EXPLORATION 2.5A Exploring the motion diagram of a dropped object

Transcription

1 -5 Acceleraton Let s turn now to moton that s not at constant elocty. An example s the moton of an object you release from rest from some dstance aboe the floor. EXPLORATION.5A Explorng the moton dagram of a dropped object Step 1 - Sketch a moton dagram for a ball that you release from rest from some dstance aboe the floor, showng ts poston at regular tme nterals as t falls. The moton dagram n Fgure.14 shows mages of the ball that are close together near the top, where the ball moes more slowly. As the ball speeds up, these mages get farther apart because the ball coers progressely larger dstances n the equal tme nterals. How do we know how much space to nclude between each mage? One thng we can do s to consult expermental edence, such as strobe photos of dropped objects. These photos show that the dsplacement from one tme nteral to the next ncreases lnearly, as n Fgure.14. Step - At each pont on the moton dagram, add an arrow representng the ball s elocty at that pont. Neglect ar resstance. The arrows n Fgure.14 represent the elocty of the ball at the arous tmes ndcated on the moton dagram. Because the dsplacement ncreases lnearly from one tme nteral to the next, the elocty also ncreases lnearly wth tme. Fgure.14: A moton dagram, and elocty ectors, for a ball released from rest at t = 0. The ball s poston and elocty are shown at 0.1-second nterals. Key dea: For an object dropped from rest, the elocty changes lnearly wth tme. Related End-of-Chapter Exercses: Another way to say that the ball s elocty ncreases lnearly wth tme s to say that the rate of change of the ball s elocty, wth respect to tme, s constant: = constant. Ths quantty, the rate of change of elocty wth respect to tme, s referred to as the acceleraton. Acceleraton s related to elocty n the same way elocty s related to poston. Aerage acceleraton: a ector representng the aerage rate of change of elocty wth respect to tme. The SI unt for acceleraton s m/s. a = = change n elocty tme nteral. (Equaton.7: Aerage acceleraton) The drecton of the aerage acceleraton s the drecton of the change n elocty. Chapter Moton n One Dmenson Page - 1

2 Instantaneous acceleraton: a ector representng the rate of change of elocty wth respect to tme at a partcular nstant n tme. The SI unt for acceleraton s m/s. A practcal defnton of nstantaneous acceleraton at a partcular nstant s that t s the slope of the elocty-ersus-tme graph at that nstant. Expressng ths as an equaton: a =, (Equaton.8: Instantaneous acceleraton) t where t s small enough that the acceleraton can be consdered to be constant oer that nteral. Note that, on Earth, objects dropped from rest are obsered to hae acceleratons of 9.8 m/s straght down. Ths s known as g, the acceleraton due to graty. EXPLORATION.5B Graphs n a constant-acceleraton stuaton The graph n Fgure.15 shows the elocty, as a functon of tme, of a bus mong n the poste drecton along a straght road. Fgure.15: A graph of the elocty of a bus as a functon of tme. Step 1 What s the acceleraton of the bus? Sketch a graph of the acceleraton as a functon of tme. The graph n Fgure.15 s a straght lne wth a constant slope. Ths tells us that the acceleraton s constant because the acceleraton s the slope of the elocty-ersus-tme graph. Usng the entre 10-second nteral, applyng equaton.8 ges: + 5 m/s ( + 5m / s) + 0 m/s a = = = =+.0 m/s. 10 s 10 s The acceleraton graph n Fgure.16 s a horzontal lne, because the acceleraton s constant. Compare Fgures.15 and.16 to the graphs for the red car n Fgure.13. Note the smlarty between the acceleraton and elocty graphs n a constant-acceleraton stuaton and the elocty and poston graphs n a constant-elocty stuaton. Fgure.16: A graph of the acceleraton of the bus as a functon of tme. Step - What on the acceleraton-ersus-tme graph s connected to the elocty? The connecton between elocty and acceleraton s smlar to that between poston and elocty - the area under the cure of the acceleraton graph s equal to the change n elocty, as shown n Fgure.17. Ths follows from Equaton.8, whch n a constant acceleraton stuaton can be wrtten as = a. Fgure.17: In 10 seconds, the elocty changes by +0 m/s. Ths s the area under the acceleraton-ersus-tme graph oer that nteral. Key deas: The acceleraton s the slope of the elocty-ersus-tme graph, whle the area under the acceleraton-ersus-tme graph for a partcular tme nteral represents the change n elocty durng that tme nteral. Related End-of-Chapter Exercses: 16 and 3. Essental Queston.5: Consder the bus n Exploraton.5B. Would the graph of the bus poston as a functon of tme be a straght lne? Why or why not? Chapter Moton n One Dmenson Page -

3 Answer to Essental Queston.5: The poston-ersus-tme graph s not a straght lne, because the slope of such a graph s the elocty. The fact that the bus elocty ncreases lnearly wth tme means the slope of the poston-ersus-tme graph also ncreases lnearly wth tme. Ths actually descrbes a parabola, whch we wll nestgate further n the next secton. -6 Equatons for Moton wth Constant Acceleraton In many stuatons, we wll analyze moton usng a model n whch we assume the acceleraton to be constant. Let s dere some equatons that we can apply n such stuatons. In general, at some ntal tme t = 0, the object has an ntal poston x r and an ntal elocty of r, whle at some (usually later) tme t, the object s poston s x r and ts elocty s r. Acceleraton s related to elocty the same way elocty s related to poston, so we can follow a procedure smlar to that at the end of Secton -3 to dere an equaton for elocty. Substtute f for n the rearrangement of Equaton.8, = a. Ths ges: f = a t = a( tf t). Generally, we defne the ntal tme t to be zero: f = atf. Remoe the f subscrpts to make the equaton as general as possble: = at. We also generally remoe the ector symbols, although we must be careful to nclude sgns. = + at. (Equaton.9: Velocty for constant-acceleraton moton) A second equaton comes from the defnton of aerage elocty (Equaton.): r r r r r x x x x x Aerage elocty = = = =. t t t If the acceleraton s constant the aerage elocty s smply the aerage of the ntal and fnal eloctes. Ths ges, after agan droppng the ector symbols: + x x =. (Equaton.10: Connectng aerage elocty and dsplacement) t Equaton.10 s sometmes awkward to work wth. If we substtute + at n for (see Equaton.9) n Equaton.10, re-arrangng produces an equaton descrbng a parabola: 1 x = x + t + at. (Equaton.11: Poston for constant-acceleraton moton) We can dere another useful equaton by combnng equatons.9 and.10 n a dfferent way. Solng equaton.9 for tme, to get t =, and substtutng the rght-hand sde of that a expresson n for t n equaton.10, ges, after some re-arrangement: = + a x. (Eq..1: Connectng elocty, acceleraton, and dsplacement) Chapter Moton n One Dmenson Page - 3

4 An mportant note about poste and negate sgns. When we make use of the equatons on the preous page, we must be careful to nclude the approprate poste or negate sgns that are bult nto each of the arables. The frst step s to choose a poste drecton. If the ntal elocty s n that drecton, t goes nto the equatons wth a poste sgn. If the ntal elocty s n the opposte drecton (the negate drecton), t goes nto the equatons wth a negate sgn. Apply a smlar rule for the fnal elocty, the acceleraton, the dsplacement, the ntal poston, and the fnal poston. For all of those quanttes, the sgn s assocated wth the drecton of the correspondng ector. Moton wth constant acceleraton s an mportant concept. Let s summarze a general, systematc approach we can apply to stuatons nolng moton wth constant acceleraton. A General Method for Solng a One-Dmensonal Constant-Acceleraton Problem 1. Pcture the scene. Draw a dagram of the stuaton. Choose an orgn to measure postons from, and a poste drecton, and show these on the dagram.. Organze what you know, and what you re lookng for. Makng a table of data can be helpful. Record alues of the arables used n the equatons below. 3. Sole the problem. Thnk about whch of the constant-acceleraton equatons to apply, and then set up and sole the problem. The three man equatons are: = + at. (Equaton.9: Velocty for constant-acceleraton moton) 1 x = x + t + at. (Equaton.11: Poston for constant-acceleraton moton) = + a x. (Equaton.1: Connectng elocty, acceleraton, and dsplacement) 4. Thnk about the answer(s). Check your answers to see f they make sense. EXAMPLE.6 Workng wth arables In physcs, beng able to work wth arables as well as numbers s an mportant skll. Ths can also produce nsghts that workng wth numbers does not. Let s say an object s dropped from rest from the top of a buldng of heght H, whle another object s dropped from rest from the top of a buldng of heght 4H. Assumng both objects fall under the nfluence of graty alone (that s, they hae the same acceleraton), compare the tmes t takes them to reach the ground. SOLUTION = 0, because the objects are dropped from rest. Take the ntal poston to be the top of the buldng n each case, so x = 0. Ths reduces Equaton.11 to x= at /. Because the acceleraton s the same n each case, the equaton tells us the poston s proportonal to the square of the tme. To quadruple the fnal poston, as we are dong, we need to ncrease the tme by a factor of two. The fall from the buldng that s four tmes as hgh takes twce as long. Ths s llustrated by the moton dagram n Fgure.18. Fgure.18: Fallng from rest for double the tme quadruples the dstance traeled. The heght of each of the four shaded regons s H, the heght of the smaller buldng n Example.6. Related End-of-Chapter Exercses: 5 and 55 Essental Queston.6: Return to the stuaton descrbed n Example.6. Compare the eloctes of the objects just before they ht the ground. Chapter Moton n One Dmenson Page - 4

5 Answer to Essental Queston.6: Usng = 0 reduces Equaton.9 to = at. Doublng the tme doubles the fnal elocty, so the object dropped from the buldng that s four tmes hgher has a fnal elocty twce as large as that of the other object. Equaton.1 ges the same result. -7 Example Problem Let s look at the arous representatons of moton wth constant acceleraton, consderng the example of a ball tossed straght up n the ar. EXPLORATION.7 A ball tossed straght up You toss a ball straght up nto the ar. The ball takes.0 s to reach ts maxmum heght, and an addtonal.0 s to return to your hand. You catch the ball at the same heght from whch you let t go, and the ball has a constant acceleraton because t s acted on only by graty. Consder the moton from the nstant just after you release the ball untl just before you catch t. Step 1 Pcture the scene draw a dagram of the stuaton. The dagram n Fgure.19 shows the ntal condtons, the orgn, and the poste drecton. We are free to choose ether up or down as the poste drecton, and to choose any reference pont as the orgn. In ths case let s choose the orgn to be the pont from whch the ball was released, and choose up to be poste. Step Organze the data. Table. summarzes what we know, ncludng alues for the acceleraton and the ntal poston. We need these alues for the constant-acceleraton equatons. Because the ball moes under the nfluence of graty alone, and we can assume the ball s on the Earth, the acceleraton s the acceleraton due to graty, 9.8 m/s drected down. Because down s the negate drecton, we nclude a negate sgn: a = 9.8m /s. Parameter Value Orgn Launch pont Poste drecton up Intal poston x = 0 Intal elocty =+ m / s Acceleraton a = 9.8 m / s Poston at t = 4.0 s x = 4s = x = 0 t Fgure.19: A dagram of the ntal stuaton. Table.: Summarzng the nformaton that was gen about the ball. Step 3 Sole the problem. In ths case, we want to draw graphs of the acceleraton, elocty, and poston of the ball, all as a functon of tme. To do ths we should frst wrte equatons for the acceleraton, elocty, and poston. The acceleraton s constant at a = 9.8m /s. Knowng the acceleraton allows us to sole for the ntal elocty. One way to do ths s to re-arrange Equaton.9 to ge = at. Because = 0 at t =.0 s (the ball s at rest for an nstant when t reaches ts maxmum heght) we get = at = 0 ( 9.8 m / s )(.0 s) = m / s. Knowng the ntal elocty enables us to wrte equatons for the ball s elocty (usng Equaton.9) and poston (usng Equaton.11) as a functon of tme. The equatons, and correspondng graphs, are part of the multple representatons of the moton shown n Fgure.0. Note that the poston ersus tme graph s parabolc, but the moton s confned to a lne. Chapter Moton n One Dmenson Page - 5

6 Step 4 - Sketch a moton dagram for the ball. The moton dagram s shown on the rght n Fgure.15. Note the symmetry of the up and down motons (ths s also apparent from the graphs). The moton of the ball on the way down s a mrror mage of ts moton on the way up. (a) Descrpton of the moton: A ball you toss straght up nto the ar reaches ts maxmum heght.0 s after beng released, takng an addton.0 s to return to your hand. It experences a constant acceleraton from the moment you release t untl just before you catch t. (b) Equatons (up s poste): Acceleraton-ersus-tme: Velocty-ersus-tme: a = 9.8 m / s = m / s (9.8 m /s ) t x = m /s t 4.9 m / s t Poston-ersus-tme: ( ) ( ) Fgure.0: Multple representatons of a ball thrown straght up, ncludng (a) a descrpton n words; (b) equatons for the ball s acceleraton, elocty, and poston; graphs gng the ball s (c) acceleraton, (d) elocty, and (e) poston as a functon of tme; and (f) a moton dagram. These dfferent perspectes show how arous aspects of the moton eole wth tme. Key deas: At the Earth s surface the acceleraton due to graty has a constant alue of g = 9.8 m /s drected down. We can thus apply constant-acceleraton methods to stuatons nolng objects dropped or thrown nto the ar. For an object that s thrown straght up the downward part of the trp s a mrror mage of the upward part of the trp. Related End-of-Chapter Exercses: 33, 61, and 6. Essental Queston.7: Consder agan the ball n Exploraton.7. The ball comes to rest for an nstant at ts maxmum-heght pont. What s the ball s acceleraton at that pont? Chapter Moton n One Dmenson Page - 6

7 Answer to Essental Queston.7: A common msconcepton s that the ball s acceleraton s zero at the maxmum-heght pont. In fact, the acceleraton s g, 9.8 m/s down, durng the entre trp. Ths s what s shown on the acceleraton graph, for one thng the graph confrms that nothng specal happens to the acceleraton at t =.0 s, een though the ball s momentarly at rest. One reason for ths s that the ball s under the nfluence of graty the entre tme. -8 Solng Constant-Acceleraton Problems Consder one more example of applyng the general method for solng a constantacceleraton problem. EXAMPLE.8 Combnng constant-acceleraton moton and constant-elocty moton A car and a bus are traelng along the same straght road n neghborng lanes. The car has a constant elocty of +5.0 m/s, and at t = 0 t s located 1 meters ahead of the bus. At tme t = 0, the bus has a elocty of +5.0 m/s and an acceleraton of +.0 m/s. When does the bus pass the car? SOLUTION 1. Pcture the scene draw a dagram. The dagram n Fgure.16 shows the ntal stuaton, the poste drecton, and the orgn. Let s choose the poste drecton to be the drecton of trael, and the orgn to be the ntal poston of the bus.. Organze the data. Data for the car and the bus s organzed separately n Table.3, usng subscrpts C and B to represent the car and bus, respectely. Car Bus Intal poston x C =+ 1 m x B = 0 Intal elocty C =+ 5 m / s B = m / s Acceleraton a C = 0 a B =+.0 m / s Table.3: Summarzng the nformaton that was gen about the car and the bus. x C Fgure.16: A dagram showng the ntal postons of the car and the bus, the poston of the orgn, and the poste drecton. 3. Sole the problem. Let s use Equaton.8 to wrte expressons for the poston of each ehcle as a functon of tme. Because we summarzed all the data n Table.3 we can easly fnd the alues of the arables n the equatons. 1 For the car: xc = xc + Ct+ act =+ 1 m + (5 m /s) t. 1 For the bus: xb = xb + Bt+ abt = 0 + (5.0 m /s) t+ (1.0 m /s ) t. The bus passes the car when the ehcles hae the same poston. At what tme does = xb? Set the two equatons equal to one another and sole for ths tme (let s call t t 1 ). + 1 m + (5 m / s) t = (5.0 m / s) t + (1.0 m / s ) t Brngng eerythng to the left sde ges: t1 t1 (1.0 m / s ) (0 m / s) 1 m = 0. Chapter Moton n One Dmenson Page - 7

8 We can sole ths wth the quadratc equaton, where c = 1 m. a = 1.0 m /s, b = 0 m /s, and b± b 4 ac + (0m/s) ± (400m /s ) + (84m /s ) + (0m/s) ± (m/s) t1 = = =. a.0 m / s.0 m / s The equaton ges us two solutons for t 1 : Ether t 1 = +1 s or t 1 = 1.0 s. Clearly the answer we want s +1 s. The other number does hae a physcal sgnfcance, howeer, so let s try to make some sense of t. The negate answer represents the tme at whch the car would hae passed the bus f the moton condtons after t = 0 also appled to the perod before t = Thnk about the answer. A nce way to check the answer s to plug t 1 = +1 s nto the two poston-ersus-tme equatons from the preous step. If the tme s correct the poston of the car should equal the poston of the bus. Both equatons ge postons of 546 m from the orgn, gng us confdence that t 1 = +1 s s correct. Note: Ths example s contnued on the accompanyng web ste, solng for the tme at whch the car and the bus hae the same elocty. Related End-of-Chapter Exercses: 54 and 56. Chapter Summary Essental Idea: Descrbng moton n one dmenson. The moton of many objects (such as you, cars, and objects that are dropped) can be approxmated ery well usng a constant-elocty or a constant-acceleraton model. Parameters used to Descrbe Moton Dsplacement s a ector representng a change n poston. If the ntal poston s x r and the fnal poston s x r f we can express the dsplacement as: r r r x = x x. (Equaton.1: Dsplacement n one dmenson) f Aerage Velocty: a ector representng the aerage rate of change of poston wth respect to tme. The MKS unt for elocty s m/s (meters per second). r x = = net dsplacement tme nteral. (Equaton.: Aerage elocty) Whle elocty s a ector, and thus has a drecton, speed s a scalar. Aerage Speed = total dstance coered = (Equaton.3: Aerage speed) tme nteral Chapter Moton n One Dmenson Page - 8

9 Instantaneous Velocty: a ector representng the rate of change of poston wth respect to tme at a partcular nstant n tme. The MKS unt for elocty s m/s (meters per second). r x =, (Equaton.4: Instantaneous elocty) t where t s small enough that the elocty can be consdered to be constant oer that nteral. Instantaneous Speed: the magntude of the nstantaneous elocty. Aerage Acceleraton: a ector representng the aerage rate of change of elocty wth respect to tme. The MKS unt for acceleraton s m/s. change n elocty a = =. (Equaton.7: Aerage acceleraton) tme nteral Instantaneous Acceleraton: a ector representng the rate of change of elocty wth respect to tme at a partcular nstant n tme. a =, (Equaton.8: Instantaneous acceleraton) t where t s small enough that the acceleraton can be consdered constant oer that nteral. The acceleraton due to graty, g, s 9.8 m/s, drected down, at the surface of the Earth. A General Method for Solng a 1-Dmensonal Constant-Acceleraton Problem 1. Pcture the scene. Draw a dagram of the stuaton. Choose an orgn to measure postons from, and a poste drecton, and show these on the dagram.. Organze what you know, and what you re lookng for. Makng a table of data can be helpful. Record alues of the arables used n the equatons below. 3. Sole the problem. Thnk about whch of the constant-acceleraton equatons to apply, and then set up and sole the problem. The three man equatons are: = + at. (Equaton.9: Velocty for constant-acceleraton moton) 1 x = x + t + at. (Equaton.11: Poston for constant-acceleraton moton) = + a x. (Equaton.1: Connectng elocty, acceleraton, and dsplacement) 4. Thnk about the answer(s). Check your answers, and/or see f they make sense. Graphs The elocty s the slope of the poston-ersus-tme graph; the dsplacement s the area under the elocty-ersus-tme graph. The acceleraton s the slope of the elocty-ersus-tme graph; the change n elocty s the area under the acceleraton-ersus-tme graph. Constant Velocty and Constant Acceleraton The moton of an object at rest s a specal case of constant-elocty moton. In constantelocty moton the poston-ersus-tme graph s a straght lne wth a slope equal to the elocty. Constant-elocty moton s a specal case of constant-acceleraton moton. In one dmenson, f the acceleraton s constant and non-zero the poston-ersus-tme graph s quadratc whle the elocty-ersus-tme graph s a straght lne wth a slope equal to the acceleraton. Chapter Moton n One Dmenson Page - 9