One Dimensional Kinematics

Transcription

1 Chaper B One Dimensional Kinemaics Blinn College - Physics Terry Honan Kinemaics is he sudy of moion. This chaper will inroduce he basic definiions of kinemaics. The definiions of he velociy and acceleraion will require he inroducion of he basic noions of calculus, mos specifically he derivaive. We will also consider in deail he simple special cases of moion wih consan velociy and consan acceleraion. Free fall will be discussed as an example of moion wih consan acceleraion. B. - The General Problem By one dimensional moion we mean moion consrained o a line. As examples, consider a car driving on a sraigh road or he verical moion of an elevaor. The problem of moion in wo or hree dimensions will be discussed in he nex chaper. Posiion as a Funcion of Time To mahemaically define posiion we need o aach a real number line (he x-axis) o he line of moion. To do his here are wo arbirary choices; we mus choose he x = 0 posiion and hen we mus choose he posiive direcion. Any D moion can be represened graphically. Time is he independen variable, so i will be he horizonal axis. We will hen consider graphs of x as a funcion of ime, where x is he verical axis. Unis: The SI uni for disances is: m = meer Velociy and he Derivaive Average Velociy x Average Velociy: v = Δx Δ = x f - x i = slope f - i x i i f x f Ineracive Figure If a car drives 30 mi in 2 hours, we can calculae a velociy of 65 mi/hr. This is no necessarily wha he speedomeer would read; he speedomeer reads he magniude of he insananeous velociy. In his case 65 mi/hr is wha we call he average velociy. We will define he average velociy by

2 2 Chaper B - One Dimensional Kinemaics where Δ (Dela) generally will represen he final value minus he iniial Δx v = Δ Δx = x f - x i and Δ = f - i. Noe ha he average velociy corresponds o wo imes i and f, and x i and x f are he posiions a he wo imes. In a graph of x vs. he average velociy has he inerpreaion as he slope of he secan line beween he wo poins ( i, x i ) and f, x f. Unis: The SI uni for velociy is: m /s Insananeous Velociy x Tangen Line slope = v x 0 Secan Lines slope = v 0 Ineracive Figure The insananeous velociy refers o a single ime. Take he posiion a o be x. We can hen consider a laer ime + Δ, where he posiion is x + Δ x. The average velociy beween hese wo imes is Δ x / Δ. To ge he insananeous velociy we le Δ become small; we do his by aking he limi as Δ 0. This gives he derivaive of calculus; insananeous velociy is he ime derivaive of posiion. d x v = d = x! Δx = lim Δ 0 Δ The graphical inerpreaion of he insananeous velociy is simple. The average velociy is he slope of he secan lines. If we consider he secan lines corresponding o differen Δ values hen as Δ 0, hese secan lines approach he angen line. The velociy a is hen he slope of ha angen line. When we refer o he slope of a graph a some ime, we mean he slope of a line angen o he graph a ha ime. Acceleraion Acceleraion is o velociy as he velociy is o he posiion. Velociy is he ime derivaive of posiion, so acceleraion is he ime derivaive of he velociy. Average Acceleraion Since he average velociy is relaed o he posiion by v = Δx / Δ we can similarly wrie he average acceleraion in erms of he velociy by We can hink of average acceleraion graphically as he slope of he secan lines of a v vs. graph. a = Δv Δ. Unis: The SI uni for acceleraion is: m s 2

3 Chaper B - One Dimensional Kinemaics 3 Insananeous Acceleraion The insananeous acceleraion (or jus acceleraion) is he ime derivaive of he velociy. This can be wrien as he second derivaive of he posiion. d v a = d = v! Δv = lim Δ 0 Δ d 2 x a = d = ẍ 2 Example B. - Graphical Analysis Consider he following graph of x vs.. For < he graph is a sraigh line. For > 4 he graph is a horizonal line. Beween and 4 he funcion is cubic wih a local maximum a = 2 and an inflecion poin a = 3. x For his graph, plo v vs. and a vs v a Discussion: For < since he posiion-ime graph is a line he velociy is a consan and consan v implies he acceleraion is zero. For > 4 he velociy has a consan zero-value and because v is consan he acceleraion is also zero. Since he graph is cubic beween and 4 he velociy graph will be quadraic and he acceleraion is a line. Because 2 is a local maximum of posiion, he velociy is zero. A he inflecion poin 3 he velociy is a is minimum and he acceleraion becomes zero. Calculus Review This secion will be a brief review of he prerequisie calculus maerial. Differeniaion d d u d v Sum Rule : (u + v) = d d + d d Power Rule : d p = p p- Noe d d = d d u Produc Rule : (u v) = d d v + u d v d

4 4 Chaper B - One Dimensional Kinemaics d u (d u / d ) v - u (d v / d ) Quoien Rule : d v = d Chain Rule : d f [ u() ] = f d [ u() ] d u() or d y d y d = d u v 2 d u d Aniderivaives and Indefinie Inegrals Anidiffereniaion is he reverse of he differeniaion procedure. If f () is a funcion hen is aniderivaive is anoher funcion F() ha saisfies d F() = f (). Generally, anidiffereniaion is a more complicaed procedure han differeniaion. Since he derivaive of a consan is d zero, if F() is an aniderivaive of f () hen F() + C is also an aniderivaive of f (), where C is an arbirary consan. The resul is sronger ha his: F() + C is he mos general aniderivaive of f (). This general aniderivaive is ofen wrien as an indefinie inegral. f () d = F() + C where F () = f () The Definie Inegral Since v = d x / d we can wrie he infiniesimal disance moved in he infiniesimal ime d as d x = v d. In a graph of v vs. he area under he curve beween and + d is d x = v d ; his is he area of a recangle of heigh v and widh d. Thus he area under he graph over he infiniesimal ime d is he infiniesimal disance raveled in ha ime. The oal disance raveled beween he imes i and f is Δx = x f - x i ; i is he sum of all he infiniesimal disances menioned above, he oal area under he curve. This is he definie inegral of calculus. Δx = i f v d Noe he logic of his noaion; in calculus d is implied o be a small Δ and he sum over an infinie number of infiniesimal hings becomes he inegral. Δ d and v Δx is he area beween i and f. (Green is posiive and red is negaive.) v i v dx=vd i +d f v f Ineracive Figure The definie inegral generally has he inerpreaion as he area under a curve. The area under y = f () beween a and b is wrien a b f () d When he funcion is negaive he conribuion o he area is aken o be negaive. Noe how he velociy and posiion are relaed. v = d x / d. The fundamenal heorem of calculus is a generalizaion of his; i gives he rule we use o evaluae definie inegrals. b b d f () d = F() = F(b) - F(a) where f () = a a d F()

5 Chaper B - One Dimensional Kinemaics 5 Thus, he definie inegral of f is he difference of an aniderivaive a he endpoins. Example B.2 - The Basic Definiions The posiion as a funcion of ime for a paricle moving in one dimension is given in SI unis by: (a) Wha is he average velociy beween s and 3s? x() = The formula for average velociy is v = Δx / Δ, so we need he posiion a each ime. x(s) = 8m and x(3s) = 0m From his we can solve for he average velociy. (b) Wha is he insananeous velociy a 2s? Δx x(3s) -x (s) 0m -8m v = Δ = = = 46 m /s 3s -s 2s To find v() we mus differeniae he posiion funcion. We hen plug in he ime. d v() = d x() = v(2s) = 40 m /s (c) Wha is he average acceleraion beween s and 3s? The formula for average acceleraion is a = Δv / Δ, so we need he velociy a each ime. From his we can solve for he average acceleraion. (d) Wha is he insananeous acceleraion a 2s? To find a() we mus differeniae he velociy funcion. v(s) = 2m /s and v(3s) = 4m /s Δv v(3s) -v (s) 4m /s -2m /s a = Δ = = = 56 m s 2 3s -s 2s We hen plug in he ime. d a() = v() = 36-6 d a(2s) = 56 m s 2 Noe ha he average acceleraion for he inerval equals he acceleraion a he midpoin of he inerval. This generally will no be he case bu because our posiion funcion is cubic i urns ou ha his mus be he case here. (e) Find he ne displacemen Δx beween s and 3s by inegraing he velociy and compare wih a direc calculaion of Δx. To find v() we mus differeniae he posiion funcion. Δx = i f v() d = s 3s d

6 6 Chaper B - One Dimensional Kinemaics = s s = 90m - (-2m) = 92 m Explicily evaluaing Δx gives he same: Δx = x(3s) -x (s) = 0m -8m = 92 m B.2 - Consan Velociy and Acceleraion Now ha we have considered he general problem of one dimensional kinemaics we can now consider special cases, firs consan velociy, hen consan acceleraion. An imporan case of consan acceleraion is free fall. Consan Velociy If velociy is a consan hen he acceleraion is zero, since he derivaive of a consan is zero. Le us now find he posiion from he velociy. Posiion is he aniderivaive of he velociy. d x = v = consan x() = v + C where C is a consan. d Define he iniial posiion x 0 o be he posiion a = 0, x 0 = x(0). Plugging his ino our expression for x() gives C = x 0 and x() = x 0 + v If we choose he convenion i = 0, f =, x i = x 0 and x f = x hen we ge Δx = x - x 0. The above expression becomes x = x 0 + v or Δx = v. This is a simple expression; for consan velociy, he disance is he produc of he velociy and ime. Consan Acceleraion If he acceleraion is a consan hen o ge he velociy we repea he procedure for going from a consan velociy o he posiion. Velociy is he aniderivaive of he acceleraion. d v d = a = consan v() = a + C where C is a consan. Define he iniial velociy v 0 o be he velociy a = 0, v 0 = v(0). Plugging his ino our expression for v() gives C = v 0 and v() = v 0 + a We need o anidiffereniae again o ge he posiion as a funcion of ime. The arbirary consan becomes he iniial posiion x 0 and we ge d x d = v() = v 0 + a x() = v a 2 + C 2 where C 2 is a differen consan. x() = x 0 + v a 2. If we choose he same convenions as in he consan velociy case and add v i = v 0 and v f = v hen he above expressions for velociy and posiion become v = v 0 + a and Δ x = v a 2. Recall how o calculae Δx from v vs. ; i is he area under he curve beween i and f.

7 Chaper B - One Dimensional Kinemaics 7 For consan acceleraion he velociy vs. ime is a sraigh line. We hen ge he area under a rapezoid wih a base of widh Δ and heighs of v i and v f. This gives Using our convenion for v 0 v and, his becomes Δx = 2 v i + v f Δ. Δx = 2 (v 0 + v). We now wan o derive an expression relaing Δ x, v 0, v and a. To do his ha he previous expression and v = v 0 + a, and hen eliminae ime. v - v 0 Δx = 2 (v 0 + v) a v 2 - v 0 2 = 2 a Δx. Wih his we have derived a se of four equaions for kinemaics wih consan acceleraion. These relae he variables, Δ x, v 0, v and a. These will be useful for a large class of problems his chaper. Consan Acceleraion Equaions v = v 0 + a Δx = 2 (v 0 + v) Δx = v a 2 v 2 = v a Δx Example B.3 - Deceleraing Car A car brakes uniformly o a sop from 30 m /s while moving 50 m. (a) Wha is he car s acceleraion while braking? The firs sep is o esablish wha we are given and wha we are looking for in erms of he variables in our consan acceleraion equaions. Here we are given he iniial velociy, he final velociy, he displacemen and we are looking for he acceleraion. v 0 = 30 m /s, v = 0, Δx = 50 m and a =? Ofen he bes way o approach he consan acceleraion equaions is by idenifying he equaions in erms of he variable i does no include. Here we are no given ime or looking for i so ha leaves us o he fourh equaion, he one ha doesn involve. (b) How long does i ake for he car o sop? 2 v v 2 = v a Δx a = - = -3 m s2 2 Δx

8 8 Chaper B - One Dimensional Kinemaics Since we have already solved for a we can use any equaion involving ime o ge. Here will solve par (b) wihou reference o par (a); his leads us o he second equaion, he one ha does no involve a. Δx = 2 (v 2 Δx 0 + v) = = 0 s v 0 Free Fall Free fall is one dimensional moion under he influence of only graviy. Assuming ha only graviy acs implies ha we are ignoring any fricion effecs. We will choose he convenion ha up is he posiive direcion. Also, we will ake y as he posiion variable; his will be consisen wih our laer usage where y is ypically aken as he upward verical variable. Galileo discovered ha he acceleraion of all bodies in he presence of graviy (ignoring air resisance) is he same. The value of he downward acceleraion is g = 9.80 m s 2 f = 32.0 s. 2 Since up is he posiive y direcion and he acceleraion is downward we ake he acceleraion o be: a = -g Using his value of a and replacing x wih y akes he consan acceleraion equaions o he free fall expressions. Free Fall Equaions v = v 0 - g Δy = 2 (v 0 + v) Δy = v 0-2 g 2 v 2 = v g Δy Example B.4 - A Dropped Ball A ball is dropped from a heigh h. (We will assume here is no air resisance for free-fall problems.) (a) Wha is is ime of fall? When solving a problem in erms of symbols, one mus wrie he answer in erms of he symbols given and in erms of physical consans. Here wha is given is h and he relevan consan is g. h is relaed o Δy and, since he ne moion is downward, Δy < 0. Since he ball is dropped we conclude is iniial velociy is zero. We are looking for. Δy = -h, v 0 = 0 and =? (Anoher way o undersand Δy = -h is o se y 0 = h, y = 0 and use Δy = y - y 0.) The variable ha we do no know or need is he final velociy v, so we are led o he hird equaion. Δy = v 0-2 g 2 -h = 0-2 g 2 Noe ha generally for a dropped objec, he ime of fall is relaed o he heigh by h = 2 g 2. Solving for and aking he posiive square roo we ge our answer. = 2 h g (b) Wha is he ball s velociy when (jus before) i his he ground? Now we are looking for v. To find his wihou reference o he resul of par (a) we should use he fourh equaion.

9 Chaper B - One Dimensional Kinemaics 9 v 2 = v g Δy We sill have Δy = -h and v 0 = 0, so we ge: v 2 = 2 g h. The moion is downward so we wan he negaive square roo when solving for v. v = - 2 g h If he problem asked for he speed insead of velociy hen you would ake he posiive square roo. Example B.5 - Upward Iniial Velociy A ball is hrown sraigh upward from he ground a 30 m /s. (a) Wha is he maximum heigh reached by he ball? We are given he iniial velociy. How do we mahemaically describe he ball s highes poin? A he very op of is verical pah he ball is insananeously a res, v = 0. The maximum heigh y max is Δy when v = 0. v 0 = 30 m /s, v = 0, y max = Δy =? Since ime is neiher given or desired we are led o he fourh equaion. (b) How long does i ake for he ball o reurn o he ground? v 2 = v g Δy y max = Δy = 2 g = 45.9 m When he ball reurns o where i began Δy = 0. Noe ha Δy is no he disance raveled; i is he ne displacemen. I is zero because Δy = y - y 0 and y = y 0. Since we are solving for ime and we sill have v 0 = 30 m /s we should use he hird equaion. Δy = v 0-2 g 2 0 = v 0-2 g = 0 and v 0-2 g = 0 I is rivially rue ha Δy = 0 when = 0; we wan he oher soluion, which becomes: 2 v 0 = g = 6.2 s (c) Solve for up, he ime for he ball o move o is highes poin. Also show ha when Δy = 0, as in he previous par, = 2 up. Since v = 0 a he op, we have v = v 0 - g 0 = v 0 - g up up = I follows ha = 2 up. This is generally he case for free-fall problems wih Δy = 0. (d) When does he ball pass a 35 m high window? v 0 2 v 0 g = 3.06 s. We now wan o find when Δy = 35 m. We sill have v 0 = 30 m /s, so he hird equaion is needed. Δy = v 0-2 g 2 35 m = (30 m /s) m s2 2 Solving for gives wo soluions. Boh soluions are required here; i passes he window wice, moving upward and hen moving downward. =.57 s and = 4.55 s

10 0 Chaper B - One Dimensional Kinemaics (The above can be solved using he quadraic formula = -b± c = 35 m.) y 45.9 m b2-4 a c 2 a, wih a = m s2, b = -30 m /s and 35 m.57 s 3.06 s 4.55 s 6.2 s Figure - y vs. graph