If you need any help, please reference youtube and specifically AP Plus Physics (Dan Fullerton) online. Enjoy the Summer. Mr. Robayo RHS STEM/Physics

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1 AP Physics C Calculus Base This is yur summer assignmen fr AP Physics C fr scieniss an engineers. Yur summer assignmen cnsiss f an eriew f ecrs an -D an -D kinemaics. Ne ha here will be calculus inle an which will inclue he eriaie fr which I hae aache a primer in he appeni fr hse ha are sill cer i. I is epece ha yu reiew he perinen maerials cering a summary f ecrs an kinemaics. Yu are hen finish he aache muliple-chice prblems an he ne free respnse quesin. This is a cllege leel curse an yu are epece use he resurces a yur ispsal figure u an learn hw sle he gien prblems. If yu nee any help, please reference yuube an specifically AP Plus Physics (Dan Fullern) nline. Enjy he Summer. Mr. Rbay RHS STEM/Physics

2 Graphing Displacemen Time Graph Velciy Time Graph Accelerain Time Graph Cnsan Velciy 0 Slpe is Fig.a Fig.b Fig.c 0 Slpe is a = 0 Area uner cure is a a 0 Unifrm Accelerain 0 Slpe is (angen line) 0 Slpe is a Area uner cure is Fig. Fig.e Fig.f Slpes f Cures are an impran analyical l use in physics. Any equain ha can be manipulae in he frma y = m+ b can be represene an analyze graphically. As an eample: = 0 + a can be rearrange slighly in = a+ 0. Cmpare his equain he equain f a line. I is apparen ha a is he slpe an ha 0 is he y-inercep (Fig.b an Fig.e). Wha equain generaes elciy in Fig..a an Fig.? Area Uner a Cure is anher impran graphical l. Muliply he y-ais (heigh) by he -ais (base) an eermine if his maches any knwn equains. Fr eample: Figures.b an.e are elciy-ime pls. Simply muliply. This is a rearrange frm f he equain =. The frm baine frm he graph is =, which means ha isplacemen is he area uner he elciy-ime pl. a a 0 Area uner cure is Calculus: Require in AP Physics C (pinal fr AP Physics B suens). Calculus is augh in mah class. These reiew shees will fcus n he Physics aspec f sluins. Calculus seps may n be shwn. Sluin will be up he suen. The equains uline in he preius page wrk well in he fllwing siuains. Linear Funcins: inling cnsan elciy an accelerain, as iagramme in he abe graphs (ecep Fig.). Nnlinear Funcins: scenaris where he prblem is seeking infrmain abu a change in a quaniy, r. Nnlinear Funcins: scenaris where he prblem is seeking an aerage elciy in an ineral. Calculus is neee fin he slpes f nnlinear funcins an he areas uner nnlinear cures.. Velciy: Slpe f he isplacemen-ime cure. = eample: = a = 0 + a = 0 + a. Accelerain: Slpe f he elciy-ime cure. a = eample: a= ( 0 + a) = a. Velciy: Area uner accelerain-ime cure. (Ne: if c = 0 cann be fun, hen yu can nly sle fr ) = a eample: = ( a) = c+ a = 0 + a 4. Displacemen: Area uner he elciy-ime cure. (Ne: if c = 0 cann be fun, hen yu can nly sle ) = eample: = ( 0 + a) = c+ 0+ a = a Reise 8/9/06 R H Jansen

3 Falling Bies: Objecs ming erically uner he influence f graiy. Earh s surface graiy is g = 9.8 m/s. This spees bjecs up (+ accelerain), bu is irece wnwar ( ). Objecs can be hrwn up, wn, r jus be rppe. They can lan belw, a he same heigh, r abe he rigin. They cme an insananeus sp a he highes pin in fligh. y = 0 y + g y = y0 + 0y+ g y = 0y + g( y y0) The psiie an negaie signs can cause ruble in hese prblems. The easies way hanle he signs is se he irecin f iniial min as psiie an hen ensure all signs are cnsisen wih his ecisin. This has ne huge benefi. I eliminae he uble sign n accelerain. When iniial elciy 0 is use anchr irecin, hen a psiie accelerain means speeing up an negaie accelerain inles slwing wn. Drppe frm res Thrwn wnwar Thrwn upwar 0 = 0, bu i will me wn iniially 0 is irece wnwar 0 is irece upwar Se wnwar as psiie irecin Se wnwar as psiie irecin Se upwar as psiie irecin = = g + + +g + + g = Eeryhing is psiie an easy Eeryhing is psiie an easy This is ifficul, an epens n where in he fligh he prblem ens. Kinemaics Prblems Inling Changes in he Magniue f Accelerain If he magniue f accelerain changes while sling a kinemaics prblem hen he prblem mus be sle in separae pars. Unlike isplacemen an elciy, he Kinemaic equains n cnain he ariables a 0 an a. Eample -: Mre han ne accelerain A car iniially a res acceleraes a 4 m/s while cering a isance f 00 m. Then he car cninues a cnsan elciy fr 500 m. Finally i slws a sp wih a ecelerain f m/s. Deermine he al ime f his isplacemen. Accelerain Phase: Cnsan Velciy Phase: = a = Decelerain Phase: = 0 + a Tal Time = = 4. s 00 ( ) = = = 707. s an ( ) ( )( ) a ( 4) ( 500) ( 8. ) = = = 7. 7 s ( ) (. ) ( ) = = = 94. s a = 0 + a = = 8. m s 500 = Reise 8/9/06 4 R H Jansen

4 Vecrs an Vecrs in Kinemaics Scalar: A physical quaniy escribe by a single number an unis. A quaniy escribing magniue nly. Vecr: Many f he ariables in physics equains are ecr quaniies. Vecrs hae magniue an irecin. Magniue: Size r een. The numerical alue. Direcin: Alignmen r rienain wih respec se lcain an sysem f rienain, such as a crinae ais. Nain: A r A The lengh f he arrw represens, an is prprinal, he ecrs magniue. The irecin he arrw pins inicaes he irecin f he ecr. Negaie Vecrs: Hae he same magniue as heir psiie cunerpar, bu pin in he ppsie irecin. If A hen A Vecr Aiin an subracin Think f i as ecr aiin nly. The resul f aing ecrs is calle he resulan R. A + B = R A + B = R When yu nee subrac ne ecr frm anher hink f he ne being subrace as being a negaie ecr. A B is really A+ e Bj = R A + ( B) = R A negaie ecr has he same lengh as is psiie cunerpar, bu is irecin is reerse. This is ery impran. In physics a negaie number es n always mean a smaller number. Mahemaically is smaller han +, bu in physics hese numbers hae he same magniue (size), hey jus pin in ifferen irecins (80 apar). Vecr Aiin: Parallelgram A A A A + B R Fig.a B B B Tip Tail A A B A B A + B R Fig.b B Bh mehs arrie a he eac same sluin since eiher meh is essenially a parallelgram. In sme prblems ne meh is aanageus, while in her prblems he alernaie meh is superir. Repring Magniue an Direcin: Cmpnen Meh: A ecr a can be repre by giing he cmpnens alng he - r y-ais. Repring a ecr his way is frmally ne by emplying he uni ecrs i an j. As an eample: ecr A in fig.a wul be A= Ai + A j, where, i + j =. There is a hir cmpnen ecr k use fr hree imensinal prblems inling he z-ais. The ecr in Fig..b shws a numerical applicain f he cmpnen meh. In his eample yu are gien he plar crinaes A= 5 a 7. Using rignmery he cmpnens can be esablishe. A = Acsθ = 5cs7 = 4 an Ay = Asinθ = 5sin 7 =. Then A can be epresse as fllws: A = 4i + j y Plar Crinaes: Vecr A in Fig..b is repre in plar crinaes, A= 5 a 7. This Fig..b is simply he lengh f a ecr an is angle measure cunerclckwise wih respec he psiie -ais. (Negaie angles are allwe an inicae ha irecin was measure clckwise frm he +-ais. If he cmpnen ecrs are gien, A = 4i+ j, Pyhagrean herem is use esablish he lengh f he paren ecr y A= A + A = 4 + = 5. A Arcangen is use fin he irecin y θ = an = an = 7. Bu, wach u! The angle arrie a by he A 4 arcangen frmula may n be he final answer. The quaran ha he final ecr lies in mus be esablishe, an an ajusmen he angle may be neee in rer prie an answer ha eens frm he +-ais. A A Fig..a A = 5 7 A A y A y Reise 8/9/06 5 R H Jansen

5 Cmpnen Aanage In Vecr Aiin:. If ecr cmpnens alng an ais are use, irecin can be specifie wih + an symbls. Vecrs A an B in Fig. hae been cnere in cmpnens. i 4 j B = 4 i+ j A = ( + ) + ( + ) ( ) ( ). This becmes aanageus if ecrs A an B nee be ae. R = A + B = = Fin he resulan f he ecrs: ( ) ( ) Fin he resulan f he y ecrs: R A B ( ) ( ) = + = =+ y y y Then cmbine hem fin R: R R Ry ( ) ( ) Fin he irecin: R y ( + ) θ = an = an = 45 R ( ) = + = + + =.4, bu his a n quaran angle an mus be ajuse: = 5. The final ecr is has a magniue f.4 an a irecin f 5.. Fining he cmpnens simplify prblems hrughu physics. In Newnian Mechanics min, frce, an mmenum fen ac a an angle he - r y-ais. Frunaely hese ecr quaniies can be resle in cmpnen ecrs alng he - r y-ais. In aiin, equains fr min, frce, an mmenum can be calculae in he -irecin inepenen f wha is happening in he y-irecin, an ice ersa. The firs eample f his will be prjecile min. In prjecile min here will be a ariey f iniial an final f elciies a angles. Scalar (D) Pruc f Tw Vecrs: The pruc (A B) f w ecrs A an B is a scalar an is equal ABcsθ. This quaniy shws hw w ecrs inerac epening n hw clse parallel he w ecrs are. The magniue f his scalar is larges when θ = 0 (parallel) an when θ = 80 (ani-parallel). The scalar is zer when θ = 90 (perpenicular). Vecr (Crss) Pruc: The crss pruc (A B) f w ecrs A an B is a hir ecr C. The magniue f his ecr is C = ABsinθ. The irecin f his ecr is eermine by he righ-han-rule. The rer ha he ecrs are muliplie is impran (n cmmuaie) If yu change he rer f muliplicain yu mus change he sign, A B = A B. The magniue f his ecr is larges when θ = 90 (perpenicular). The ecr is zer when θ = 0 (parallel) r when θ = 80 (ani-parallel). Eamples: In he fllwing eamples ecr A = an B =. The irecin f ecr A will ary. Vecrs A & B Scalar (D) Pruc Vecr (Crss Pruc) ( )( ) cs0 = 6 ( )( ) sin0 = B A + +4 Fig. ( )( ) cs0 = 5. ( )( ) sin0 =.0 ( )( ) cs60 = ( )( ) sin60 = 5. ( )( ) cs90 = 0 ( )( ) sin90 = 6 ( )( ) cs0 = ( )( ) sin0 = 5. ( )( ) cs50 = 5. ( )( ) sin50 = ( )( ) cs80 = 6 ( )( ) sin80 = 0 Anher way: When cs (parallel) appears in a frmula yu nee w ecrs ha are parallel (+) r ani-parallel ( ). Jus use yur rig skills fin a cmpnen f ne f he ecrs ha pins in he same irecin as he her ecr. In he eamples abe, fin he cmpnen f ecr B ha is parallel ecr A. When sin (perpenicular) appears in a frmula, fin he cmpnen f ecr B ha is perpenicular ecr A. Reise 8/9/06 6 R H Jansen

6 Min in Tw Dimensins Relaie Velciy: Min in w imensins, bh a cnsan elciy. A cmmn eample is ha f a ba crssing a rier. In figure., a ba leaes perpenicular he shre wih a elciy f B. The riers curren, R, carries i wnsream a isance y. In figure. he ba aims a an angle f θ upsream in Curren ming wnsream Curren ming wnsream rer en up sraigh acrss. There is a riangle frme by he sli elciy ecrs, an anher frme by he ashe isplacemen ecrs. These riangles are B similar riangles. The resulan elciy f he ba will be a cmbinain f he bas θ R wn elciy an curren. This ecr is Σ y labele Σ. This is hw fas he ba will y B appear be ming as seen frm a R θ sainary bserain pin n shre. When Σ calculaing he ime crss he sream use Fig. Fig. he elciy ecr an isplacemen ecr ha pin in he same irecin. In figure., B =. In figure., = Prjecile Min: Min in ne imensin inles accelerain, while he her is a cnsan elciy. In he -irecin he elciy is cnsan, wih n accelerain ccurring in his imensin. In he y-irecin he accelerain f graiy slws upwar min an enhances wnwar min. Vecr Cmpnens in Prjecile Min: The -irecin an he y-irecin are inepenen f each her. Fig. Nw 0 is a an angle. 0 +θ y + -θ - y Sle fr 0 : 0 0csθ Sle fr 0y : 0 0sinθ y = hen use 0 in he kinemaic equains sle fr. = hen use 0y in he falling by equains sle fr y. an y are cmpnen ecrs. T fin, use Pyhagrean Therem y = + an arcangen θ = an The highes pin in he fligh: = 0 an y = 0. If he prblem ene here hese cniins wul apply. y Hriznal Launches The launch angle is θ = 0 0 0cs 0cs 0 = θ = = 0 0 = 0 = sinθ = sin 0 = 0 0 = 0 0y 0 0 The abe mah is n really necessary. Inspecin f he Fig.4 shws ha 0 is + irece sraigh wn he -ais wih n y-cmpnen ecr isible a all. The en f he -θ prblem is similar he prblem epice in Fig., abe. Fig.4 Crinae Ais Sysem pries he necessary rienain hanle he fllwing ariables an heir apprpriae signs: launch angle, iniial elciies in & y, final elciies in & y, final laning heigh, an final erall elciy. Orienain maers an hus he crinae ais becmes a pwerful l, as epice n he ne page. y y Reise 8/9/06 7 R H Jansen

7 Iniial Launch + 0, + 0y During he prblem (A he p = an y = 0) Final Laning = 0 + 0y + 0, + 0y = 0 y + - 0y 's - y + 0, - 0y Fig.5 Iniial isplacemen = y = Falling bies: θ = ±90 0 = 0cs 90 = 0 0 0sin 90 y = = 0 Hriznal launch: θ = 0 0 0cs 0 = = 0 0y = 0sin 0 = 0 s quaran launch: +θ 0 = 0csθ will be + 0y = 0sinθ will be + 4 h quaran launch θ 0 = 0csθ will be + 0y = 0sinθ will be Fig.6 If i lans a he same heigh as i sare (y = y 0 ), hen up = wn. There are w s fr eery y. The shrer is fr he upwar rip. The lnger is fr he wnwar rip. Sle fr maimum heigh w ways. Frm grun up where y = 0. y = 0y + g( y y0). Or he easy way. Sar a he p an preen i is a falling by. esn maer since ime is cnrlle by he y-irecin. An a he p y is zer. Hweer, his sles fr half f he al fligh. y = g Mus uble ime! Fig.7 + Always y Lans lwer han y 0 y = 0 Lans same heigh as y 0 + y Lans higher han y 0 a = 0 N a in he -irecin = 0 Wha is i ing a he en f he prblem in he y-irecin? ± y I is usually ming wnwar a he en f he prblem. S y is usually negaie The final mus be resle. = + y If y= y0 = & = 0 Prjecile Min Sraegies. Hriznal Launch: Since 0y = 0 an 0 = 0, hen use y = g an = 0.. When ime () r range () is gien: Sar wih = 0 an hen y = y g. y. N an n : Time is he key falling by & prjecile prblems. Tw sraegies are useful when ime is missing. y 0 y s y = y g y n Quaraic Equain = r 0 s g( y y ) n y 0 y r 0 y = + sle fr y which is usually negaie. 0y 0 = + g use - y frm abe ge. = use frm abe sle fr range. Oher Prjecile Min Facs Any w launch angles ha a 90 will arrie a he same laning sie if fire n leel grun. Eamples: 5 an 75, 0 an 60, an 40 an 50. Maimum range (maimum isance in he -irecin) is achiee by launching a 45 ( = 90 ). Maimum aliue (maimum isance in he y-irecin) is achiee by firing sraigh up, a 90. Circular Min Frequency: Hw fen a repeaing een happens. Measure in reluins per secn. Peri: The ime fr ne reluin,. Time is in he numerar. T = f Velciy: In unifrm circular min he magniue (spee) f he bjec is n changing. Hweer, he irecin is cnsanly changing, an his means a change in elciy (a ecr cmpse f bh magniue an irecin). In circular min ne can escribe he rae f min as eiher a spee r as a angenial Fig.8 elciy π r =. This elciy is an insananeus elciy an i is irece angen he cure. T Cenripeal Accelerain: The bjec is cninually urning war he cener f he circle, bu neer ges here ue is angenial elciy. This cenripeal (cener seeking) change in elciy, is a cenripeal accelerain ac = r Reise 8/9/06 8 R H Jansen a c

8 AP Physics C Pracice Prblems: Vecrs Muliple Chice Quesins. The cmpnens f ecr A are gien as fllws: A = 0.5 Ay = 5. Wha is he magniue f he ecr? A. 0.5 B. 5. C. 8.5 D. 5.7 E The cmpnens f ecr A are gien as fllws: A = 5.6 Ay = -4.7 Wha is he angle beween ecr A an psiie irecin f ais? A. 0 B. 80 C. 90 D. 7 E. 0. The cmpnens f ecrs A an B are gien as fllws: A = 5. B = -.6 Ay = -5 By = -4. Wha is he magniue f ecr sum A + B A. 5. B..5 C. -9. D. 9.6 E The magniue f ecr A is 8.6. Vecr lies in he furh quaran an frms an angle f 7 wih he -ais. Wha are he cmpnens f ecr A? A. A = 8.6 Ay = -8.6 B. A = -6.9 Ay = 5. C. A = -6.9 Ay = -5. D. A = 6.9 Ay = 5. E. A = 6.9 Ay = -5.

9 5. Fin he magniue f ecr C = A - B. Use all he infrmain presene by he graph. A. 5.7 B. 6.9 C. 7.4 D. 8.6 E Fin he pruc f w ecrs A B. Use all he infrmain presene by he graph. A. 8.6 B..5 C. -.6 D E Tw ecrs are gien as fllws: A = -i 5j + k B = -4i j k Wha is he angle beween he ecrs? A. 4 B. 67 C. 4 D. E. 94

10 Vecrs A an B are shwn. Vecr C is gien by C = B - A. Please refer his figure fr prblems The magniue f C is clses a).9 b) 5.9 c) 6.8 ) 7.7 e) The angle, measure frm he -ais ecr C, in egrees, is clses : a) 0 b) 4 c) 67 ) 70 e) 8 0. The cmpnens f ecr Z are gien as fllws: Z = 0.7 Z y = 8. Wha is he magniue f he ecr? a) 7.8 b) 9.5 c) 4. ) 6 e).6. The cmpnens f ecr Q are gien as fllw: Q =.5 Q y = 8.6 Wha is he measure f he angle, in egrees, ha he resulan ecr makes wih he -ais? a) 8.4 b) 47.9 c) 56. ) 6 e) 74.7

11 . The cmpnens f ecrs U an V are gien as fllw: U = -8.6 V =0.7 U y = 9.4 V y = 4. Wha is he magniue f he ecr sum U + V? a) 9.8 b).7 c) 4.6 ) 5. e) 6.9. Which f he fllwing saemens is rue? a) A scalar quaniy can be ae a ecr b) I is pssible fr he magniue f a ecr equal zer een hugh ne f is cmpnens is nn-zer c) Scalar quaniies are pah epenen, while ecrs are n. ) Scalar quaniies an ecr quaniies can bh be ae algebraically e) A scalar cnains magniue an irecin while a ecr es n. Quesins 4-6: Tw ecrs are gien as fllws: A = i + 6j 5k B = i + j + k 4. The ecr pruc A B equals: a) - b) 0 c) 4 ) 9 e) 0 5. The ifference beween ecrs A an B is: a) i + 9j 4k b) i + j 6k c) i + j 6k c) 5i + 9j 4k ) 6i + 8j 5k 6. The magniue f he sum f he ecrs A an B is ms nearly: a) 6.8 b) 7.4 c) 9.0 ) 0.4 e)

12 7. The cmpnens f ecr E are as fllws: E = -4.8 E y = -.6 Wha is he measure f he angle, in egrees, frme by ecr E an +-ais? a) b)4. c) 4. ) 45.9 e) 95.7 Vecrs A an B are shwn. Vecr C is gien by C = A + B. Refer his figure fr prblems Wha is he magniue f ecr C? a) 4.7 b).9 c) 4. ) 6.7 e) 7. Three Vecrs are gien as shwn. Refer his figure fr numbers 9-.

13 9. In he figure abe, he magniue an irecin f he ecr pruc A B are clses : a) 0, irece u f he plane. b) 0, irece in he plane. c), irece u f he plane. ), irece in he plane. e), irece n he plane. 0. The magniue an irecin f he ecr pruc C B are clses a), irece in he page. b), irece u f he page. c), irece n he plane. ) 9, irece in he page. e) 9, irece u f he page.. The scalar pruc f A B is clses : a) 5 b) 0 c) 7 ) e) 5. Tw ecrs are gien as fllws: A = -i 5j + k B = -5i j k Fin he magniue f he fllwing ecr: A B. A. B. 4 C. 8 D. 6 E.

14 Answer Key.. C. A. D 4. E 5. D 6. C 7. B 8. D 9. A 0. E. A. B. C 4. D 5. B 6. E 7. A 8. C 9. D 0. E. A. E

15 AP Physics C Muliple Chice Pracice Kinemaics An bjec slies ff a rf 0 meers abe he grun wih an iniial hriznal spee f 5 meers per secn as shwn abe. The ime beween he bjec's leaing he rf an hiing he grun is ms nearly (A) s (B) s (C) s (D) s (E) 5 s Quesins - A ime = 0, car X raeling wih spee 0 passes car Y. which is jus saring me. Bh cars hen rael n w parallel lanes f he same sraigh ra. The graphs f spee ersus ime fr bh cars are shwn abe. Which f he fllwing is rue a ime = 0 secns? (A) Car Y is behin car X. (B) Car Y is passing car X. (C) Car Y is in frn f car X. ( D) Bh cars hae he same accelerain. (E) Car X is acceleraing faser han car Y. Frm ime = 0 ime = 40 secns, he areas uner bh cures are equal. Therefre, which f he fllwing is rue a ime = 40 secns? (A) Car Y is behin car X. (B) Car Y is passing car X. (C) Car Y is in frn f car X. (D) Bh cars hae he same accelerain. (E) Car X is acceleraing faser han car Y. 4 A by ming in he psiie irecin passes he rigin a ime = 0. Beween = 0 an = secn, he by has a cnsan spee f 4 meers per secn. A = secn, he by is gien a cnsan accelerain f 6 meers per secn square in he negaie irecin. The psiin f he by a = secns is (A) +99 m (B) +6 m (C) -6 m (D) -75 m (E) -99 m 9

16 5 Which f he fllwing pairs f graphs shws he isance raele ersus ime an he spee ersus ime fr an bjec unifrmly accelerae frm res? Disance Spee Disance Spee (A) (B) Disance (C) Spee Disance (D) Spee Disance (E) Spee 6 An bjec release frm res a ime = 0 slies wn a fricinless incline a isance f meer uring he firs secn. The isance raele by he bjec uring he ime ineral frm = secn = secns is (A) m (B) m (C) m (D) 4m (E) 5 m 7 Tw peple are in a ba ha is capable f a maimum spee f 5 kilmeers per hur in sill waer, an wish crss a rier kilmeer wie a pin irecly acrss frm heir saring pin. If he spee f he waer in he rier is 5 kilmeers per hur, hw much ime is require fr he crssing? (A) 0.05 hr (B) 0. hr (C) hr (D) 0 hr (E) The pin irecly acrss frm he saring pin cann be reache uner hese cniins. 8 Vecrs V an V shwn abe hae equal magniues. The ecrs represen he elciies f an bjec a imes, an, respeciely. The aerage accelerain f he bjec beween ime an was (A) zer (B) irece nrh (C) irece wes (D) irece nrh f eas (E) irece nrh f wes 9 A prjecile is fire frm he surface f he Earh wih a spee f 00 meers per secn a an angle f 0 abe he hriznal. If he grun is leel, wha is he maimum heigh reache by he prjecile? (A) 5 m (B) 0 m (C) 500 m (D),000 m (E),000 m 0 A paricle mes alng he -ais wih a nncnsan accelerain escribe by a =, where a is in meers per secn square an is in secns. If he paricle sars frm res s ha is spee an psiin are zer when = 0, where is i lcae when = secns? (A) = m (B) = 6m (C) = 4 m (D) = m (E) = 48 m 0

17 AP Physics C Free Respnse Pracice Kinemaics 98M. A paricle mes alng he parabla wih equain y = shwn abe. a. Suppse he paricle mes s ha he -cmpnen f is elciy has he cnsan alue = C; ha is, = C i. On he iagram abe, inicae he irecins f he paricle's elciy ecr an accelerain ecr a a pin R, an label each ecr. ii. Deermine he y-cmpnen f he paricle's elciy as a funcin f. iii. Deermine he y-cmpnen f he paricle's accelerain. b. Suppse, insea, ha he paricle mes alng he same parabla wih a elciy whse -cmpnen is gien ½ by = C/(+²) i. Shw ha he paricle's spee is cnsan in his case. ii. On he iagram belw, inicae he irecins f he paricle's elciy ecr an accelerain ecr a a pin S, an label each ecr. Sae he reasns fr yur chices. 5

18 Applicains f he Deriaie The ms cmmn applicain f he eriaie ha we use in AP Physics is he psiin elciy - accelerain applicain. (By he way, in AP Physics we use represen psiin, n, which is wha we use his pas year.) Since elciy is he slpe f a psiin s ime an accelerain is he slpe f elciy ersus ime, hen we can say (in calculus-speak ) ha elciy is he eriaie f psiin an accelerain is he eriaie f elciy. There are a bunch f her uses fr he eriaie in AP Physics, bu we ll ge hem uring he schl year. S, fr he ime being, Gien: () psiin as a funcin f ime Then: / = elciy an / = accelerain Here s a ip ha may help yu remember wheher r n use a eriaie: If a quaniy can be epresse as a quien in is basic frm, hen ha same quaniy can be fun by aking a eriaie. Fr eample, = / an I = q/ (basically) Then, in he wrl f calculus, = / an I = q/ 6

19 Eample A paricle s psiin is gien by: = 4 -, where is in meers an is in secns. (a) Fin he elciy f he paricle a ime = secns. (b) Fin he ime a which he paricle sps mmenarily (c) Fin he paricle s accelerain a ime = 6 secns. Sluin: = / = - an a = / = 4 (a) = () - = 06 m/s (b) if = 0, hen 0 = = = /6 = 0.4 secns (c) a 6 = 4(6) = 44 m/s 7

20 Prblem Se. The isance in meers ha a paricle can me in ime secns is gien by: = A wha ime es he paricle sp ming?. A paricle prjece erically reaches an eleain, in meers, gien by: y = 50-5, where is in secns. Hw fas is he paricle ming when i reaches an eleain f 75 meers?. A paricle s psiin in meers is gien by : = -, where is in secns. (a) Fin he paricle s elciy a ime 5 secns. (b) Fin he paricle s accelerain a ime secns. 4. A paricle s psiin in meers is gien by: = - 4, where is in secns. Fin he paricle s accelerain a each insan fr which he spee is zer. 8

21 5. A paricle s psiin is gien by: = - Fin he fllwing: (a) The aerage spee beween = 0 secn an = secns. (Ne: ag = / always, fr any min) (b) The insananeus elciy a = secns (c) The insananeus spee a = secns () The aerage accelerain beween = 0 secns an = secns (Ne: a ag = / always, fr any min) (e) The insananeus accelerain a = secns (f) Skech a graph f s by ing he fllwing: Fin when = 0 (where he graph sars) Fin when = 0 (where crsses he ais) Fin where he slpe (he eriaie!) = 0 9

22 (his ells yu where he graph urns arun ) 0

23 AP Physics C Essenial Calculus Eamples f he calculus skills ienifie n he Cllege Bar AP Physics C Equain Shee. Differeniain - Pwer Rule ( ) n = n n Eample # Gien he funcin, y = y + + g Fin he firs eriaie, ( y) + ( ) y = y y = g y = + g = + g ( ) + g Fin he secn eriaie, ( ) = = 0 + g = g a = g Eample # un + ( g ) ( ) = nu n u.. Gien = a = a ( ) = u u ; u = a ; = a ( ) un =! u $ # & = " % u a ( ) = a u = a a = a

24 Differeniain Chain Rule f = f u u Eample # Gien he funcins = a () an = a (). Fin he accelerain, a =. Frm he equains gien i is bius he accelerain is a. Nw wrie an equain fr he elciy as a funcin f hriznal psiin. Sle equain () fr ime an subsiue in equain ().! = a # " a $ & = % a Again fin he accelerain, a =. Since is n a funcin f ime, apply he chain rule. a = = = = a ( ) a = a a a = =! # " a $ & a % a = a Differeniain Trig Funcins ( sin ) = cs ( cs ) = sin ( an ) = sec The eriaies f ALL rig funcins saring wih c are negaie. u ( sinu) = csu ( csu) Eample Fin he firs eriaie ( A sin( ω + φ) ) = sinu u Le u u = ω + φ an = ω ( sinu) = A ( sin( Aω + φ) ) = A cs( Aω + φ)( ω) = Aωcs( Aω + φ)

25 Differeniain The Pruc Rule Always use he pruc rule; neer he quien rule. ( u) Eample = u Differeniae + f = u + Epress f as pruc; ( ) f = + Le u = an = ( + ) ( f) = ( u) = u + u ( f) = ( u) = ( + ) ( ) + ( + ) ( f) = ( u) ( f) = ( u) ( f) = ( u) = = = + ( ) + ( + ) ( + ) ( + ) + ( + ) ( + )

26 Differeniain e ( ) e = e ( e ) u = Eample e Differeniae k ( e ) u = e u k e Le u k u u ( e ) k k = e ( k) k k ( e ) = ke Differeniain ln ( ln) ( lnu) = = u u u = an = k

27 Inegrain Pwer Rule n = n + Eample # + n+,n + k = k = k This is an inefinie inegral. Prblems ypically require limis f inegrain. k = k = k k = 9k 8 k 9 = k Eample # n n+ u u = u,n n + Le u = + an u = + ( ) ( + ) = + + ( + ) = ( + ) = ( + ) Eample # ( ) + Le u = + an u = Frm he equain u = ; nee u = ; missing. Muliple by. = ( ) = = + ( + ) ( + ) 4( + )

28 Inegrain Trig Funcins sin = cs cs = sin Inegrain e e = e u e u = e u Eample # k ke Le ke k = u = k an u = k u e u = e k Eample # k e Le u = k an u = k Frm he equain u = ; nee u = -k; missing -k. Muliple by. e k k = k k ke k = k e k

29 Inegrain ln = ln Eample # Inegrae where represens elciy = ln This is an inefinie inegral. Prblems ypically require limis f inegrain. i f Recall = ln e ln = f i = ln f ln i = ln f i u = lnu u Eample # kq Inegrae where q represens charge + kq ( ) Le u = + q an u = kq kq ( kq) = + u = ln+ q u Eample # q Le k q ( ) u = q an u = q Frm he equain u = q; nee u = -q; missing -. Muliple by. ( k q) ( ) q q = = lnk q This is an inefinie inegral. k q Prblems ypically require limis f inegrain. qf q qf = lnk q 0 = ln k qf ln k 0 0 k q ( ) Recall e ln = e ln = k q k ( ) ( ( )) = [ ln( k q ) ln( k) ] = ln f

30 Differenial Equains B f ( ) + Cf ( ) = 0 B an C are cnsans Rearrange s he secn eriaie erm has n cefficiens. f ( ) + C B f ( ) = 0 Suppse he cnsans B an C equal. f ( ) + f ( ) = 0 Wha funcin when iffereniae wice an ae iself equals zer? f ( ) = sin( ) OR f ( ) = cs( ) Nw le B an C f ( ) = sin( ) f ( ) = cs( ) Le f ( ) = sin( ω) an f ( ) = ω sin( ω) Subsiue he abe equains in he general frm f he ifferenial equain. f ( ) + C B f ( ) = 0 ω sin( ω) + C sin( ω) = 0 B Sle fr ω ω = C B ω = C B Final frm f he sluin. f ( ) = sin! # " C $ B & OR f % ( ) = A sin! # " C $ B & %