Chapter 2. Pythagorean Theorem. Right Hand Rule. Position. Distance Formula

Transcription

1 Chapter Moton n One Dmenson Cartesan Coordnate System The most common coordnate system or representng postons n space s one based on three perpendcular spatal axes generally desgnated x, y, and z. Any pont P may be represented by three sgned numbers, usually wrtten (x, y, z) where the coordnate s the perpendcular dstance rom the plane ormed by the other two axes. Rght Hand Rule Pont the ngers o your rght hand down the poste X axs Turn your ngers nward towards the poste Y axs Your thumb s now pontng n the poste Z axs Pythagorean Theorem a + b = c x + y = z Dstance Formula 3-D -D -D Poston Dened n terms o a rame o reerence One dmensonal, so generally the x- or y- axs Pont Partcle

2 Scalar Quanttes Scalar quanttes are completely descrbed by magntude only Examples: 35 meters 40 mles per hour 0 klograms Vector Quanttes Vector quanttes need both magntude (sze) and drecton to completely descrbe them Examples 35 mles per hour north 0 meters east We can draw ectors Represented by an arrow, the length o the arrow s proportonal to the magntude o the ector Head o the arrow represents the drecton We can denty ector arables: Generally prnted n bold ace type Dsplacement Measures the change n poston Represented as Δx ( horzontal) or Δy ( ertcal) Vector quantty + or - s generally sucent to ndcate drecton or one-dmensonal moton Unts are meters (m) Dstance Measures how ar somethng has traeled Scalar quantty Ignore sgns take the absolute alues o eer measurement and add them together. Unts are meters (m) n SI, centmeters (cm) n cgs or eet (t) n US Customary Dsplacement Dstance Dstance may be, but s not necessarly, the magntude o the dsplacement Blue lne shows the dstance Red lne shows the dsplacement

3 Dsplacements s Dstance The dea o arables Do not thnk that a arable always stands or the same concept. You must ealuate what you are lookng at. Example Dsplacement s represented by (d, x, s, y, z ) Velocty It takes tme or an object to undergo a dsplacement The aerage elocty s rate at whch the dsplacement occurs Δx x x aerage = = generally use a tme nteral, so to let t = 0 V(bar) = V aerage Velocty contnued Drecton wll be the same as the drecton o the dsplacement (tme nteral s always poste) + or - s sucent Unts o elocty are m/s Other unts may be gen n a problem, but generally wll need to be conerted to these Speed The aerage speed o an object s dened as the total dstance traeled dded by the total tme elapsed total dstance Aerage speed = total tme s = d t Speed s a scalar quantty Speed, cont Aerage speed totally gnores any aratons n the object s actual moton durng the trp May be, but s not necessarly, the magntude o the elocty The total dstance and the total tme are all that s mportant SI unts are m/s same unts as elocty 3

4 Slope Slope = Rse Run = Δy Δx = y x y x Graphcal Interpretaton o Velocty Velocty can be determned rom a poston-tme graph Aerage elocty equals the slope o the lne jonng the ntal and nal postons Compare unts o slope Instantaneous elocty s the slope o the tangent to the cure at the tme o nterest The nstantaneous speed s the magntude o the nstantaneous elocty Aerage Velocty, Constant The straght lne ndcates constant elocty The slope o the lne s the alue o the aerage elocty Aerage Velocty Instantaneous Velocty Instantaneous Velocty The lmt o the aerage elocty as the tme nteral becomes nntesmally short, or as the tme nteral approaches zero lm 0 Δx The nstantaneous elocty ndcates what s happenng at eery pont o tme 4

5 Instantaneous Velocty on a Graph The slope o the lne tangent to the poston-s.- tme graph s dened to be the nstantaneous elocty at that tme The nstantaneous speed s dened as the magntude o the nstantaneous elocty Unorm Velocty Unorm elocty s constant elocty I the nstantaneous eloctes are always the same All the nstantaneous eloctes wll also equal the aerage elocty Acceleraton Changng elocty (non-unorm) means an acceleraton s present Acceleraton s the rate o change o the elocty a a erage Δ = = Acceleraton Negate acceleraton s drectonal Deceleraton s just acceleraton n the negate drecton. Deceleratons does not exst Unts are m/s² (SI), cm/s² (cgs), and t/s² (US Cust) Relatonshp Between Acceleraton and Velocty Unorm elocty (shown by red arrows mantanng the same sze) Acceleraton equals zero Tme Interals and dsplacement Objects that are mong equal dstances n equal tmes are mong at constant elocty Objects that are not mong equal dstances n equal tmes are acceleratng n the poste or negate drecton 5

6 Relatonshp Between Velocty and Acceleraton Relatonshp Between Velocty and Acceleraton Velocty and acceleraton are n the same drecton Acceleraton s unorm (blue arrows mantan the same length) Velocty s ncreasng (red arrows are gettng longer) Acceleraton and elocty are n opposte drectons Acceleraton s unorm (blue arrows mantan the same length) Velocty s decreasng (red arrows are gettng shorter) Velocty and Acceleraton Equatons Rested aerage a aerage Δx = = x Δ = = x Aerage Acceleraton Vector quantty Compare dmensons o elocty s tme graph When the sgn o the elocty and the acceleraton are the same (ether poste or negate), then the speed s ncreasng When the sgn o the elocty and the acceleraton are n the opposte drectons, the speed s decreasng Instantaneous and Unorm Acceleraton The lmt o the aerage acceleraton as the tme nteral goes to zero When the nstantaneous acceleratons are always the same, the acceleraton wll be unorm The nstantaneous acceleratons wll all be equal to the aerage acceleraton Graphcal Interpretaton o Acceleraton Aerage acceleraton s the slope o the lne connectng the ntal and nal eloctes on a elocty-tme graph Instantaneous acceleraton s the slope o the tangent to the cure o the elocty-tme graph 6

7 Aerage Acceleraton Area Under Cures The area under a elocty s tme cure s equalent to the dsplacement. Usng the area o a trangle ormula A=/ bh We can determne the dsplacement rom a elocty tme graph. Area Under Cure Practce Deraton o Knematc equaton a aerage = a = a + = Graphcal Deraton o Knematcs Knematc Equaton V = o + a * t 7

8 Deraton o Knematcs Equaton We know Aerage Velocty Δd = Aerage Velocty s also o + = But only when acceleraton s constant Deraton o Knematcs Equaton Thereore puttng t together: Δd + = 0 Manpulatng ths equaton Δ d = ( 0 + ) Keep ths or Knematc 3 Deraton o Knematcs Equaton Now substtute Knematc n or d = 0 ) ( + ( + at ) Δ 0 Now smply the equaton and you get: Δd = o * t + at Knematcs Alternate Deraton -I an object s not acceleratng then: d = o * t Remember that d=area under s t graph area _ o _ Trangle = / ( b)( h) area _ o _ Trangle = / ( t)( Combnng o t we get d = o * t + area _ o _ Trangle d = o * t + ( o )* t o ) Knematcs Alternate Deraton a t + o = d = ( o) * t Combnng the Equaton Aboe we get: Δd = o * t + at Deraton o Knematcs Equaton 3 We start wth: d = ( + 0 ) t From Knematc we get d = ( t = ( ) / a Puttng them together we get: + )( 0 ) / a 0 8

9 Deraton o Knematcs Equaton 3 Smplyng the equaton we get ad = 0 Solng or we get: = + aδd o IMPORTANT Make Sure the Table.3 (Equaton o moton or unorm acceleraton) makes t onto your equaton sheet Quadratc Equaton Sometmes we are gong to nd that we hae a quadratc trnomal equaton we can t sole. So we us the quadratc solng equaton to sole t. The equaton should be put nto the orm: ax + bx + c = 0 Then sole t wth Arstotle Broke moton nto parts Natural Moton Thngs always try to return to ther proper place Objects allng speed proportonate to ther weght Volent Moton Pushng Pullng b ± x = b 4ac a Earth Centered Unerse Tme Lne Arstotle BC Eratosthenes 76 BC 00 BC Coperncus /9/473-5/4/543 Kepler /7/ Galleo Newton /4/643 3/3/77 Free Fall All objects mong under the nluence o only graty are sad to be n ree all All objects allng near the earth s surace all wth a constant acceleraton Galleo orgnated our present deas about ree all rom hs nclned planes The acceleraton s called the acceleraton due to graty, and ndcated by g 9

10 Acceleraton due to Graty Symbolzed by g g = 9.8 m/s² g s always drected downward toward the center o the earth Free Fall -- an object dropped Intal elocty s zero Let up be poste Use the knematc equatons Generally use y nstead o x snce ertcal o = 0 a = g Free Fall -- an object thrown downward a = g Intal elocty 0 Wth upward beng poste, ntal elocty wll be negate Free Fall -- object thrown upward Intal elocty s upward, so poste The nstantaneous elocty at the maxmum heght s zero a = g eerywhere n the moton g s always downward, negate = 0 Thrown upward, cont. The moton may be symmetrcal then t up = t down then = - o The moton may not be symmetrcal Break the moton nto arous parts generally up and down Non-symmetrcal Free Fall Need to dde the moton nto segments Possbltes nclude Upward and downward portons The symmetrcal porton back to the release pont and then the nonsymmetrcal porton 0

11 Combnaton Motons Addtonal Problems How ar does the plane go n the rst second? How about once t starts breakng? How long does t take the trooper to catch up?