOTION ALONG A STRAIGHT LINE moion. The acceleraion is always a g 9.8 m/s, negaive and hus downward. The velociy, however, changes, as indicaed by Eqs. - and -6: during he ascen, he magniude of he posiive velociy decreases, unil i momenarily becomes zero. Because he omao has hen sopped, i is a is maximum heigh. During he descen, he magniude of he (now negaive) velociy increases. CHECKPOINT 5 (a) If you oss a ball sraigh up, wha is he sign of he ball s displacemen for he ascen, from he release poin o he highes poin? (b) Wha is i for he descen, from he highes poin back o he release poin? (c) Wha is he ball s acceleraion a is highes poin? Sample Problem Time for full up-down fligh, baseball oss In Fig. -, a picher osses a baseball up along a y axis, wih an iniial speed of m/s. (a) How long does he ball ake o reach is maximum heigh? KEY IDEAS () Once he ball leaves he picher and before i reurns o his hand, is acceleraion is he free-fall acceleraion a g. Because his is consan, Table - applies o he moion. () The velociy v a he maximum heigh mus be. Calculaion: Knowing v, a, and he iniial velociy v m/s, and seeking, we solve Eq. -, which conains Fig. - A picher osses a baseball sraigh up ino he air. The equaions of free fall apply for rising as well as for falling objecs, provided any effecs from he air can be negleced. Ball v = a highes poin During ascen, a = g, speed decreases, and velociy becomes less posiive y During descen, a = g, speed increases, and velociy becomes more negaive y = hose four variables. This yields v v a (Answer) (b) Wha is he ball s maximum heigh above is release poin? Calculaion: We can ake he ball s release poin o be y.we can hen wrie Eq. -6 in y noaion, se y y y and v (a he maximum heigh), and solve for y.we ge y v v a m/s 9.8 m/s. s. ( m/s) ( 9.8 m/s ) (Answer) (c) How long does he ball ake o reach a poin 5. m above is release poin? Calculaions: We know v, a g, and displacemen y y 5. m, and we wan, so we choose Eq. -5. Rewriing i for y and seing y give us y v g, 7.3 m. or 5. m ( m/s) ( )(9.8 m/s ). If we emporarily omi he unis (having noed ha hey are consisen), we can rewrie his as Solving his quadraic equaion for yields.53 s and.9 s. (Answer) There are wo such imes! This is no really surprising because he ball passes wice hrough y 5. m, once on he way up and once on he way down. Addiional examples, video, and pracice available a WileyPLUS

15 hall-isv_c_3-37hr.qxd 6--9 :7 Page 7 - GRAPHICAL INTEGRATION IN MOTION ANALYSIS 7 PART - Graphical Inegraion in Moion Analysis When we have a graph of an objec s acceleraion versus ime, we can inegrae on he graph o find he objec s velociy a any given ime. Because acceleraion a is defined in erms of velociy as a dv/d, he Fundamenal Theorem of Calculus ells us ha (-) The righ side of he equaion is a definie inegral (i gives a numerical resul raher han a funcion), v is he velociy a ime, and v is he velociy a laer ime.the definie inegral can be evaluaed from an a() graph, such as in Fig. -a.in paricular, v v a d. a d area beween acceleraion curve and ime axis, from o. (-3) If a uni of acceleraion is m/s and a uni of ime is s, hen he corresponding uni of area on he graph is ( m/s )( s) m/s, which is (properly) a uni of velociy. When he acceleraion curve is above he ime axis, he area is posiive; when he curve is below he ime axis, he area is negaive. Similarly, because velociy v is defined in erms of he posiion x as v dx/d, hen x x v d, (-4) where x is he posiion a ime and x is he posiion a ime. The definie inegral on he righ side of Eq. -4 can be evaluaed from a v() graph, like ha shown in Fig. -b. In paricular, v d area beween velociy curve and ime axis, from o. (-5) If he uni of velociy is m/s and he uni of ime is s, hen he corresponding uni of area on he graph is ( m/s)( s) m, which is (properly) a uni of posiion and displacemen. Wheher his area is posiive or negaive is deermined as described for he a() curve of Fig. -a. a Area This area gives he change in velociy. (a) Fig. - The area beween a ploed curve and he horizonal ime axis, from ime o ime, is indicaed for (a) a graph of acceleraion a versus and (b) a graph of velociy v versus. v (b) Area This area gives he change in posiion.

16 hall-isv_c_3-37hr.qxd 6--9 :7 Page 8 8 CHAPTER MOTION ALONG A STRAIGHT LINE Sample Problem Graphical inegraion a versus, whiplash injury Whiplash injury commonly occurs in a rear-end collision where a fron car is hi from behind by a second car. In he 97s, researchers concluded ha he injury was due o he occupan s head being whipped back over he op of he sea as he car was slammed forward. As a resul of his finding, head resrains were buil ino cars, ye neck injuries in rearend collisions coninued o occur. In a recen es o sudy neck injury in rear-end collisions, a voluneer was srapped o a sea ha was hen moved abruply o simulae a collision by a rear car moving a.5 km/h. Figure -3a gives he acceleraions of he voluneer s orso and head during he collision, which began a ime. The orso acceleraion was delayed by 4 ms because during ha ime inerval he sea back had o compress agains he voluneer. The head acceleraion was delayed by an addiional 7 ms. Wha was he orso speed when he head began o accelerae? KEY IDEA We can calculae he orso speed a any ime by finding an area on he orso a() graph. Calculaions: We know ha he iniial orso speed is v a ime, a he sar of he collision. We wan he orso speed v a ime ms, which is when he head begins o accelerae. Combining Eqs. - and -3, we can wrie v v area beween acceleraion curve and ime axis, from o. (-6) For convenience, le us separae he area ino hree regions (Fig. -3b). From o 4 ms, region A has no area: area A. From 4 ms o ms, region B has he shape of a riangle, wih area area B (.6 s)(5 m/s ).5 m/s. From ms o ms, region C has he shape of a recangle, wih area area C (. s)(5 m/s ).5 m/s. Subsiuing hese values and v ino Eq. -6 gives us v.5 m/s.5 m/s, or v. m/s 7. km/h. (Answer) Commens: When he head is jus saring o move forward, he orso already has a speed of 7. km/h. Researchers argue ha i is his difference in speeds during he early sage of a rear-end collision ha injures he neck. The backward whipping of he head happens laer and could, especially if here is no head resrain, increase he injury. a (m/s ) 5 Torso Head (ms) (a) a 5 A (b) B 4 Fig. -3 (a) The a() curve of he orso and head of a voluneer in a simulaion of a rear-end collision. (b) Breaking up he region beween he ploed curve and he ime axis o calculae he area. C The oal area gives he change in velociy. Addiional examples, video, and pracice available a WileyPLUS

17 hall-isv_c_3-37hr.qxd 6--9 :7 Page 9 QUESTIONS 9 PART Posiion The posiion x of a paricle on an x axis locaes he paricle wih respec o he origin, or zero poin, of he axis. The posiion is eiher posiive or negaive, according o which side of he origin he paricle is on, or zero if he paricle is a he origin. The posiive direcion on an axis is he direcion of increasing posiive numbers; he opposie direcion is he negaive direcion on he axis. Displacemen in is posiion: The displacemen x of a paricle is he change x x x. (-) Displacemen is a vecor quaniy. I is posiive if he paricle has moved in he posiive direcion of he x axis and negaive if he paricle has moved in he negaive direcion. Average Velociy When a paricle has moved from posiion x o posiion x during a ime inerval, is average velociy during ha inerval is v avg x (-) The algebraic sign of v avg indicaes he direcion of moion (v avg is a vecor quaniy). Average velociy does no depend on he acual disance a paricle moves, bu insead depends on is original and final posiions. On a graph of x versus, he average velociy for a ime inerval is he slope of he sraigh line connecing he poins on he curve ha represen he wo ends of he inerval. Average Speed The average speed s avg of a paricle during a ime inerval depends on he oal disance he paricle moves in ha ime inerval: s avg (-3) Insananeous Velociy The insananeous velociy (or simply velociy) v of a moving paricle is x v lim : x x. oal disance. dx d, (-4) where x and are defined by Eq. -. The insananeous velociy (a a paricular ime) may be found as he slope (a ha paricular ime) of he graph of x versus. Speed is he magniude of insananeous velociy. Average Acceleraion Average acceleraion is he raio of a change in velociy v o he ime inerval in which he change occurs: a avg v (-7). The algebraic sign indicaes he direcion of a avg. Insananeous Acceleraion Insananeous acceleraion (or simply acceleraion) a is he firs ime derivaive of velociy v() and he second ime derivaive of posiion x(): a dv d d x d. (-8, -9) On a graph of v versus, he acceleraion a a any ime is he slope of he curve a he poin ha represens. Consan Acceleraion The five equaions in Table - describe he moion of a paricle wih consan acceleraion: v v a, (-) x x v a, v v a(x x ), x x (v v), x x v a. These are no valid when he acceleraion is no consan. (-5) (-6) (-7) (-8) Free-Fall Acceleraion An imporan example of sraighline moion wih consan acceleraion is ha of an objec rising or falling freely near Earh s surface. The consan acceleraion equaions describe his moion, bu we make wo changes in noaion: () we refer he moion o he verical y axis wih y verically up; () we replace a wih g, where g is he magniude of he free-fall acceleraion. Near Earh s surface, g 9.8 m/s ( 3 f/s ). Figure -4 gives he velociy of a paricle moving on an x axis. Wha are (a) he iniial and (b) he final direcions of ravel? (c) Does he paricle sop momenarily? (d) Is he acceleraion posiive or negaive? (e) Is i consan or varying? Figure -5 gives he acceleraion a() of a Chihuahua as i chases v Fig. -4 Quesion. a A B C D E F G H Fig. -5 Quesion.

18 hall-isv_c_3-37hr.qxd 3--9 : Page 3 3 CHAPTER MOTION ALONG A STRAIGHT LINE a German shepherd along an axis. In which of he ime periods indicaed does he Chihuahua move a consan speed? 3 Figure -6 shows four pahs along which objecs move from a saring poin o a final poin, all in he same ime inerval. The pahs pass over a grid of equally spaced sraigh lines. Rank he pahs according o (a) he average velociy of he objecs and (b) he average speed of he objecs, greaes firs. 4 Figure -7 is a graph of a paricle s posiion along an x axis versus ime. (a) A ime, wha is he sign of he paricle s posiion? Is he paricle s velociy posiive, negaive, or a (b) s, (c) s, and (d) 3 s? (e) How many imes does he paricle go hrough he poin x? 5 Figure -8 gives he velociy of a paricle moving along an axis. Poin is a he highes poin on he curve; poin 4 is a he lowes poin; and poins and 6 are a he same heigh. Wha is he direcion of ravel a (a) ime and (b) poin 4? (c) A which of he six numbered poins does he paricle reverse is direcion of ravel? (d) Rank he six poins according o he magniude of he acceleraion, greaes firs. v 4 Fig. -6 Quesion 3. x 3 4 Fig. -7 Quesion Fig. -8 Quesion A, a paricle moving along v an x axis is a posiion x m. The signs of he paricle s iniial velociy v (a ime ) and consan acceleraion a are, respecively, for four siuaions: (), ; (), ; (3), ; (4),. In which siuaions will he paricle (a) sop momenarily, (b) pass hrough he origin, and (c) never pass hrough he origin? 7 Hanging over he railing of a bridge, you drop an egg (no iniial velociy) as you hrow a second egg downward. Which curves in Fig. -9 give he velociy v() for (a) he dropped egg and (b) he hrown egg? (Curves A and B are parallel; so are C, D, and E; so are F and G.) 8 The following equaions give he velociy v() of a paricle in four siuaions: (a) v 3; (b) v 4 6; (c) v 3 4; (d) v 5 3. To which of hese siuaions do he equaions of Table - apply? 9 In Fig. -, a cream angerine is hrown direcly upward pas hree evenly spaced windows of equal heighs. Rank he windows according o (a) he average speed of he cream angerine while passing hem, (b) he ime he cream angerine akes o pass hem, (c) he magniude of he acceleraion of he cream angerine while passing hem, and (d) he change v in he speed of he cream angerine during he passage, greaes firs. A G F E Fig. -9 Quesion 7. 3 D Fig. - Quesion 9. C B SSM Tuoring problem available (a insrucor s discreion) in WileyPLUS and WebAssign Worked-ou soluion available in Suden Soluions Manual WWW Worked-ou soluion is a Number of dos indicaes level of problem difficuly ILW Ineracive soluion is a Addiional informaion available in The Flying Circus of Physics and a flyingcircusofphysics.com sec. -4 Average Velociy and Average Speed During a hard sneeze, your eyes migh shu for.5 s. If you are driving a car a 9 km/h during such a sneeze, how far does he car move during ha ime? Compue your average velociy in he following wo cases: (a) You walk 73. m a a speed of. m/s and hen run 73. m a a speed of.85 m/s along a sraigh rack. (b) You walk for. min a a speed of. m/s and hen run for. min a 3.5 m/s along a sraigh rack. (c) Graph x versus for boh cases and indicae how he average velociy is found on he graph. 3 SSM WWW An auomobile ravels on a sraigh road for 4 km a 3 km/h. I hen coninues in he same direcion for anoher 4 km a 6 km/h. (a) Wha is he average velociy of he car during he full 8 km rip? (Assume ha i moves in he posiive x direc- ion.) (b) Wha is he average speed? (c) Graph x versus and indicae how he average velociy is found on he graph. 4 A car ravels up a hill a a consan speed of 35 km/h and reurns down he hill a a consan speed of 6 km/h. Calculae he average speed for he round rip. 5 SSM The posiion of an objec moving along an x axis is given by x 3 4 3, where x is in meers and in seconds. Find he posiion of he objec a he following values of : (a) s, (b) s, (c) 3 s, and (d) 4 s. (e) Wha is he objec s displacemen beween and 4 s? (f) Wha is is average velociy for he ime inerval from s o 4 s? (g) Graph x versus for 4 s and indicae how he answer for (f) can be found on he graph. 6 The 99 world speed record for a bicycle (human-powered vehicle) was se by Chris Huber. His ime hrough he measured

19 hall-isv_c_3-37hr.qxd 6--9 :7 Page 3 PROBLEMS 3 PART m srech was a sizzling 6.59 s, a which he commened, Cogio ergo zoom! (I hink, herefore I go fas!). In, Sam Whiingham bea Huber s record by 9. km/h. Wha was Whiingham s ime hrough he m? 7 Two rains, each having a speed of 3 km/h, are headed a each oher on he same sraigh rack. A bird ha can fly 6 km/h flies off he fron of one rain when hey are 6 km apar and heads direcly for he oher rain. On reaching he oher rain, he bird flies direcly back o he firs rain, and so forh. (We have no idea why a bird would behave in his way.) Wha is he oal disance he bird ravels before he rains collide? 8 Panic escape. Figure - shows a general siuaion in which a sream of people aemp o escape hrough an exi door ha urns ou o be locked. The people move oward he door a speed v s 3.5 m/s, are each d.5 m in deph, and are separaed by L.75 m.the arrangemen in Fig. - occurs a ime. (a) A wha average rae does he layer of people a he door increase? (b) A wha ime does he layer s deph reach 5. m? (The answers reveal how quickly such a siuaion becomes dangerous.) L L L d d d Locked door Fig. - Problem 8. 9 ILW In km races, runner on rack (wih ime min, 7.95 s) appears o be faser han runner on rack ( min, 8.5 s). However, lengh L of rack migh be slighly greaer han lengh L of rack. How large can L L be for us sill o conclude ha runner is faser? To se a speed record in a measured (sraigh-line) disance d, a race car mus be driven firs in one direcion (in ime ) and hen in he opposie direcion (in ime ). (a) To eliminae he effecs of he wind and obain he car s speed v c in a windless siuaion, should we find he average of d/ and d/ (mehod ) or should we divide d by he average of and? (b) Wha is he fracional difference in he wo mehods when a seady wind blows along he car s roue and he raio of he wind speed v w o he car s speed v c is.8? You are o drive o an inerview in anoher own, a a disance of 3 km on an expressway. The inerview is a 5 A.M. You plan o drive a km/h, so you leave a 8 A.M. o allow some exra ime. You drive a ha speed for he firs km, bu hen consrucion work forces you o slow o 4 km/h for 4 km. Wha would be he leas speed needed for he res of he rip o arrive in ime for he inerview? Traffic shock wave. An abrup slowdown in concenraed raffic can ravel as a pulse, ermed a shock wave, along he line of cars, eiher downsream (in he raffic direcion) or upsream, or i can be saionary. Figure - shows a uniformly spaced line of cars moving a speed v 5. m/s oward a uniformly spaced line of slow cars moving a speed v s 5. m/s. Assume ha each faser car adds lengh L. m (car lengh plus buffer zone) o he line of slow cars when i joins he line, and assume i slows abruply a he las insan. (a) For wha separaion disance d beween he faser cars does he shock wave remain saionary? If he separaion is wice ha amoun, wha are he (b) speed and (c) direcion (upsream or downsream) of he shock wave? L d v L d L L L Car Buffer Fig. - Problem. 3 ILW You drive on Inersae from San Anonio o Houson, half he ime a 55 km/h and he oher half a 9 km/h. On he way back you ravel half he disance a 55 km/h and he oher half a 9 km/h. Wha is your average speed (a) from San Anonio o Houson, (b) from Houson back o San Anonio, and (c) for he enire rip? (d) Wha is your average velociy for he enire rip? (e) Skech x versus for (a), assuming he moion is all in he posiive x direcion. Indicae how he average velociy can be found on he skech. sec. -5 Insananeous Velociy and Speed 4 An elecron moving along he x axis has a posiion given by x 6e m, where is in seconds. How far is he elecron from he origin when i momenarily sops? 5 (a) If a paricle s posiion is given by x 4 3 (where is in seconds and x is in meers), wha is is velociy a s? (b) Is i moving in he posiive or negaive direcion of x jus hen? (c) Wha is is speed jus hen? (d) Is he speed increasing or decreasing jus hen? (Try answering he nex wo quesions wihou furher calculaion.) (e) Is here ever an insan when he velociy is zero? If so, give he ime ; if no, answer no. (f) Is here a ime afer 3 s when he paricle is moving in he negaive direcion of x? If so, give he ime ; if no, answer no. 6 The posiion funcion x() of a paricle moving along an x axis is x 4. 6., wih x in meers and in seconds. (a) A wha ime and (b) where does he paricle (momenarily) sop? A wha (c) negaive ime and (d) posiive ime does he paricle pass hrough he origin? (e) Graph x versus for he range 5 s o 5 s. (f) To shif he curve righward on he graph, should we include he erm or he erm in x()? (g) Does ha inclusion increase or decrease he value of x a which he paricle momenarily sops? 7 The posiion of a paricle moving along he x axis is given in cenimeers by x , where is in seconds. Calculae (a) he average velociy during he ime inerval. s o 3. s; (b) he insananeous velociy a. s; (c) he insananeous velociy a 3. s; (d) he insananeous velociy a.5 s; and (e) he insananeous velociy when he paricle is midway beween is posiions a. s and 3. s. (f) Graph x versus and indicae your answers graphically. sec. -6 Acceleraion 8 The posiion of a paricle moving along an x axis is given by x 3, where x is in meers and is in seconds. Deermine (a) he posiion, (b) he velociy, and (c) he acceleraion of he paricle a 3.5 s. (d) Wha is he maximum posiive coordinae reached by he paricle and (e) a wha ime is i reached? (f) Wha is he maximum posiive velociy reached by he paricle and (g) a v s