Chapter 2. Motion along a straight line. 9/9/2015 Physics 218


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1 Chper Moion long srigh line 9/9/05 Physics 8
2 Gols for Chper How o describe srigh line moion in erms of displcemen nd erge elociy. The mening of insnneous elociy nd speed. Aerge elociy/insnneous elociy Aerge ccelerion/insnneous ccelerion How o use equions nd inerpre grphs of posiion s ime; elociy s ime nd ccelerion s ime for srigh line moion. How o sole problems for srigh line moion wih consn ccelerion, like free fll. How o nlyze moion in srigh line when he ccelerion is NOT consn. 9/9/05 Physics 8
3 Displcemen Vecor displcemen 9/9/05 Physics 8 3
4 Aerge Velociy erge 9/9/05 Physics 8 4
5 Insnneous Velociy Noe: his corresponds of lim 0 he posiion ersus d d o he slopeof ime grph he ngen o hecure 9/9/05 Physics 8 5
6 Figure.7
7 Figure.8  grph Moion Digrm
8 Aerge nd Insnneous Accelerion 9/9/05 Physics lim 0 nd in he limi where d d d d d d d d e
9 Moion wih consn ccelerion 9/9/05 Physics 8 9 e e e 0 (for consn cclerion only) 0 hen his becomes for ) (
10 Remembering h he re under cure is reled o he inegrl of h funcion 0 e 9/9/05 Physics 8 0
11 Using inegrl noion In generl for consn ccelerion we know d d d d 0 9/9/05 Physics 8
12 Ne sep o find he posiion 9/9/05 Physics ) ( ) ( for consn ccelerion we know In generl d d d d d d
13 For consn ccelerion In generl ( ) ( ) for consn ccelerion we know /9/05 Physics 8 3
14 Noe: Freely Flling Bodies Uniform ccelerion of griy ner he erh s surfce. 9.8 m/s So we cn re free fll s cse of consn ccelerion. (This will include moion up nd down n inclined plne wihou fricion.) 9/9/05 Physics 8 4
15 An emple ) Find he posiion nd elociy of he moorcycle rider.0 sec. b) Where will she be when her elociy is 30 m/s? 9/9/05 Physics 8 5
16 Pr A) 3.0 sec (3.0m/s 6.0 m 0 0 )(.0 s) 0.0 m (3.0 m/s m/s(.0 s) 5.0 m )(.0 s) 0 m/s 3.0 m 6.0 m/s 5.0 m 9/9/05 Physics 8 6
17 Pr B) 3.0 where will she be when her elociy is 30.0 m/s? so use his ime o now find (3.0m/s 66.7 m 0 0 (3.0 m/s ( ) / 3.0 )(6.67 s) 66.7 m 0 )() 0 m/s 6.67 s, 30.0 m/s 0.0 m/s(6.67 s) 5.0 m 38.4 m 5.0 m 9/9/05 Physics 8 7
18 Anoher emple wih wo moing objecs Suppose h our preious emple, he moorcyclis is rying o cch up wih friend reling in cr long he sme highwy, bu who lef he sring poin 30 sec before her nd ws reling consn elociy of m/s. 9/9/05 Physics 8 8
19 We need o find he ime where he posiion of he wo objecs re so we mus sole for he ime in his epression moorcycle cr (3.0m/s he sme. (3.0m/s 0.0 m/s () m )() 0 )() m/s() 5.0 m 0.0 m/s() 5.0 m.0 m/s () m 9/9/05 Physics 8 9
20 Soling qudric equion of he form : () b() c 0 b b 4c 9/9/05 Physics 8 0
21 Soling (3.0m/s qudric equion of )().0 ( 3 4( ) 3 )( 355) heform :.0 m/s() m 0 6., 4.7 s 9/9/05 Physics 8
22 Grphicl soluion (posiion s ime) 900 Wh is he significnce of he TWO soluions?? cr /9/05 Physics 8
23 Now for freely flling bodies Remember his is cse of consn ccelerion when deling wih moion ner he surfce of he erh. 9/9/05 Physics 8 3
24 Figure.4
25 How high will he bll rise fer being hrown upwrd? How long will i ke o rech he op of he rjecory? 9/9/05 Physics 8 5
26 Our woequions y y y y for elociy nd posiion re : 0 y y y /9/05 Physics 8 6
27 Figure.5
28 Clculus Summry The Deriie of lim 0 f ( ) f ( funcion, ) df d f() is where defined s follows : Grphiclly his corresponds o he slope of he ngen o he cure f() he poin. 9/9/05 Physics 8 8
29 For simple polynomilfuncion like f ( ) df d for lim generl df d we cn crry ou his limiing 0 n polynomilof n f ( ) n process nd we find we find 9/9/05 Physics 8 9
30 The inegrl of funcion he definie inegrl We cn define second operion on hese polynomilfuncions inesiging he re under n rbirry cure. This is will cll he inegrl noe i wih he following symbol. b of he funcion f ( ) d wh we f ( ) from ob, nd we will 9/9/05 Physics 8 30
31 Figure.8
32 The re f ( ) b Then for b f ( ) d under simple polynomilcure like m b, "srigh line" cn be worked ou esily. f ( ) d m( b ) generl b k n d polynomilof n ( mb m)( b ) k n n he form, kb n m( b f ( ) k for n ny posiieineger n, ) 9/9/05 Physics 8 3
33 The indefinie inegrl f ( ) d k n d n n k consn of inegrion 9/9/05 Physics 8 33
34 The connecion beween differeniion nd inegrion The operions of inegrion nd differeniion of when pplied in series d d o he sme funcion reurn you oh funcion. df ( ) f ( ) d f ( ) nd he d f ( ) d c funcion 9/9/05 Physics 8 34
35 Clculus summry The deriie of polynomil f ( ) df ( ) n kn d The inegrl of polynomil f ( ) nd f ( ) d df ( ) d n k n d f ( ) c c 9/9/05 Physics 8 35 k k n n
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