Kinemaics Inroducion Moion is fundamenal o our lives and o our hinking. Moving from place o place in a given amoun of ime helps define boh who we are and how we see he world. Seeing oher people, objecs or animals moving and being able o imagine where hey came from, where hey re going and how long i ll ake hem o ge here is naural o us. All animals, no jus humans, do calculaions abou he moion of hemselves and he world around hem. Wihou ha abiliy we could no survive. A basic undersanding of moion is deep in our minds and was here long before we could wrie or alk abou physics. I should be no surprise ha he firs, and mos fundamenal, quesions of physics relae o moion. Many of he firs wriings of physics are on his opic and dae back housands of years. The sudy of moion is called kinemaics. I comes from he Greek words kinema, which means moion. Almos everyhing we learn in physics will involve he moion of objecs. So, kinemaics mus be undersood well in order o undersand oher opics we will be sudying in he fuure. Time and Disance Everyone knows wha ime and disance are unil hey re asked o define hem. Go ahead; ry o define wha ime is wihou using he idea of ime iself in your definiion. Here are some definiions we ve heard before. I m sure you ll come up wih some new ones as well. Time is he amoun of ime ha goes by. Time is how long i akes for somehing o happen. Time is how long I have o wai. The problem wih hese definiions is ha hey use he word ime in he definiion, or imply is use. In he firs definiion, if you don know wha ime is, how can you use i o define ime? In he second wo definiions, he phrase how long is jus anoher way of saying he amoun of ime. They don qualify as definiions since if you didn know wha ime was in he beginning; you sill don. See if you can define disance. We hink you ll run ino he same problem. We all believe ha we know wha hese erms mean, bu i s impossible o define hem. We move hrough ime and space as naurally as a fish moves in he waer. Time and space are all around us, bu we can really say wha hey are. They are oo fundamenal o be defined. We have o ake hem as givens. You and I have a sense ha we know wha hey are. Tha sense comes direcly from our minds and bodies; bu we can really define hem beyond ha. Our sense of ime and disance mus have evolved in us long before we could hink abou hem. All animals need a basic percepion abou ime and disance in order o survive. The mos primiive one celled animals move abou hrough ime and space and surely hey don have a definiion of wha hose conceps represen. The sense of ime and disance predaes our abiliy o hink and perhaps ha s why we can use our minds o define hem; bu we can work wih hem. Kinemaics - 1 v 1.0 2009 by Goodman & Zavoroniy
We can measure he flow of ime wih clocks and he disance using a ruler. These don represen definiions; bu hey do allow us o compare differen inervals of ime and space wih one anoher. The abiliy o measure ime and disance represens a saring poin for physics. For insance, le s say he amoun of ime i akes for me o run once around a rack is 2 minues. Tha means ha he minue hand of my sop wach will go around wo imes while I run around he rack one ime. Tha doesn ell me wha ime is, bu i does ell me ha hose wo processes ook he same amoun of i. So I can compare he ime i akes for somehing o happen o he ime i akes for somehing else o happen. Similarly, I don need o know wha disance is in order o compare he disance beween wo objecs wih he disance beween wo oher objecs. I can say ha he disance from he heel o he oe of my foo is he same as he disance from one end o he oher of a one foo ruler. So I can say ha my foo is one foo long, even wihou a definiion of wha lengh means. Unis of Time and Disance In order ha people can compare heir measuremens wih hose aken by ohers; an inernaional sysem of measuremens was agreed upon. The Sysem Inernaional (SI) is used by virually all scieniss in he world. In ha sysem, he basic uni of lengh is he meer and he basic uni of ime is he second. Disance is measured in meers (m). Time is measured in seconds (s). Lengh measuremens are made by comparing he disance beween any wo locaions o he disance beween he wo ends of a rod ha is defined o be one meer long. Time measuremens are made by measuring he ime beween evens wih he ime i akes for a second o ick off on a clock Scieniss use he meer, insead of he foo, for measuring disances because i s simpler. Using he SI sysem o measure lenghs in meers, is no more accurae han using he English sysem measuring lenghs in fee. However, i urns ou o be a lo easier. Tha s because he mahemaics of dealing wih 12 inches in a foo and 5280 fee in a mile is jus a lo more difficul han he meric sysem of 100 cenimeers in a meer and 1000 meers in a kilomeer. When you sar solving problems, you ll be happy no having o deal wih fee, miles and inches. Consan Speed If everyhing in he world jus sood sill, we d need o measure ime and disance separaely and we d be done. Bu ha d be a prey boring world. Many of he mos ineresing hings involve moion; objecs moving from one locaion o anoher in a cerain amoun of ime. How fas hey do his is he speed of he objec. Speed is no a fundamenal propery of he world, like disance and ime, bu is a human invenion. I is defined as he raio of he disance raveled divided by he ime i ook o ravel ha disance. Speed Disance Time or Kinemaics - 2 v 1.0 2009 by Goodman & Zavoroniy
s d The equal sign wih hree parallel lines jus poins ou ha his is a definiion. We made up he word speed and hen defined i o mean he raio of disance and ime. There s no way ha his could be proven righ or wrong, hrough experimen or any oher means, since we jus made i up. When we use he formula, we ll usually jus wrie i wih a normal equals sign, bu we should remember ha i s jus our definiion. Speed and disance don depend on he direcion raveled. So if you walk wo miles o school and hen reurn back home, he oal disance you raveled is four miles. If i ook you one hour o do ha, your average speed was four miles per hour. In his secion we re assuming ha your speed is consan. In laer secions, we ll alk abou cases where your speed changes. Unis of Speed The unis of speed can be derived from is formula: s = d The SI uni of disance is meer (m) and ime in second (s). Therefore, he uni of speed is m/s. Problem Solving When solving physics problems here is a series of seps ha should be followed. In he early problems ha we ll be doing, i ll be possible o skip some seps and sill ge he correc answer. Bu ha won give you a chance o pracice he mehods ha you ll need o solve more difficul problems. I s wise o learn how o swim in he shallow end of he pool, bu if all you do is sand up here, i won be much help when he waer ges deep. So, please use he following approach on all he problems you solve righ from he beginning. I ll pay off in he end. 1. Read he problem carefully and underline, or make noe of, any informaion ha seems like i may be useful. 2. Read he problem hrough again, bu now sar wriing down he informaion ha will be of value o you. Idenify wha is being asked for and wha is being given. 3. If appropriae, draw a skech. 4. Idenify a formula ha relaes o he informaion ha you ve been given o he informaion you ve been o asked o solve for. 5. Rearrange he formula so i s solved for he variable you re looking for. This means, ge ha variable o be alone on he lef side of he equals sign. 6. Subsiue in he values you ve been given, including unis. 7. Calculae he numerical resul. 8. Solve for he unis on he righ side of he equaion and compare hose o he unis ha are appropriae for wha you re solving for. For insance, if you re solving for disance, he unis should be in meers no meers per second. 9. Reread he problem and make sure ha your answer makes sense. I s been shown ha successful physics sudens read each problem a leas hree imes. Example 1: Riding your bike a a consan speed akes you 25 seconds o ravel a disance of 1500 meers. Wha was your speed? Kinemaics - 3 v 1.0 2009 by Goodman & Zavoroniy
We ve been given disance and ime and we need o find speed. s =? d = 1500 m = 25 s We can direcly use he equaion s = d. Tha gives he relaionship beween he hree variables and is already solved for he variable ha we re looking for. Afer wriing down he formula we jus have o subsiue in he values wih unis. s = d s = 1500m 25s s = 60 m/s Noe ha no only is 1500 divided by 25 equal o 60 bu also ha m divided by s yields m/s which is he correc uni for speed. By always doing he same mahemaical operaions on he unis as well as he numbers you should end up wih he correc unis in your answer. This is a good way o check if you did he problem correcly. Le s now looks a an example where he formula can be used direcly. Example 2: How far will you ravel if you are driving a a consan speed of 25 m/s for a ime of 360s? We ve been given speed and ime and we need o find disance. s = 25m/s d =? = 360s We ll use he same formula (s=d/) since i relaes he known values (speed and ime) o he unknown value (disance). Bu, in his case while we need o solve for disance (d), he formula ha we have (s = d ) is solved for speed (s). We mus firs use algebra o rearrange he formula so ha we solve for d. Once you reach his poin, you subsiue he values ino he formula. We ll make use of hree rules o rearrange he formula. 1. If he variable ha we are solving for is in he numeraor and isn alone, hen i is mahemaically conneced o oher numbers and/or variables. We can ge i alone by performing he inverse operaion on each of he oher variables or numbers. For insance, if d is divided by ; we can ge d alone by muliplying by (since muliplicaion is he opposie of division). 2. We can do anyhing we wan o one side of an equaion as long as we also do i o he oher side (excep dividing by zero). So if we muliply he righ side of he equaion, d/, by we also have o muliply he lef side of he equaion, s, by. Kinemaics - 4 v 1.0 2009 by Goodman & Zavoroniy
3. We can always swich he righ and lef sides of an equaion. Le s use his approach o solve he equaion, s = d, for d. s = d s = ( d ) s = d d = s Since we are solving for d, and d is being divided by, we mus muliply d/ by. Bu we can only do ha if we muliply s by. So muliply boh sides by. Cancel on he righ since i s in he numeraor and he denominaor and / =1 Swich he d o he lef side Subsiue in values for s and d = (25 m ) (360s) s d = 9000m Noe ha no only is 25 imes 360 equal o 9000 bu also ha meers per second imes seconds is equal o meers since he seconds will cancel ou. Tha gives us unis of meers, which makes sense since we are solving for a disance. Example 3: How much ime will i ake you o ravel 3600 m if you are driving a a consan speed of 20 m/s? We ve been given he disance and he speed, and we need o find he ime. s = 20m/s d = 3600m =? In his case we are solving for ime () bu he formula we have (s = d/) is solved for speed (s). We mus firs use algebra o rearrange he formula so ha i is solved for. Only a ha poin should values be subsiued ino he formula s = d/. We need o add one addiional rule o rearrange he formula in his case. 4. The unknown for which we are solving mus be in he numeraor, no he denominaor. So if we are solving he formula s = d/ for, our firs sep mus be o move o he numeraor on he lef insead of leaving i in he denominaor on he righ. To do his, we need o muliply boh sides of he equaion by, giving us s = d. Then we can proceed jus as we did above. Kinemaics - 5 v 1.0 2009 by Goodman & Zavoroniy
s = d Muliply boh sides by o cancel he on he righ and ge i on he lef s = ( d ) Cancel on he righ since / = 1 s = d s = d s s = d s Since is no alone, because i is muliplied by s, we mus divide boh sides by s Cancel s on he lef side since s/s = 1 Subsiue in values for d and s = 3600m 20 m s = 180s The unis may be a bi harder o undersand in his case. We have meers divided by meers per second. Bu you may recall from fracions ha dividing by a fracion is he same as muliplying by is reciprocal. (Dividing by 1/3 is he same as muliplying by 3.) So dividing by m/s is he same as muliplying by s/m. This makes i clear ha he meers will cancel ou, when we muliply s/m by m, and we are lef wih seconds, an appropriae uni for ime. Insananeous Speed There s an old joke abou a person who s pulled over for speeding. The police officer ells he speeder ha he was going 60 miles per hour in a fory mile per hour zone. The speeder s response is ha he couldn have been going sixy miles per hour since he d only been driving for fifeen minues. The reason ha argumen doesn work is ha speed is a raio of disance and ime. There are an infinie number of ways ha you can calculae a speed of en meers per second. Some are shown in he able below. Disance Time Speed (m) (s) (m/s) 1000 100 10 500 50 10 100 10 10 10 1 10 1 0.1 10 0.1 0.01 10 0.01 0.001 10 0.001 0.0001 10 Kinemaics - 6 v 1.0 2009 by Goodman & Zavoroniy
You can see ha a he boom of he char ha if you ravel one housandh of a meer in one en housandh of a second you are raveling a a speed of en meers per second. Tha ime and disance can be made as small as you like. When he ime over which he speed is measured is very small, he speed ha is calculaed is called he insananeous speed. This is he speed ha you read on your speedomeer or ha a policeman reads on his radar or laser gun. Average Speed While raveling along, your varies; o go up and down along he way. You migh even sop for a while o have lunch. Your insananeous speed a some momen during your rip and your average speed for he oal rip are ofen no he same. Your average speed is calculaed by deermining he oal disance ha you raveled and dividing by he oal ime ha i ook you o ravel ha disance. Example 4: You ride your bike home from school by way of your friend s house. I akes you 7 minues o ravel he 2500m o your friend s house. You hen spend 10 minues here. You hen ravel he 3500 m o your house in 9 minues. Wha was your average speed for your oal rip home? Your average speed will be obained by dividing he oal disance raveled by he oal ime i ook o ravel ha disance. In his case, he rip consised of hree segmens. The firs segmen (I) is he ride o your friend s house, he second segmen (II) was he ime a your friend s house and he hird segmen (III) was your ride home from your friend s house. In he char below, he speed is calculaed for each segmen, even hough ha is no necessary o ge he answer ha was requesed, he average speed for your oal rip. All calculaed figures are shown in bold ype. Segmen Disance Time Speed (m) (s) (m/s) I 2500 420 6.0 II 0 600 0.0 III 3500 540 6.5 Toal/ Average 6000 1560 3.8 For insance he speed for he firs segmen is given by: s =? d = 2500m = 7 minues = 420 seconds Noe ha we need o conver he given ime in seconds in order o use SI unis. Since here are 60 seconds in a minue, his requires muliplying seven minues by he fracion (60 seconds / 1 minue). This leaves us wih 420 seconds. s = d s = 2500m 420s s = 6.0 m/s Kinemaics - 7 v 1.0 2009 by Goodman & Zavoroniy
For he second segmen, your speed was zero since you were wihin he house. Bu even hough you weren moving, ime was going by. So he 10 minues, or 600 seconds, sill couns owards he oal elapsed ime. The hird segmen is calculaed in he same manner as was he firs. s =? d = 3500m = 9 minues = 540 seconds s = d s = 3500m 540s s = 6.5 m/s The average speed is calculaed by aking he oal disance, 6000m, and dividing i by he oal ime, 1560s, o ge an average speed of 3.8 m/s. While i wasn necessary in his case o calculae he speed for each inerval, i s imporan o noe ha he average speed is no he average of he speeds. The average of 6.0 m/s, 0.0 m/s and 6.5 m/s is 4.2 m/s. Bu his is no he correc answer. The correc answer can only be obained by firs finding he oal disance and dividing ha by he oal ime, by doing his you ge he answer of 3.8 m/s. Example 5: You run a disance of 210 m a a speed of 7 m/s. You hen jog a disance of 200 m in a ime of 40s. Finally, you run for 25s a a speed of 6 m/s. Wha was he average speed of your oal run? Your average speed will be obained by dividing he oal disance raveled by he oal ime i ook o ravel ha disance. In his case, he rip consised of hree segmens. In he char below, differen calculaions are required for each segmen in order o obain he average speed for your oal rip. Segmen Disance Time Speed (m) (s) (m/s) I 210 30 7.0 II 200 40 5.0 III 150 25 6.0 Toal / Average 560 95 5.9 The ime for he firs segmen is given by: s = 7.0 m/s d = 210m =? Kinemaics - 8 v 1.0 2009 by Goodman & Zavoroniy
s= d s = d = d s = 210m 7 m s = 30 s We don really need o calculae your speed for he second segmen, bu we ll do i anyway. s =? d = 200m = 40s s = d s = 200m 40s s= 5.0 m/s The disance needs o be calculaed for he hird segmen. s = 6.0 m/s d =? = 25s s = d s = d d = s d = (6.0 m/s) (25s) d = 150 m The average speed is calculaed by aking he oal disance, 560m, and dividing i by he oal ime, 95s, o ge 5.9 m/s. Posiion, Displacemen and Velociy So far our analysis has no required, or even allowed, us o know anyhing abou he direcion of he moion under sudy. Bu in real life, direcion is usually very imporan. Wheher you re driving 60 miles per hour norh or 60 miles per hour souh, i makes a grea deal of difference as o where you end up. Scalars are quaniies ha are defined only by heir magniude; he numerical value. Speed, ime and disance are all examples of scalars. When we speak of 40 m/s, 20 minues or 3 miles we re no giving any informaion abou direcion. Vecors are quaniies ha are defined by boh he magniude and direcion. So, insead of saying ha I raveled a disance of 400m, I would say ha I raveled 400m norh; I am now defining vecor. The vecor ha is or defined by combining disance wih direcion is called displacemen. The symbol for displacemen is Δx. We ll alk more abou ha symbol a lile Kinemaics - 9 v 1.0 2009 by Goodman & Zavoroniy
laer, bu you can use i in he meanime. Also, in order o keep rack of wha s a scalar and wha s a vecor, we ll always show vecors in a bold ypeface. There are imporan differences when we work wih scalars and vecors. This differences can be mos easily seen by using disance and displacemen as examples. For insance, while disance are always posiive, since hey have no direcion associaed wih hem, displacemen can be posiive or negaive. Tha means ha if I were o ake a rip which involved going 200m norh and hen 200m souh I ge very differen answers for he oal disance I raveled and my oal displacemen. I ge my oal disance by adding 200m o 200m and geing 400m. Tha s he oal disance ha I walked. On he oher hand, my displacemen represens he sum of he wo displacemens. My iniial displacemen norh is equal and opposie o my final displacemen souh, so hey will cancel each oher ou. If I hink of norh as he posiive direcion, he firs displacemen would be +200m, while my second displacemen would be -200m, since i s reveling o he souh. The sum of +200m and -200m is zero. Tha s because he direcion of he moion maers wih displacemen while i doesn apply o disance. As a resul, displacemen ells you how far you are from where you sared. In his case, I am zero disance from where I sared, since I end up back where I sared off. Example 6: You drive 1500m norh and hen 500m souh. Deermine boh he oal disance you raveled and your oal displacemen from where you sared. The disance raveled is jus he sum of he wo disances, 1500m and 500m, 2000m. In order o deermine your oal displacemen we need o firs define our direcions. Le s call movemen o he norh posiive and movemen o he souh negaive (which direcion we call posiive won affec our answer as long as we re consisen.). Tha means ha for he firs par of he rip your displacemen is +1500m and for he second par of he rip your displacemen is - 500m. Your oal displacemen is he sum of hose wo, +1000m. Since we decided ha we d call he norh direcion posiive, your final displacemen is 1000m norh. The las sep of convering from +1000m o 1000m norh is imporan in ha our choice of + or was arbirary so we need o ranslae back o he original direcions we were given in he problem. The imporan poin here is ha he answers are differen and have differen uses. The disance you raveled, 2000m, ells you somehing abou how ired you may be because i ells you he oal disance you had o move yourself during his rip. Your displacemen, 1000m norh, ells you where you are a his poin in your ravels relaive o where you sared. The same difference exiss beween speed and velociy. The symbol for velociy is v and he symbol for average velociy is vavg. The average velociy is deermined by dividing your oal displacemen by he ime i ook for ha displacemen. This is similar o how we calculaed average speed by dividing he oal disance raveled by he oal ime i ook o ravel ha disance. s d while Kinemaics - 10 v 1.0 2009 by Goodman & Zavoroniy vavg x
Example 7: If he ravel in Example 6 was done a consan speed and required a oal ime of 500s, deermine he average speed and he average velociy. s =? d = 2000m = 500s s = d s = 2000m 500s s = 4.0 m/s vavg =? Δx = 1000m norh = 500s vavg = x 1000m Norh v = 500s v = 2m/s Norh Noe ha he numerical answers are differen and ha he answer for velociy includes a direcion while he answer for speed does no. Coordinae Sysems The displacemen of an objec ells us how is posiion has changed. In order o beer undersand wha ha means we need a way of defining posiion; we need a coordinae sysem. The requiremens of any coordinae sysem are an origin and an orienaion. In oher words, you need o pick a zero from which you ll be making measuremens and you need o know he direcion in which you will be measuring. The simples ype of coordinae sysem is onedimensional, in which case he coordinae sysem becomes a number line, as shown below. The origin is locaed a zero, negaive posiions are o he lef of he origin and posiive posiions are o he righ. We can idenify differen locaions on he number line as such as x0, x1 and x2. In he diagram above, x0 is locaed a 0, x1 is locaed a +5m and x2 is locaed a -5m. We can now refine our definiion of displacemen, he change in he posiion of an objec, as being he difference beween an objec s final posiion, x, and is iniial posiion, xo. I now becomes clear why he symbol for displacemen is Δx. The Greek leer dela, Δ, means he change in so Δx Kinemaics - 11 v 1.0 2009 by Goodman & Zavoroniy
can be read as dela x or he change in x. Symbolically his becomes; Δx x - xo Example 8: An objec moves from an iniial posiion of +5m o a final posiion of +10m in a ime of 10s. Wha displacemen did i undergo? Wha was is average velociy? x = +10m xo = + 5m Δx =? Δx = x - xo Δx = (+10m) (+5m) Δx = +5m vavg =? Δx = +5m = 10 s vavg = x v = +5m 10s v = +0.5m/s Example 9: An objec moves from an iniial posiion of +5m o a final posiion of -10m in a ime of 0.25s. Wha displacemen did i undergo? Wha was is average velociy? x = -10m xo = + 5m Δx =? Δx = x - xo Δx = (-10m) (+5m) Δx = -15m vavg =? Δx = -15m = 0.25 s vavg = x vavg = 5m.25s vavg = -60m/s Once again, noice ha he answer includes a magniude, 15m, as well as a direcion, -. Kinemaics - 12 v 1.0 2009 by Goodman & Zavoroniy
Vecors, such as displacemen or velociy, can be depiced as an arrow. The lengh of he arrow represens he magniude of he vecor and he direcion i is poined represens he direcion of he vecor. Vecors can be added eiher graphically or algebraically. (Even if you re solving a problem algebraically i s helpful o also skech he addiion graphically so you can make cerain ha your answer makes sense.) The way o add vecors graphically is o draw he firs vecor saring a he origin of he problem. I mus be drawn o scale and poined in he correc direcion. The second vecor should be drawn in he same manner, bu saring where he firs one ended. The sum of he wo vecors is simply a hird vecor which sars where he firs vecor sared and ends where he las vecor ended. In oher words, he soluion is a hird vecor ha connecs he beginning of he firs o he end of he las vecor drawn. The arrow ip of ha vecor should poin away from he locaion from which i sared. The following example is solved boh graphically and algebraically. Example 10: Beginning a a locaion ha s 400 m eas of your home, you ravel 500m eas and hen 300m wes. How far are you now from your home? Wha displacemen have you experienced during your rip? If his ravel ook a oal ime of 20s, wha was your average velociy? The graphical soluion, shown below, sars by drawing a sufficienly large eas-wes axis. If your home is locaed a x = 0m, hen your iniial posiion, xo, is 400m eas of ha. Draw a vecor ha describes he firs par of your rip by drawing an arrow ha begins a he locaion 400m o he eas of your house, is 500m long and poins owards he eas. Tha ip of ha arrow should hen be 900m eas of your house. Then draw he vecor for he second par of your rip by drawing an arrow ha begins a he ip of he firs arrow, 900m eas of your house, is 300m long and is poined o he wes, owards your house. The ip of ha arrow should now lie a a locaion 600m o he eas of your house; ha is your final locaion, x. Your displacemen is he difference beween your final and iniial posiions. This is obained, graphically, by drawing an arrow ha sars a your iniial posiion and ends a your final posiion. The lengh of his arrow, which can be physically measured or read off he scale, is he magniude of your displacemen. The direcion of he arrow is he direcion of your displacemen. As is shown below, i can be seen ha your displacemen is 200m eas. Your average velociy is your oal displacemen divided by he oal ime i ook o undergo ha displacemen. In his case, we graphically deermined ha your displacemen is 200m eas and we were old ha your ravel ime was 20s. So, vavg =? Δx = 200m eas = 20s vavg = x 200m eas vavg = 20s vavg = 10m/s eas Kinemaics - 13 v 1.0 2009 by Goodman & Zavoroniy
The same problem can be solved algebraically, alhough an iniial skech is sill a good idea. The key sep o an algebraic soluion is o conver direcions o be eiher posiive or negaive. In his case, we can define eas as posiive and wes as negaive (The choice won maer as long as we re consisen hroughou he problem.) Your iniial posiion hen becomes +400m (400m eas), your iniial ravel is +500m (500m eas) and he las leg of your rip is -300m (300m wes). You can now jus add hese ogeher o ge your final posiion, +600m. This ranslaes ino a final posiion of 600m eas of your house. Your displacemen is jus he change in your posiion. x = +600m xo = +400m Δx =? Δx = x - xo Δx = +600m (+400m) Δx = +200m Δx = 200m eas noe ha his las sep is required since our choice of posiive or negaive was arbirary The calculaion of average velociy can be done jus as i was above for he graphical soluion. Insananeous Velociy and Acceleraion The mos boring world would be one in which he posiions of all objecs were consan...nohing would move: velociy would have no meaning. Forunaely our world is a lo more ineresing han ha. Objecs are changing heir posiions all he ime, so velociy is an imporan concep. Bu our world is even more ineresing, objecs are also changing heir velociy all he ime: hey are speeding up, changing direcion and/or slowing down. Jus as change in posiion over ime leads o he idea of velociy, change in velociy over ime leads o he concep of acceleraion. In he same manner ha we defined insananeous speed as he speed measured during a very shor period of ime, we can now define insananeous velociy as he velociy measured during a very shor period of ime. The symbol, v will be used for insananeous velociy. In a world wih acceleraion, he idea of insananeous velociy is very imporan since an objec s velociy may ofen be changing from momen o momen. v x for a very shor period of ime...an insan We can now define acceleraion as he change in velociy over ime. a v or a v v 0 Kinemaics - 14 v 1.0 2009 by Goodman & Zavoroniy
Unis of Acceleraion The unis of acceleraion can be derived from is formula: a v The SI uni of velociy is meers/second (m/s) and of ime is he second (s). Therefore, he uni of acceleraion is (m/s)/s or m/s/s. This is he same as (m/s) x (1/s) since dividing by s is he same as muliplying by 1/s. This resuls in m/s 2 which, while no having any inuiive meaning, is a lo easier o keep rack of han m/s/s, meers per second per second, he alernaive way of wriing he unis for acceleraion. Since velociy is a vecor, i has a magniude and a direcion. For he res of his chaper, we ll be focused on acceleraions ha change he magniude of an objec s velociy. However, in laer chapers a key aspec of acceleraion will involve changing he direcion of an objec s velociy. These are boh examples of acceleraion. Bu le s firs sar wih acceleraions ha change only he magniude of an objec s velociy. Example 11 An objec is raveling a a velociy of 20m/s norh when i experiences an acceleraion over 12s ha increases is velociy o 40m/s in he same direcion. Wha was he magniude and direcion of he acceleraion? Le s solve his algebraically by defining velociies owards he norh as posiive and owards he souh as negaive. Then, v = +40m/s vo = +20m/s = 12s a =? a v a v v 0 a = (+40m s ) (+20m s ) 12s a = +20m s 12s a = +1.7m/s 2 Example 12 Wha will an objec s velociy be a he end of 8.0s if is iniial velociy is +35m/s and i is subjec o an acceleraion of -2.5m/s 2? v =? vo = +35m/s = 8.0s a = -2.5m/s 2 Kinemaics - 15 v 1.0 2009 by Goodman & Zavoroniy
a v a v v 0 a = v - vo vo + a = v v = vo + a v = +35m/s + (-2.5m/s 2 ) (8.0s) v = +35m/s + (-20m/s) v = +15m/s Solve for v: Firs, muliply boh sides by Then add vo o boh sides Swich so ha v is on he lef side of he = Subsiue in values and solve Free Fall You now know enough o be able o undersand one of he grea debaes ha marked he beginning of wha we now call physics. The erm physics was being used by he ancien Greeks more han 2000 years ago. Their philosophy, much of i described in he book iled Physics by Arisole, included some ideas ha sood unil Galileo made some imporan argumens and measuremens ha showed he ancien Greek physics o be of limied value. The physics of ancien Greece included he idea ha all objecs were made up of a combinaion of four elemens (he fifh elemen was reserved for objecs ha were beyond he earh). The four elemens o be found in our world were earh, waer, air and fire. Each of hese elemens had heir naural place. If you removed an elemen from is naural place, i would, when released, immediaely move back o ha place; and i would do so wih is naural (consan) velociy. Their view of he world could be hough of a se of concenric circles wih each of he elemens occupying a layer. Earh occupied he cener of he circle, so rocks, which are predominanly made of earh would naurally move down, owards he cener of our world. Above earh was waer, which would fill he area above he rocks, like a lake or an ocean above he land ha forms he lake or ocean bed. Above waer is air, which is seen everywhere in our world, above boh earh and waer. Finally, fire rises up hrough he air, searching for is naural locaion above everyhing else. All objecs were considered o be a mixure of hese four elemens. Rocks were predominanly earh: so if you drop a rock i falls as i ries o ge back o is naural locaion a he cener of he earh. In so doing, i will pass hrough waer and air: If you drop a rock in a lake, i sinks o he boom. Fire passes upwards o he highes locaion, so if you make a fire, i always passes upwards hrough he air. One conclusion ha his led o is ha objecs which were made of a higher percenage of earh would feel a greaer drive o reach heir naural locaion. Since earh is he heavies of he elemens, his would mean ha heavier objecs would fall faser han ligher objecs. Also, hey would fall wih a naural consan velociy. Tha philosophy sood for more han 2000 years unil Galileo, in he 1600 s made a series of argumens, and conduced a series of experimens, ha proved ha neiher of hose wo Kinemaics - 16 v 1.0 2009 by Goodman & Zavoroniy
conclusions was accurae. He showed ha he naural endency of all unsuppored objecs is o fall owards he cener of he earh wih he same acceleraion: 9.8m/s 2. Tha number 9.8m/s 2 is used so ofen ha i as given is own symbol: g. In modern erms, his conclusion can be saed as follows. All unsuppored objecs fall owards he cener of he earh wih an acceleraion of g: 9.8m/s 2. This saemen requires some explanaion and some caveas. 1. Unsuppored means ha nohing is holding he objec up. So if you release somehing and nohing is sopping i from falling, hen i is unsuppored. In ha case, all objecs will experience he same acceleraion downwards. I does no depend on how heavy he objec is: all objecs fall wih ha same acceleraion. 2. Suppor can also come from air resisance. So a parachue provides suppor by caching air ha slows down he sky diver. In ha case, he parachuis is no an unsuppored objec: he or she is suppored by air resisance. Bu his is generally rue o a lesser exen. So a feaher or an uncrumpled piece of paper also receives suppor from he air: so hey don fall wih a consan acceleraion eiher. Galileo s conclusion is an idealizaion, i assumes ha we can ignore air resisance, which is never compleely rue near he earh (or airplanes and parachues would have a hard ime of i) bu will work for he problems we ll be doing. 3. His conclusion does no depend on he moion of he objec. So baseballs hrown oward home plae, dropped, or hrown sraigh up all fall wih he same acceleraion owards he cener of he earh. This is an area of grea confusion for sudens, so you ll be reminded of i ofen. Whenever nohing is sopping an objec from falling, i will accelerae downwards a 9.8m/s 2, regardless of is overall moion. In his book, we will assume ha air resisance can be ignored unless i is specifically saed o be a facor. Example 13 An objec is dropped near he surface of he earh. Wha will is velociy be afer i has fallen for 6.0s? v =? vo = 0 = 6.0s a = g = -9.8m/s 2 All unsuppored objecs have an acceleraion of 9.8m/s 2 downwards a = v a = v v 0 Solve for v: Firs, muliply boh sides by Kinemaics - 17 v 1.0 2009 by Goodman & Zavoroniy
a = v - vo vo + a = v v = vo + a v = 0 + (-9.8m/s 2 ) (6.0s) v = -59m/s Then add vo o boh sides Swich so ha v is on he lef side of he = Subsiue in values and solve The Kinemaics Equaions So far we have wo definiions of moion ha we will be using he foundaion for our sudy of moion: vavg x v and a. We need o add jus one more equaion o complee our foundaion; and hen we can sar building he se of equaions ha we ll be using o solve a range of problems hroughou his book. The las equaion ells how o calculae an objec s average velociy if we know is iniial and final velociy. I urns ou ha under he condiion of consan acceleraion an objec s average velociy is jus he average of is iniial and final velociies. This average is compued jus by adding he wo velociies, v and v0, ogeher and dividing by 2: vavg = v v 0 2 vavg = ½ (v0 + v) Or, since dividing by 2 is he same as muliplying by ½ This will be rue whenever he acceleraion is consan. However, ha condiion of consan acceleraion will hold rue no only for his course, bu for mos all he high school or universiy physics ha you will ake. I d be possible o solve all problems involving he locaion, velociy and acceleraion of an objec jus using he wo definiions and he compuaion for average velociy shown above. However, in physics i s someimes bes o do some harder work up fron so as o make our work easier laer on. In his case, we ll use he above hree equaions o creae a se of kinemaics equaions ha are easier o work wih. We ll firs derive hose equaions algebraically; hen we ll derive hem using a graphical approach. Then we ll pracice working wih hem. Le s sar wih by using our definiion of acceleraion o derive an equaion ha will ell us an objec s velociy as a funcion of ime. a v a = v v 0 a = v - vo vo + a = v v = vo + a Subsiue: Δv = v - vo Muliply boh sides by Add vo o boh sides Rearrange o solve for v Kinemaics - 18 v 1.0 2009 by Goodman & Zavoroniy
This equaion ells us ha an objec s velociy a some laer ime will be he sum of wo erms: is velociy a he beginning of he problem, vo, and he produc of is acceleraion, a, and he amoun of ime i was acceleraion,. If is acceleraion is zero, his jus says ha is velociy never changes. If is acceleraion is no zero, his equaion ells us ha he objec s velociy will change more as more ime goes by and i will change faser if he size of is acceleraion is greaer. Ofen in physics we wan o know he final posiion or velociy of an objec afer a cerain amoun of ime. The bolded equaion above gives us a direc way of calculaing velociies a laer imes given iniial condiions: This is a key kinemaics equaion. Example 14 Wha will an objec s velociy be a he end of 15s if is iniial velociy is -15m/s and i is subjec o an acceleraion of +4.5m/s 2? v =? vo = -15m/s = 15s a = +4.5m/s 2 v = vo + a v = -15m/s + (+4.5m/s 2 ) (15s) v = -15m/s + 68m/s v = +53m/s Example 15 How long will i ake an objec o reach a velociy of 86m/s if is iniial velociy is 14m/s and i experiences an acceleraion of 1.5m/s? v = 86m/s vo = 14m/s =? a = +1.5m/s 2 v = vo + a Solve for : Firs subrac vo from boh sides v - vo = a Then divide boh sides by a v v 0 = a Finally swich sides so is on he lef = v v 0 a Now subsiue in he values and solve = 86m s 14m s 1.5 m s 2 = 72m s 1.5 m s 2 = 48s Kinemaics - 19 v 1.0 2009 by Goodman & Zavoroniy
Noe ha jus as 72 divided by 14 equals 48, so oo does (m/s) / (m/s 2 ) equal seconds. This can be seen if one remembers ha dividing by a fracion is he same a muliplying by is reciprocal: so (m/s) / (m/s 2 ) is he same as (m/s) (s 2 /m). In his case, i can be seen ha he meers cancel ou as does one of he seconds in he numeraor, leaving only seconds in he numeraor; which is he correc uni for ime. Example 16 Wha acceleraion mus an objec experience if i is o aain a velociy of 40m/s o he norh in a ime of 18s if i s sars ou wih a velociy of 24m/s owards he souh? For his problem, le s define norh as posiive and souh as negaive. Then, v = +40m/s vo = -24m/s = 18s a =? v = vo + a v - vo = a v v 0 = a a = v v 0 Solve for : Firs subrac vo from boh sides Then divide boh sides by Finally swich sides so a is on he lef Noe ha his is jus our original definiion for acceleraion. We could have jus used ha definiion, bu i s easy enough o recover from he equaion we ll be using...eiher way works. Now subsiue in he values and solve a =) 40m s 24m s 18s a = 64m s 18s a = 3.6 m/s 2 a = 3.6 m/s 2 owards he norh Noe ha in subsiuing -24m/s for vo we pu i ino is own parenheses. Tha s so we don lose he negaive sign...a common misake. Now we can see ha -(-24m/s) equals +24m/s Since he answer is posiive, he acceleraion mus be owards he norh based on our original decision ha norh was posiive Noe ha jus as 64 divided by 18 equals 3.6, so oo does (m/s) / s equal m/s 2. This can be seen if one remembers ha dividing by a fracion is he same a muliplying by is reciprocal: so (m/s) / s is he same as (m/s) x (1/s). We now have a useful equaion for deermining how an objec s velociy will vary wih ime, given Kinemaics - 20 v 1.0 2009 by Goodman & Zavoroniy
is iniial velociy and is acceleraion. We need o derive a similar expression ha will ell us where an objec is locaed as a funcion of ime given is iniial posiion and velociy and is acceleraion. We need o combine hree of our equaions ogeher o do ha v = vo + a vavg x x 0 The equaion we jus derived from he definiion of acceleraion The definiion of average velociy vavg = ½ (v + v0) The equaion for average velociy in he case of consan acceleraion Since we have wo equaions for average velociy, vavg, hey mus be equal o each oher. vavg = vavg [ x x 0 ]= [½ (v + v0)] We can hen subsiue in he wo differen equaions for vavg from above: one on he lef side of he equals sign and he oher on he righ Le s solve his for x: firs muliply boh sides by o ge i ou of he denominaor on he lef x - x0 = ½ (v + v0) Then add x0 o boh sides o ge x by iself x = x0 +½ (v + v0) Disribue ino he parenheses on he righ x = x0 + ½ v + ½v0 Now subsiue in our new equaion for v: v = vo + a x = x0 + ½ (vo + a) + ½v0 Disribue ½ ino he parenheses x = x0 + ½ vo + ½a 2 + ½v0 Combine he wo ½ vo erms x = x0 + vo + ½a 2 This is anoher of he key kinemaics equaions. I allows us o deermine where an objec will be as ime goes by based on a se of iniial condiions. In his case, here are hree erms: x0 ells us where he objec sared; vo ells us how fas i was moving iniially and how long i s been raveling; and ½a 2, ells us how much is acceleraion has affeced he disance i has raveled. The reason ha in he las erm is squared is ha no only does i s velociy change more as ime goes by, i also has had more ime for ha change in velociy o affec how far i s gone. Example 17 A car is a res when i experiences an acceleraion of 2.0m/s 2 owards he norh for 5.0s. How far will i ravel during he ime i acceleraes? For his problem, le s define norh as posiive and souh as negaive. Also, since we re no old where he car sars, le s jus define is iniial posiion as he origin for his problem, zero. In ha case, he disance i ravels will jus be is posiion, x, a he end of he problem. Then, x0 = 0 Kinemaics - 21 v 1.0 2009 by Goodman & Zavoroniy
x =? vo = 0 = 5.0s a = 2.0m/s 2 x = x0 + vo + ½a 2 The equaion is already solved for x so we jus have o subsiue in numbers. However, a good firs sep is o cross ou he erms ha will clearly be zero, in his case he firs and second erms. (Since vo = 0, anyhing imes vo will also be zero.) x = ½a 2 Tha vasly simplifies he equaion and avoids some algebra misakes. Now we can subsiue numbers in for he variables. x = ½(2.0m/s 2 )( 5.0s) 2 Make sure o square 5.0s before muliplying i by anyhing else x = (1.0m/s 2 )(25s 2 ) x = 25m The car will ravel 25m o he norh during he ime ha i acceleraes Example 18 An objec acceleraes from res. How long will i ake for i o ravel 40m if is acceleraion is 4m/s 2? Since we re no old where he objec sars, le s jus define is iniial posiion as he origin for his problem, zero. In ha case, he disance i ravels, 40m, will jus be is posiion, x, a he end of he problem. Also, since i s iniial a res ha means ha is iniial velociy is zero. Then, x0 = 0 x = 40m vo = 0 =? a = 2.0m/s 2 x = x0 + vo + ½a 2 x = ½a 2 2x a = 2 Le s cross ou he erms ha will clearly be zero, in his case he firs and second erms. (Since vo = 0, anyhing imes vo will also be zero.) Now le s solve his for : Muliply boh sides by 2 and divide boh sides by a Now ake he square roo of boh sides, o ge insead of 2, and swich o he lef Kinemaics - 22 v 1.0 2009 by Goodman & Zavoroniy
= 2x a = 2(40m) 4 m s 2 Now we can subsiue in he values = 80m 4 m s 2 = 20s 2 = 4.47s The firs equaion ha we derived allows us o deermine he velociy of an objec as a funcion of ime if we know is acceleraion. The second equaion allows us o deermine he posiion of an objec as a funcion of ime if we know is iniial posiion and velociy and is acceleraion. Someimes we use boh equaions o solve one problem. Example 19 A plane mus reach a speed of 36m/s in order o ake off and is maximum acceleraion is 3.0m/s. How long a runway does i require? Solving his problem requires us o use boh our kinemaics equaion. Firs, le s figure ou how much ime i mus accelerae o reach akeoff velociy. Then, le s figure ou how far i will ravel in ha ime. x0 = 0 x =? vo = 0 v = 36m/s =? a = 3.0m/s 2 v = vo + a v - vo = a Solve for = v v 0 a = (36m s 0) 3 m s 2 = 12s Now we can add ha new piece of informaion o wha we knew before: x0 = 0 x =? vo = 0 v = 120m/s Kinemaics - 23 v 1.0 2009 by Goodman & Zavoroniy
= 12s a = 3.0m/s 2 We can now use he second equaion o solve for he locaion of he plane when i akes off. Tha will be he minimum required lengh of he runway. x = x0 + vo + ½a 2 Eliminaing he zero erms x = ½a 2 x = ½(3.0m/s 2 )(12s) 2 x = (1.5m/s 2 )(144s 2 ) x = 216m The final kinemaics equaion ha we ll develop combines he firs wo so ha we can deermine he velociy of an objec as a funcion of is posiion, raher han as a funcion of ime. Tha would allow us o have solved Example 18 in jus one sep. Essenially we derive his equaion by doing exacly wha we did in Example 18, bu wihou insering in any numbers, we leave everyhing as variables. The resul is a soluion ha we can use in he fuure o save a lo of work. Le s firs solve our velociy versus ime equaion for ime. v = vo + a v - vo = a a = v - vo = v v 0 a Then we ll use ha expression for ime in he equaion ha ells us an objec s posiion as a funcion of ime. Tha will eliminae ime from ha equaion. x = x0 + vo + ½a 2 Now we ll subsiue in [ v v 0 ] wherever we see a. The a brackes jus help us see wha we ve done. Review his nex equaion carefully o see ha wherever here used o be here s now [ v v 0 ]. a x = x0 + vo[ v v 0 a ] + ½a(v v 0 a )2 Le s subrac x0 from boh sides, disribue he vo ino he second bracke and square he conens of he hird bracke. x - x0 = [ v 2 0y v 0 ] + ½a(v - vo) 2 / a 2 a We can now cancel one of he a s in he las erm and use he fac ha (v - vo) 2 = v 2-2v vo + vo 2 x - x0 = [ v 2 0y v 0 ] + ½(v 2-2v vo + vo 2 ) a Kinemaics - 24 v 1.0 2009 by Goodman & Zavoroniy
a(x - x0) = vov - vo 2 + ½(v 2-2v vo + vo 2 ) Since a is in he denominaor of boh erms on he righ we can simplify a bi by muliply all he erms by a Now le s disribue he ½ ino he las parenheses on he righ a(x - x0) = vov - vo 2 + ½v 2 - v vo + ½vo 2 By combining he wo vvo erms, hey cancel ou. A he same ime we can combine - vo 2 and ½vo 2 o ge -½vo 2 a(x - x0) = ½v 2 - ½vo 2 2a(x - x0) = v 2 - vo 2 v 2 - vo 2 = 2a(x - x0) v 2 - vo 2 = 2ad Now muliply boh sides by 2 o cancel ou he ½ s Swiching he erms from lef o righ complees his derivaion This is someimes wrien by subsiuing d for (x - x0) since ha is jus he disance ha he objec has raveled and i s easier o read This equaion les us deermine how he velociy of an objec will change as is posiion changes. Having done all his work now, will save us work laer. In Example 19, le s ake a look a how we would use his equaion o solve he same problem as was posed in Example 18 Example 20 As was he case in Example 18, a plane mus reach a speed of 36m/s in order o ake off and is maximum acceleraion is 3.0m/s 2. How long a runway does i require? x0 = 0 x =? vo = 0 v = 36m/s a = 3.0m/s 2 v 2 - vo 2 = 2a(x - x0) v 2 = 2ax x = v2 2a Simplify by eliminaing he zero erms, vo & x0 Solve for x by dividing by 2a and swiching sides x = 362 2(3 m s 2) x = 216m Kinemaics - 25 v 1.0 2009 by Goodman & Zavoroniy
Problem Solving wih he Kinemaics Equaions The Kinemaics Equaions x = x0 + vo + ½a 2 v = vo + a v 2 - vo 2 = 2a(x - x0) All kinemaics problems can be solved by using eiher one or wo of he above equaions. I is never necessary o use all hree equaions o answer one quesion. The bigges quesion sudens have in working wih hese equaions is which one o use? Firs, you need o relax and undersand ha you can ge he wrong answer by using he wrong equaion; you jus won ge an answer a all. You ll jus find ha you re missing he informaion ha you need o solve he problem using ha equaion. A ha poin, you should realize ha you need o use a differen equaion, or read he problem again o see if you re missing a piece of informaion ha you need bu have overlooked. For insance, if a problem indicaes or implies ha an objec was a res a he beginning of he problem ha means ha is iniial velociy was zero. Someimes his is obvious... someimes i isn. For insance, if I drop somehing, he implicaion is ha i had a zero iniial velociy, bu ha isn explicily saed; i s implied. Physics will help you learn o read very carefully o undersand wha he auhor mean when hey wroe he problem or described he siuaion. So, he firs sep in solving he problem is o read i very carefully. The second sep is o read i again. This ime wriing down he informaion you ve been given in erms of he variables wih which you ll be working. For insance, ranslaing dropped ino vo = 0. One of he pieces of informaion ha you ll be given is wha you re supposed o be looking for: wha s he quesion. If he auhor asks Wha is is final velociy? ha is ranslaed as v =? and becomes anoher of he facs o add o your lis of facs ha you ll use o solve he problem. The nex sep is o deermine which of he kinemaics equaions relae your collecion of facs o one anoher. Each equaion represens a relaionship beween a differen se of facs: picking he correc equaion is jus a maer of deermining which equaion relaes o his specific se of facs. If you pick he wrong one, no harm will be done (excep some wased ime) since you ll find you jus don have he righ facs o use ha equaion. If he problem ha you re working wih doesn have ime as one of is facs, hen you ll be using he hird of he equaions lised above: v 2 - vo 2 = 2a(x - x0); i s he only one ha doesn include ime as a facor. If ime is included, you ll be using one of he firs wo. In ha case, you jus need o deermine which of hose firs wo equaions o use. If he problem deals wih he posiion of he objec as a funcion of ime, hen you d use he firs equaion: x = x0 + vo + ½a 2. If i is dealing wih he velociy of he objec as ime goes by, you ll use he second equaion: v = vo + a. I s really as simple as ha. Example 21 In his example, we re jus going o decide which equaion(s) will be needed o solve each problem. Kinemaics - 26 v 1.0 2009 by Goodman & Zavoroniy
1. A ball is subjec o an acceleraion of -9.8 m/s 2. How long afer i is dropped does i ake o reach a velociy of -24m/s? 2. A ball is released from res and subjec o an acceleraion of -9.8 m/s 2. How far will i ravel before i reaches a velociy of 24m/s? 3. A dropped ball is subjec o an acceleraion of -9.8 m/s 2. How far will i ravel in he firs 5.0s? 4. You hrow an objec upwards from he ground wih a velociy of 20m/s and i is subjec o a downward acceleraion of 9.8 m/s 2. How high does i go? 5. You hrow an objec upwards from he ground wih a velociy of 20m/s and i is subjec o a downward acceleraion of -9.8 m/s 2. How much laer does i momenarily coming o a sop? 6. You hrow an objec upwards from he ground wih a velociy of 20m/s and i is subjec o an downward acceleraion of 9.8 m/s 2. How high is i afer 2.0s? Take a second o wrie down he facs in each problem and hen deermine which equaion you would use. Then compare your resuls o hose shown below. 1. an acceleraion of -9.8 m/s 2 means a =-9.8 m/s 2 How long afer means =? i is dropped means vo = 0 o reach a velociy of -24m/s means v = -24m/s Since ime,, is a facor, we need o only choose beween he firs wo equaions. Since velociy, v, is a facor, i mus be he second equaion: v = vo + a 2. released from res means vo = 0 an acceleraion of -9.8 m/s 2 means a =-9.8 m/s 2 How far will i ravel means x0=0 and x=? reaches a velociy of 24m/s means v = -24m/s Since ime,, is no a facor, we need o use he hird equaion v 2 - vo 2 = 2a(x - x0) 3. A dropped ball means vo = 0 an acceleraion of -9.8 m/s 2 means a =-9.8 m/s 2 How far will i ravel means x0=0 and x=? in he firs 5.0s means = 5.0s Since ime is a facor, we need o only choose beween he firs wo equaions. Since posiion, x, is a facor, i mus be he firs equaion: x = x0 + vo + ½a 2 4. upwards from he ground wih a velociy of 20m/s means vo = + 20m/s and x0 = 0 a downward acceleraion of Kinemaics - 27 v 1.0 2009 by Goodman & Zavoroniy