L%hh LHH` X% " ӠӠΠǠ ˮԠˠǠӬĠӠ٠٠ϠŮ+JJJJ ,,,L Lک ٥ҥ>ӥ? ( ㆭ  ؆` | A""""""QQQYV <"<Q '(h((QQQ`@GHOHO $"!$q}T@` `@@``p0 8p@@`00 |0@` `pppPox`D3 < p?|@@@|@`f ppn|>0p  qcs6<, 3['Nxxxpp0` ```@vl  ?`@:3q  AcqcCyO@  0p`@@q @p?|`3cC ? ?>|xp`p`@Acwsssssss__????????~^^ ?~~~~~~~?>~||xpp`@~| xyy{~|@`pp xpp`@yyyyyyyx~ ~x`|xpxqa` cqx|||||||cCC?>>>??<<|||~?`ppxx||<||xxpp`|||||||| |?|p`@@@@`p|? OOGgccqpxsssssssppxxx|<<~?>~||xxxxppp```@@@`?~|x|~|||||||p@@p|~><|xp`@>|x|xpx}yyyyyyyAA?~|xpp`ppx|~? ?? 33@``@@glloLL LlOgloloG @`@glolo@@gLLL GaamGGlg`GGlg`G  xL| x@@xMLLxMyyM MyxAxMx|Yq @lllLglllgG LlL`@GLOLo033366m~0>3>33~G<`G}rccb``acccq{{{{8008????~ |lL COoooog CgoooocCFFLLXXPpqa88`}}}}>????~ | ~FCCFL ____|GG|GGccp{{{{ ;1q``qp`aas{{{{8008 0y| |}}}}AAQg((h' $C@0j ed`y < I1@@q q s8EE9EE8> ` `$DC0ANPQN@'h(' WPG((H@stq q8AyEx:NPQ@GhHHG#$p# @ (*Y9AyEy: QQQO~C%%C~_DDD qAAAaq9`ppxxx|xxx|AsGN8088 CcˆʎõĵL õ ĵµ aµ`` L̦µ_bJLuLz`  ȟ QlXJ̥KlV  ȟ QlV eօ3L e3L &RL &QL d L4 Ne)n `@-eff L f`L . tQLѤ LҦL` OPu d L Ne)noon 8ɍ` ^f\õL ^NR  RΩLҦ)\Z ʽ LHv 3h`0h8` [L NС õ`A@` ŵL^L iõ`  \ 濭0 \  ȟ Q ^\lZl^?cqH şch`fhjõĵ@OAP`u@`@&`QR`E Ls  @DAE@u`8` %@ @A@`@`@A`Mµ ) LЦ`8@AWc@8@-@HAȑ@hHȑ@ȑ@hHȑ@Ȋ@ch8&ȑ@Hȑ@Ah@LHȑ@ȑ@ htphso`hMhL`9V8U897T6S67`INILOASAVRUCHAIDELETLOCUNLOCCLOSREAEXEWRITPOSITIOOPEAPPENRENAMCATALOMONOMOPRINMAXFILEFINBSAVBLOABRUVERIF!pppp p p p p`" t""#x"p0p@p@@@p@!y q q p@  LANGUAGE NOT AVAILABLRANGE ERROWRITE PROTECTEEND OF DATFILE NOT FOUNVOLUME MISMATCI/O ERRODISK FULFILE LOCKESYNTAX ERRONO BUFFERS AVAILABLFILE TYPE MISMATCPROGRAM TOO LARGNOT DIRECT COMMANč$3>L[dmx- ( t Ϡ@跻~!Wo*9~~~~ɬƬ~_ j ʪHɪH`Lc (L ܫ㵮赎 ɱ^_ J QL_Ls贩紎 DǴҵԵƴѵӵµȴ 7 ַ :ŵƴѵǴҵȴµ納贍﵎ٵ്ᵭⳍڵL^ѵ-I `  4 ò-յ!  8صٵ紭ﵝ 7L (0+BC  7L HH`LgL{0 HH` õL H hBL BH [ h`Lo õ ڬL B ڬ LʬH hB@ յյ [L (ȴ) ȴ 7L L ( L (ȴL{ƴѵ洩ƴǴҵ 7 ^* B0 HȱBh ӵԵ 8 L8 ݲ` ܫ  / / ED B / / ]ƴS0Jȴ ȴ)  紅D贅E B ƴ  / 0L Ν `HD٤DEEhiHLGh ` ŵBѵ-` ѵB-` ܫ XI볩쳢8 DH E𳈈췍Ȍ X0 · JLǵBȵC`,յp` 䯩 R-յյ`յ0` K R-յյ`ɵʵӵԵ` 4 K ( ѵҵLBȱBL8` DBHBH : ַ޵BȭߵBhhӵԵ RBܵmڵ޵ȱBݵm۵ߵ` 䯩LR˵̵ֵ׵`êĪLR E( 8` R` ELRŪƪ`췌 յյI뷭鷭귭ⵍ㵍跬ª 뷰` Lf ݵܵߵ޵ ^`8ܵ i B8` 4L ֵȱB׵ ܯ䵍൭嵍 ` DȑB׵Bֵ  ַ յյ`굎뵎쵬 뵎쵌``õĵBCõĵ`µµ`L õBĵCصص Qƴ0"Bƴ 󮜳` 0۰ϬBƴ8`i#`ЗLw!0>ﵭ` m ﳐ 7i볍 8 ЉLw`H h ݲL~ `浍국䵍뵩嵠Jm赍嵊mjnnn浈m浍浭m䵍䵐`"L ŵ8ŵH ~(` d ֠z# u`,155+JX270:Y84:9200:X132:Y9:9300:g1:135,80250,60:249,61243,64:249,61243,59:243,61X254:Y64:3:9000: 2:132,82102,146102,141:102,146106,142 X97:Y158:3:9100: (#X,YX,Y5:X1,Y6X3,Y6:X4,Y5X4,Y:X,Y":::"TYPE 1 TO REVIEW THIS PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A1230R::"PLEASE WAIT WHILE PART II IS LOADED.":20:15:"-LOADING-":D11500:\(4);"BLOAD CHAIN,A520":520"VECTOR 3"?@3:15,81266,81:133,12133,155:134,12134EES (S OF E)":@ "THAT'S THE ANSWER. WE'RE DONE!":24:10050z*:16304,0:3:135,82220,126:9800:9900:24:100504::8000:135,82220,126:9800:9900>22:" HERE IT IS, SIMPLIFIED."::10050LH::6:Z$;", YOU'VE FINISHED PART I. ************************":B#8'255:D11500:::(7):W#B'"";A$:10:%"T$1,2932:V132:V230:9020:C"$8,395:V13:V21:9040:n"%6,2732:26,31:25,30:26,33:25,34:"%10,315:6,11:7,12:4,11:3,12:"'"':100:8"'" ************************":"$'" VECTOR RESOLUTION"##.':"12,V24:V13,V21:V14,V2:b!d#V2,V24V1:V2,V24V12:V11,V13V22:V11,V2:V11,V24:n!#T021|!#T6,31T!#!#24,2710!#10,1427!#V132:V25:9000:!#27,17:27,21:27,25:27,29"#8,10:11,10:14,10:17,10:20,10:23,NU" ? '#p (#V1,V2:V1,V21:V11,V22:V11,V23:V12,V24:V13,V23:V13,V22:V14,V21:V14,V2: <#V1,V2:V11,V21:V12,V22:V13,V23:V14,V24:V1,V24:V11,V23:V13,V21:V14,V2:!P#V1,V2:V11,V21:V12,V22:V12,V23:V ONE"::"MORE TIME. IF YOU'RE STILL HAVING ":T"TROUBLE, SEE YOUR INSTRUCTOR."20:"TO REVIEW THIS PROGRAM, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A1250:10:"BYE FOR NOW, ";Z$;"!":X13:(7)::20:"END" X12500:::(4);"RUN MEUNDS":::ij" V(N) = V * COS(THETA)":" V(N) = 400 * COS(40)"::" V(N) = 306 POUNDS"::xt23:10050~:6:"HOW DID YOU DO? IF YOU GOT THE RIGHT"::"ANSWERS, YOU'RE DONE WITH THIS LESSON.":/"IF NOT, YOU SHOULD RUN THE PROGRAMRB"IF THE VECTOR IS 400 POUNDS AND"::"THE ANGLE THETA IS 40 DEGREES."::L" FIND V(P) AND V(N). THEN,":16:4:10050V:6:"HERE ARE THE ANSWERS:":: `" V(P) = V * SIN(THETA)":" V(P) = 400 * SIN(40)"::" V(P) = 257 POE COSINE?":::"ANSWER P OR N:";::B$8B$"N"3640$:(7):21:"SORRY. REMEMBER THAT THE COMPONENT":"NEXT TO THE ANGLE USES THE COSINE.".24:10050::36008::5:"GREAT, ";Z$;". YOUR'RE GETTING"::"THE IDEA! NOW SOLVE THE PROBLEM":CULAR TO THE AXES."" 24:100504 :20,2:20,3@ T013R .65T21,T3e .65T10,16Tk u 18,2 13,1629:27,2916 13,169:9,1116 :21:"FINALLY, THE COMPONENTS ARE FOUND."24:10050':21:"WHICH ONE USES TH N":"AXES WHICH ARE AT RIGHT-ANGLES."5 24:10050= 15M V122:V214W 9060 :21:"AND, HERE'S OUR ANGLE THETA."> 24:10050H 1R 15,25:13,22:11,19:23,25:25,22:27,19\ :21:"NATURALLY, WE NEED TO SKETCH LINES":"PERPENDI18,0& T124!0 .65T19,T': 5D 33,3425CN 22,2535iX 27,3235:27,3238:36,28:37,29p 3| T025 .65T4,25T  4,525 22,253 27,321:27,293:2,27:2,29& :21:"HERE ARE THE AXES WE'LL USE--THE P &E--THE STEPS ARE THE SAME."::q "*** AS YOU DO THIS EXAMPLE, TRY TO"::" PREDICT EACH STEP!!":24:10050z :2 2,3019:20,29:21,28:18,29:17,28 V118:V233:9000 :21:"HERE'S THE VECTOR WE'LL CONSIDER." 24:10050 12::" V(Y) = 5.74 METERS"::9 "IT'S EASY!":24:10050 :3:"LET'S LOOK AT ANOTHER COMMON SITUATION."::"WE'LL CONSIDER THE COMPONENTS OF": "A VECTOR WHEN THE AXES ARE NOT THE X-"::"AND Y-AXES. IT REALLY DOESN'T MAKE ANY": "DIFFERENC3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:g$X,Y6X4,Y6:X4,Y5:X3,Y5X,Y1:X,Y1:X,YX4,Y:%X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:':100:8'" ***********************":=$'" :X4,Y5X4,Y:X,Y3X4,Y3:u#X,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y:#X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y1:X4,Y5:(T$X,Y6X,Y4:X1,Y3X10,90229,90:21,110181,30XJ3:X224:Y102:9200:X49:Y10:9400:X170:Y27:9300:2:65,89200,76196,74196,79200,763:X206:Y81:9000: 2:65,89176,51169,51171,55176,51 3:X181:Y50:9100: (#X,YX,Y5:X1,Y6X3,Y6:"::1R" 'CROSS PRODUCTS - M/D FORM'"::V\"YOU'LL FIND IT HELPFUL.":::f"TYPE 1 TO REVIEW THIS PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A1250p::D13:(7)::20:"END"zD11500:::(4);"RUN MENU"?@1:61,361,156:AT IS 40 * 14 * COS(40)?":A*"ENTER A CORRECT ANSWER! ";L::s4"** THE CORRECT ANSWER IS 429 FT-LBS **"::>" DID YOU LEARN ANYTHING?":20:10050H:5:"THAT'S THE END OF THIS PROGRAM."::Z$;", YOU REALLY SHOULD WATCH"::"THE PROGRAM ?"::"ENTER THE NUMBER OF DEGREES: ";L9L40760:(7):"YOU'RE SLEEPING, ";Z$;"!"::"THE ANGLE IS 40 DEGREES.":770::"GREAT, ";Z$;"!"::"NOW SOLVE AND ENTER YOUR ANSWER: ";L :3L427L430820 (7):"NOPE. WHT'S GO ON.":24:10050l::"WE'LL DO ONE MORE TOGETHER. THIS TIME"::"WE'LL USE A VERTICAL PLANE."::"FIND . IF:"::" = 40 POUNDS DUE WEST":" = 14 FT, 40 DEGREES UP FROM WEST"::+"WHAT IS THE ANGLE BETWEEN &1,123135,123:135,81220,81:217,79217,83:;':100:8e'" ************************":$'" VECTOR ADDITION - PART II"::" ************************".'255:D11000:::(7):B'"";A$:D5,Y3:)%X2,YX,Y2X,Y4X2,Y6:J%X,YX2,Y2X2,Y4X,Y6:p%X,Y3X4,Y3:X2,Y1X2,Y5:H&221,126215,126219,121221,126:X226:Y130:9000:X238:9700:X242:9100:X172:Y92:9400:-&0:134,82134,126:5:133,82133,126:13:q#X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y1:X4,Y5:T$X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:$X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4,Y1:X1,YX3,Y:X1,Y3X9:243,61#X254:Y64:3:9000:S 2:132,82102,146102,141:102,146106,142m X97:Y158:3:9100:(#X,YX,Y5:X1,Y6X3,Y6:X4,Y5X4,Y:X,Y3X4,Y3:#X,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y";Z$;".":12:20:"GOOD BYE!":20:10050H:T13:(7)::20:"END"kX12500:::(4);"RUN MENU"q?@3:15,81266,81:133,12133,155:134,12134,155JX270:Y84:9200:X132:Y9:9300: 1:135,80250,60:249,61243,64:249,61243,5****************"/.'255:D1900:::(7):DB'"";A$:D9000:X238:9700:X242:9100:X172:Y92:9400:&0:134,82134,126:5:133,82133,126:131,123135,123:135,81220,81:217,79217,83:':100:8'" ************************":$'" VECTOR ADDITION - PART I"::" ********6:X2,Y2X2,Y:f$X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:%X2,YX,Y2X,Y4X2,Y6:%X,YX2,Y2X2,Y4X,Y6:%X,Y3X4,Y3:X2,Y1X2,Y5:0H&221,126215,126219,121221,126:X226:Y130:3X4,Y3:`#X,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y:#X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y1:X4,Y5:T$X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y" -12 - 4 + 3"::255:840:"HERE'S ONE LAST DOT PRODUCT. SOLVE IT!":::"FIND . IF = 2 + 3 - 1":::" AND = 3 + 1 + 2"::"WHEN YOU'VE SOLVED THE PROBLEM, ":10050::"ANSWERJ> + 4":" -3 + 2 - 2":" -12 + 8 - 8":z"X = 6(0) - 4 + 4<-J>":" -3<-K> + 2(0) - 2":" -12 + 8<-I> - 8(0)":-"X = -2 - 8 - 4": = -10 - 16 - 1":|R:"TO SEE THE COMPLETE SOLUTION, INPUT 1."::"IF YOU WERE CORRECT, INPUT 2 (1 OR 2)?";A\A2910f:"SOLUTION:"::" = 2 - 1 - 4":" = 3 - 2 + "::3^p"X = 6 - 4X IF = 2 - 1 - 4"::" AND = 3 - 2 + 2"::>"WHEN YOU'VE SOLVED THE PROBLEM, ":10050H::"ANSWER: X 3 + 1":" = 4 + 6 - 5"::D13000::3". = 8 + 12 - 10":" - 12 - 18 + 15":" + 4 + 6 - 5": ". = 8 - 18 - 5 = -15 (A SCALAR)":::255&*"DO> + 15<-I> - 5(0)":e"X = 6 - 15 - 4":" -10 - 12 - 12":"X = -9 - 14 - 24":::255:"THAT'S IT!"::10050= :"NOW WE'LL SOLVE THE SAME PROBLEM":"FOR .:"::" = 2 -> + 1"::D11200::3"X = 8 - 12 + 4":" +12 - 18 + 6":" -10 + 15 - 5":"X = 8(0) - 12 + 4<-J>":" +12<-K> - 18(0) + 6":" -10 = 2 - 3 + 1"::" = 4 + 6 - 5":"WRITE IT DOWN & FIND X ."::10050:"HERE WE GO!"::" = 4 + 6 - 5":" = 2 - 3";A$: :X4,Y1:X4,Y5:XT$X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:$X,Y6X4,Y6:X4,Y5:X3,Y5X,Y1:X,Y1:X,YX4,Y:%X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:=%X,Y5:X1,Y4:181:Y50:9100:P(#X,YX,Y5:X1,Y6X3,Y6:X4,Y5X4,Y:X,Y3X4,Y3:#X,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y:#X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:(4);"RUN MENU"?L@1:61,361,156:10,90229,90:21,110181,30J3:X224:Y102:9200:X49:Y10:9400:X170:Y27:9300:2:65,89200,76196,74196,79200,763:X206:Y81:9000: 2:65,89176,51169,51171,55176,51 3:XTHE DIRECTION IS NORTH! **":24:10050:6:"THAT'S ALL FOR NOW. SEE YOUR INSTRUCTOR"::"IF YOU'RE STILL HAVING TROUBLE."::"TYPE 1 TO REIVEW THIS PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A1320:D13:(7)::20:"END"D11500::"X = 62 * 15 * SIN(40) = 393 NT-M"::"WHAT IS THE DIRECTION OF X?"::"ENTER N, S, E, W, UP, OR DOWN: ";T$:T$"N"950(7):"INCORRECT. SINCE & ARE IN THE"::"X-Z PLANE, X IS EITHER +Y OR -Y.":&:" ** :T\"WHAT IS THE ANGLE BETWEEN & ?"::"ENTER THE ANGLE IN DEGREES: ";T:bfT25890p(7):"SLEEPING? TRY AGAIN."::860z:"GOOD, ";Z$;".":24:10050:3:"NOW SOLVE THE PROBLEM. ENTER YOUR"::"ANSWER WITHOUT UNITS. ";T/:DOT PRODUCTS - M/D FORM"::" ***********************"\.'255:D11000:::(7):qB'"";A$:D"TYPE 1. TYPE 2 TO CONTINUE: ";A:A1790:3:"HERE'S A ROTATIONAL PROBLEM:":::"A WHEEL, INITIALLY AT REST, IS GIVEN AN ""ANGULAR ACCELERATION. AFTER 10 SECONDS ":"IT IS ROTATING AT 2 RAD/SEC. WHAT WAS "; $"THE ACCELERATION? HOW MANY510::"#3 IS FINE, BUT #2 IS EASIER.":::10050q:" V = V(0) + A*T"::"SUBSTITUTING THE DATA--":" V = 0 + 3*5"::"THEN--":5::"V = 15 M/S (MOVING TO THE RIGHT)"::::"EASY! IF YOU WANT TO REVIEW THIS PROBLEM")7.5960V"WRONG, ";Z$;"."::" Y = 0 + 1/2*3*25 = 37.5 METERS"::10050:970"GOOD!":12::"Y = 37.5 METERS":::10050:3:"NOW FOR THE SECOND PART--TO FIND THE ":"VELOCITY. WHICH EQUATION DO YOU WANT?";A::A1990"O.K!";. BUT EQUATION #3 ":"NEEDS THE FINAL VELOCITY 'V', SO WE WILL"e"USE EQUATION #1:";:31:10050:" Y = V(0)*T + 1/2*A*T^2"::"NOW WE SUBSTITUTE THE DATA--":" Y = 0 * 5 + 1/2*3*T^2"::"WHAT'S THE ANSWER (WITHOUT UNITS)?";A::A3YOU THINK WOULD ":"BE MOST HELPFUL TO FIND THE DISTANCE: ";A::A1890Xp"GOOD!";z11::"#1 IS THE BEST";::31:10050:2:"SINCE WE'RE LOOKING FOR DISTANCE (X), WE":"WOULD LOOK AT EQUATIONS #1 & #3 (SINCE "?"#2 DOES NOT INCLUDE X):J>"NOW LET'S FILL IN THE CHART WITH WHAT WE":"KNOW...";:31:10050XH::8000R7:27:"5":9:19:"0":11:22:"+3"\14:"NOTICE THAT WE HAVE ONLY TWO UNKNOWNS! "::"NOW LOOK AT THE EQUATIONS. INPUT THE":Jf"NUMBER OF THE EQUATION :3:"NOW HERE'S OUR FIRST PROBLEM:":: "AN OBJECT, INITIALLY AT REST, IS GIVEN ":"A POSITIVE ACCELERATION OF +3 M/S^2. HOW"*"FAR HAS IT GONE AFTER 5 SECONDS? HOW ":"FAST IS IT GOING AT THAT TIME?"::4"COPY THE PROBLEM.";:31:10050::8200P"COPY THE EQUATIONS. WE'LL REFER TO THEM ":"BY NUMBER.";:31:10050:5:"THAT'S THE DISCUSSION ON METHOD. IF YOU ":"WOULD LIKE TO REVIEW IT BEFORE WE BEGIN " "TO SOLVE PROBLEMS, TYPE 1. TYPE 2 TO ":"CONTINUE: ";A:A1300(BEGIN ANY"::"CALCULATIONS, CHECK TO SEE THAT YOU HAVE""OBTAINED ALL THE INFORMATIION IN THE"::"PROBLEM STATEMENT. FOR EXAMPLE, A BODY":"'INITIALLY AT REST' MEANS V(0) = 0."::10050:3:"NOW WE'RE READY FOR THE EQUATIONS."::10050:TORS AND SCALARS QUIZ"::" ************************"[.'255:D11000:::(7):pB'"";A$: "V"ī7013+dMM1:(7):25:"ERROR!"::3e:](#::"THIS IS THE END FOR NOW. BYE!"m2#D12000:<#X13:(7):::21:"END"F#D11500:::(4);"RUN MENU"P#':100:8'" ************************":<$'" VEC**********"*#.'255:D11000:::(7):?#B'"";A$:S PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A1160P":T13:(7)::20:"END"s"D11500:::(4);"RUN MENU"y"?"'"':100:8"'" ************************": #$'" DOT AND CROSS PRODUCTS - UNIT VECTORS"::" **************":" = -2 + 5 - 1"w!::"ANSWERS:"::"

. = -25"::"

X = -7 + 0 + 14":!::"SEE YOUR INSTRUCTOR IF YOU'RE STILL"::"HAVING TROUBLE. THAT'S ALL FOR NOW! "!::100500":10:"TYPE 1 TO REVIEW THI6 + 9 - 3":" 2 + 3 - 1":" 4 + 6 - 2": ". = 6(1) + 3(1) -2(1) "::255:930"!:"FINALLY, HERE'S 2 FOR YOU."::"FIND

. AND

X IF:"::"

= 4 - 3 + 2: . = 7 (A SCALAR)"::s:"TO SEE THE COMPLETE SOLUTION, INPUT 1"::"TO CONTINUE, INPUT 2 (1 OR 2)?";AA2990:"HERE'S THE SOLUTION:"::" = 3 + 1 + 2":" = 2 + 3 - 1"::D11200::3g ". = "EXCELLENT!";[/14::"USE EQUATION #1":::"BECAUSE #2 AND #3 REQUIRE YOU TO KNOW":|/"THE FINAL VELOCITY.":::/"NOW YOU SOLVE FOR THE FINAL POSITION."::"INPUT YOUR ANSWER WITHOUT UNITS:";A/A161730B0::"VERY GOOD! THE POS THE PROBLEM AND FILL IN"::"THE CHART.":::10050C.T:2:8000o.^7:27:"8":9:18:"+10":11:22:"-3".h15:"BY NOW THIS SHOULD BE EASY!"::10050::.r"WHICH EQUATION ARE YOU GOING TO USE TO"::"FIND THE POSITION? ";A::3:A11670/|C-,:3:"HERE'S A PROBLEM WITH A NEGATIVE"::"ACCELERATION:"::-6"AN OBJECT WITH A VELOCITY OF 10 M/S IS ":"GIVEN A NEGATIVE ACCELERATION OF":-@"-3 M/S^2. WHERE IS IT AND HOW FAST IS"::"IT GOING AFTER 8 SECONDS?":::4.J"AS USUAL, COPY";A::A141560X,:"SOLUTION--"::" V = V(0) + A*T"::" V = 6 + 2*4":,6::"V = 14 M/S":::10050:1570,12:"OF COURSE! V = 14 M/S","20:"IF YOU WANT TO REVIEW THIS PROBLEM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A113604 OR -10 SECONDS"::10050::x+"NATURALLY, WE CHOOSE THE +4 SECONDS"::"ANSWER. YOU SHOULD VERIFY THE ANSWER BY":+"PUTTING IT BACK INTO EQUATION #1"::10050,:3:"WHAT IS THE FINAL VELOCITY? SOLVE"::"EQUATION #2, AND INPUT YOUR ANSWER: *2*T^2"::10050:\*:"NOTICE THAT THIS IS A QUADRATIC--"::" T^2 + 6*T - 40 = 0":*"SOLVE IT BY FACTORING AND INPUT YOUR"::"ANSWER (WITHOUT UNITS):";A*:3:"FACTORING--"::" (T + 10)(T - 4) = 0":+"FINALLY--"::" T = + EQUATIONS. WHICH ":"ONE IS BEST FOR FINDING THE TIME?";A:::A11440X)"SURE!";)11::"#1 CAN BE USED DIRECTLY.":::10050):3:"LET'S SOLVE IT--"::" Y = V(0)*T + 1/2*A*T^2":*"SUBSTITUTE THE DATA--"::" 40 = 6*T + 1/2"AN ACCELERATION OF +2 M/S^2. HOW LONG":Y(d"DOES IT TAKE TO TRAVEL 40 METERS?":::(n"TRY TO FILL OUT THE CHART. PRESS"::" TO CHECK YOUR WORK.";A$(x:2:8000(5:26:"40":9:19:"6":11:22:"+2"J)15:"O.K? NOW LOOK AT THE:6::"REVOLUTIONS = 1.59":::31:10050'F:10:"TO REVIEW THIS LAST PROBLEM, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A11040'P:3:"HERE'S ANOTHER PROBLEM. IT'S A LITTLE"::"MORE COMPLICATED:"::*(Z"AN OBJECT TRAVELLING AT 6 M/S IS GIVEN ":"FOR THE NUMBER OF REVOLUTIONS AND NOT":S&"NUMBER OF RADIANS?";:31:10050::&("REMEMBER HOW TO FIND REVOLUTIONS FROM"::"RADIANS?";:31:10050::&2" REVOLUTIONS = RADIANS/2*PI":::"THEREFORE--":*'<" REVOLUTIONS = 10/2*3.14":::" OMEGA^2 = OMEGA(0)^2 + 2*A*THETA":%" (2)^2 = (0)^2 + 2*0.2*THETA"::" THETA = 4/0.4 = 10 RADIANS"::10050:1300% 10:"VERY GOOD!";:12:"THETA = 10 RADIANS":::10050)&:3:"DID YOU NOTICE THAT THE PROBLEM ASKED"::"NOW FOR THE DISTANCE. WHAT EQUATION DO ":"YOU WANT TO USE? ";A::A21240Z$"O.K!";$10::"#1 OR #3 ARE GOOD.":::10050::$"YOU USE #3 AND SOLVE FOR THE DISTANCE."::10050$:3:"INPUT YOUR ANSWER: ";A::A101290+%::"NOPE!":"ALPHA. INPUT YOUR ANSWER (NO UNITS): ";A::A.21210o#3:"WRONG, ";Z$;"."::"HERE'S THE SOLUTION--":#" 2 = 0 + ALPHA*10"::6::"ALPHA = 2/10 = 0.2 M/S^2":::10050:1220#6:"GREAT! ALPHA = 0.2 M/S^2"::10050M$13:;!"~11::"#2 IS THE BEST"::}""WE CHOOSE #2, SINCE #1 AND #3 NEED THE"::"POSITION. REMEMBER THAT WE'RE WORKING IN"""THE ROTATIONAL SYSTEM, SO--"::" OMEGA = OMEGA(0) + ALPHA*T"::10050::9#"YOU SUBSTITUTE THE DATA AND SOLVE FOR": GET IT RIGHT, ";Z$;"?"::"IF NOT, THINK ABOUT WHAT YOU DID WRONG.":!`"ANALYZING YOUR ERRORS IS A GOOD WAY TO ":"LEARN. O.K?"::31:10050!j:3:"NOW SELECT THE BEST EQUATION TO USE FOR ":"FINDING THE ACCELERATION: ";A::A21150"t"YES." REVOLUTIONS ":"DID IT MAKE IN THE FIRST 10 SECONDS?":: .:"COPY THE PROBLEM AND TRY TO FILL IN THE ":"CHART YOURSELF. WHEN YOU'RE READY TO": 8"CHECK YOUR WORK, ";:10050 B:2:8000 L7:26:"10":9:19:"0";:27:"2"F!V15:"DID YOU::"T = 5 SEC"::::10050]:3:"INPUT THE INITIAL VELOCITY (NO UNITS):";A::A0810t "RIGHT, ";Z$;".";*25::"V(0) = 0 M/S":::"BECAUSE THE ROCK IS DROPPED, NOT THROWN."4:"NOW FILL IN YOUR CHART AND ";:10050>:8000 H17:SERVES THAT IT STRIKES THE":"GROUND 5 SECONDS LATER. HOW TALL IS THE":"CLIFF? (ALSO FIND THE IMPACT VELOCITY.)":::"WRITE DOWN THE PROBLEM AND ";:10050::"INPUT THE FINAL TIME (WITHOUT UNITS): ";A::A5780"GOOD, ";Z$;"."; 25'" ONE DIMENSIONAL KINEMATICS"::" ************************"j=.'255:D11000:::(7):=B'"";A$:=L'" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:X014<$9:10,129X:T1100:T:0:10,119X:F< $12:18,1419T<$X301<$9:10,1235(2XX2):T1100:T:0:10,1135(2XX2):<$9:10,1237<"$22:10060:A19200<,$<':100:8<'" ************************":K=$18,1412;#X04O;#7:10,12122XX2:T160:T:0:10,11122XX2:h;#22:10060:A19100n;#;#::9:10,120:21:"A NEGATIVE ACCELERATION ADDED TO A"::"CONSTANT VELOCITY. PRESS .";A$;#12:14,180<#T1800:::(7):0:10,11X22X:+:F#22:10060:A190001:P#:#::7:10,120:21:"A POSITIVE ACCELERATION ADDED TO A"::"CONSTANT VELOCITY. PRESS .";A$:#13:14,180:#T1800:::(7):X05:#7:10,122X:T1100:T:0:10,112X: ;#13: V - FINAL":::" ACCELERATION A - ACCELERATION":::10050::9(#::12:10,120:21:"AN OBJECT AT REST IS GIVEN A POSITIVE"::"ACCELERATION. PRESS TO START.";A$9-#7:14,18092#T1800:::(7):X05:<#12:10,12X22X:2 + 2*A*X"::10050::^8l :"HERE ARE THE VARIABLES:":::" POSITION X(0) - INITIAL":8v " X - FINAL":::" TIME T(0) - INITIAL":8 " T - FINAL":::" VELOCITY V(0) - INITIAL":H9 " " 2. DETERMINE KNOWN VARIABLES"::" 3. ORGANIZE DATA":7" 4. APPLY EQUATIONS--YOU MUST KNOW THEM!":" 5. SOLVE FOR UNKNOWNS"::10050::7 :"EQUATIONS:"::" 1. X = V(0)*T + 1/2*A*T^2"::8 " 2. V = V(0) + A*T":::" 3. V^2 = V(0)^ T * 0 * *":" ------*-------*-------*"6^" V * * *":" ------*-------*-------*"6h" A * *":" ------*---------------*"::6:" 1. ESTABLISH CONVENTIONS":87LATER, ";Z$;". BYE!"#5v20:10050C5:T13:(7)::20:"END"W5(4);"RUNMENU"]5?5@:" *INITIAL* FINAL *":" ------*-------*-------*"5J" X * 0 * *":" ------*-------*-------*"=6T" TE BEFORE IT":;4N"STOPS?":16:"WHAT'S YOUR ANSWER? ";Au4X20:"THE CORRECT ANSWER IS 2400 RADIANS.":24:100504b:6:"THAT'S THE END OF THIS PROGRAM. YOU"::"SHOULD ALSO WATCH:"::5l" KINEMATICS & GRAVITATIONAL ACCELERATION":::"SEE YOU REVERSED.":22:"TYPE 1 TO REVIEW THIS PROBLEM. TYPE 2"::"TO CONTINUE: ";A:A115803::3:"TRY ONE LAST PROBLEM ON YOUR OWN:":::"A WHEEL ROTATING AT 120 RAD/SEC IS GIVEN"4D"AN ACCELERATION OF -3 RAD/S/S. THRU HOW ":"MANY RADIANS DOES IT ROTA:3:"FINDING THE FINAL VELOCITY IS EASY."::"WHAT ANSWER DO YOU GET?";A:::A141820s2"GOOD, ";Z$;".":2"SINCE V = V(0) + A*T"::" V = 10 + (-3)*8":2&7::"V = -14 M/S":::"WHICH IS WHAT WE EXPECT IF THE OBJECT":V30"HAS 2":G1"SUBSTITUTE THE DATA--"::" X = 10*8 + 1/2*(-3)*8^2":s1"THEN--"::" X = 80 - 96":1"FINALLY--"::6::"X = -16 METERS"::1"WHICH MEANS THE OBJECT REVERSES AND"::"ENDS UP TO THE LEFT OF START.":110050\2ITION IS -16 METERS,"::"WHICH MEANS TO THE LEFT OF THE START.":0"SO, THE OBJECT HAS REVERSED DIRECTION."::10050:18000::"THAT'S NOT RIGHT. LET'S SEE WHERE YOU"::"WENT WRONG."::100501:3:"SOLUTION--"::" X = V(0)*T + 1/2*A*T^:10:"IF YOU WANT TO REVIEW THIS PROBLEM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A11090(:3:"HAVE YOU NOTICED THAT IF YOU'RE CAREFUL"::"WITH SIGN CONVENTIONS, YOU GET THE RIGHT"("ANSWERS? IT'S REALLY NEAT! ";:10050$)::"IF YOU CT THE"::"VELOCITY OF THE BALL AT THE TOP IS ZERO,"j'"AND USE EQUATION #3. LET'S DO IT."::10050::'" V^2 = V(0)^2 - 2*G*Y"::" 0 = 15^2 - 2*9.8*Y":'"OR,":" 225 = 19.6*Y":'"THUS,":8::"Y = 11.5 M"::::10050\(06 SECONDS":::10050n&Z::"HOW WOULD YOU FIND THE MAXIMUM HEIGHT"::"OF THE BALL? ";:10050:&d:"THERE ARE TWO METHODS. YOU COULD DIVIDE"::"THE TIME IN HALF AND USE EQUATION #1."::&n100504'x:3:"OR, YOU COULD RECOGNIZE THAITY! THIS IS TRUE":::"ONLY WHEN Y = Y(0)!! ",::28:10050q%2::"THEN, WE CAN USE EQUATION #2 DIRECTLY:":%<" V = V(0) - G*T"::10050%F:3:"SUBSTITUTING VALUES,"::" -15 = +15 - 9.8*T":&P" 9.8*T = 30"::11::"T = 3."::"WE COULD PUT THE POSITION, Y = 0, AND":$"THE INITIAL VELOCITY, V(0) = 15, INTO"::"EQUATION #1 AND SOLVE A QUADRATIC. BUT":$"AN EASIER METHOD IS TO RECOGNIZE THAT"::"THE FINAL VELOCITY WILL BE THE NEGATIVE":=%("OF THE INITIAL VELOC11Y2:#Y05.58#2:19,11Y2:0:19,11Y2:L#(7):2:19,36s#T11200::7:Y05:32,11Y2:#22:"TIME MARKERS SHOW THE ACCELERATION."::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11200.$ ::3:"SOLUTION TO FIND THE TIME:T.":/""NOW FILL IN THE CHART AND ";:10050"8000:14:27:"0":18:18:"+15":24:"NOW LET'S WATCH. ";:10050"::9000:2:19,36:22:" PRESS TO THROW THE BALL.";A$":T1700:"Y50.5#2:19,11Y2:0:19, VELOCITY: ";A::A1511401!j"YES, ";Z$;".";v!t25::"V(0) = +15 M/S":::"BECAUSE IT'S GOING UP.":::10050!~:3:"INPUT THE FINAL POSITION:";A::A01170!"EXCELLENT!";"25::"Y = 0 M":::"BECAUSE IT RETURNS TO THE INITIAL POIN ";A:A1730] B:3:"HERE'S ANOTHER PROBLEM:"::"A GIRL THROWS A BALL VERTICALLY INTO": L"THE AIR WITH A SPEED OF 15 M/S. HOW"::"LONG DOES IT TAKE TO REACH ITS STARTING": V"POSITION? COPY THE PROBLEM & ";:10050!`15:"INPUT THE INITIALNG,"::" (T + 5)(T - 4) = 0":X."AND,":10::"T = -5 OR 4 SECONDS":::8"OF COURSE, WE CHOOSE THE POSITIVE 4 SEC.":"IT'S NOT TOO HARD, ";Z$;"! ";:10050 =:10:"TYPE 1 IF YOU WANT TO REVIEW THIS"::"PROBLEM. TYPE 2 TO CONTINUE: Y = V(0)*T - 1/2*G*T^2":n" -98 = -4.9*T - .5*9.8*T^2"::" -98 = -4.9*T - 4.9*T^2"::10050:3:"REARRANGE:"::" 4.9*T^2 + 4.9*T - 98 = 0":"AND DIVIDE BY 4.9:"::" T^2 + T - 20 = 0"::10050:)$"FACTORI"ARE DOWN)!":::"COPY THE CHART IF YOU GOT IT WRONG."::10050:3:"NOW YOU SOLVE THIS PROBLEM USING THE"::"FIRST EQUATION. HINT: YOU'LL HAVE TO":"SOLVE A QUADRATIC EQUATION.":::"WHEN YOU'RE DONE, ";:10050:::"SOLUTION:"::" WHEN"::"WILL IT HIT THE GROUND IF HE THROWS IT":t"AT 4.9 M/S?":23:"TRY TO FILL IN THE CHART AND ";:10050:8000:4:26:"-98":8:17:"-4.9"15:"BE SURE TO SEE THAT THE (-) SIGNS ARE"::"NEEDED (BOTH VELOCITY & FINAL POSITION":BV = 0 - 9.8*5 = -49 M/S":::10050:5:"DO YOU SEE THAT -49 M/S MEANS THE ROCK"::"IS MOVING DOWN? GOOD!"::10050::"LET'S CHANGE THE PROBLEM A LITTLE."::"SUPPOSE A MAN THROWS A BALL DOWN FROM":4"THE TOP OF A 98 METER BUILDING.AL SOLUTION IS:"::7::"Y = -122.5 M"::z"WHERE THE (-) SIGN MEANS THE ROCK ENDS"::"UP BELOW ITS STARTING POSITION.":10050:::"TO FIND THE FINAL VELOCITY USE THE":"SECOND EQUATION:"::" V = V(0) - G*T"::10050(:7::"27:"5":19:19:"0":24:10050R:5:"FOR THE SOLUTION, WE USE EQUATION #1:"::" Y = V(0)*T - 1/2*G*T^2"::10050::\"SUBSTITUTING:"::" Y = 0*5 - .5*9.8*5^2":f"OR,"::" Y = 0 - .5*9.8*25"::10050:+p:3:"THE FIN"ACCELERATES FROM ZERO TO THE CONSTANT"::"VALUE AT 'A'.";:31:10050\ 34,0:133768:"NOW LET'S APPLY THESE IDEAS TO"::"A COMPLETE PROBLEM.":: "IF YOU WOULD LIKE TO REVIEW BEFORE WE"::"GO ON, TYPE 1. TYPE 2 TO CONTINUE: ";A:A11:100507:4:"0")34,9:133769:"NOW HERE'S A QUESTION: "::"WHAT IS THE VELOCITY AT THE ORIGIN?":"INPUT YOUR ANSWER: ";A::"SINCE THE SLOPE AT THE ORIGIN IS ZERO,"::"THE VELOCITY MUST BE ZERO. THE OBJECT":J2:X515:12,36X::$6':100:8N6'" ************************":6$'"KINEMATICS & GRAVITATIONAL ACCELERATION"::" ************************"6.'255:D11000:::(7):6B'"";A$:" T * 0 * *":" ------*-------*-------*"5^" V * * *":" ------*-------*-------*"5h" A * -9.8 M/S^2 *":" ------*---------------*"::5(#12:5,3437:5,3438:62#1EMS. BYE,BYE!":20:10050<4:D13:(7)::20:"END"_4D11500:::(4);"RUN MENU"e4?4@:" *INITIAL* FINAL *":" ------*-------*-------*"4J" Y * 0 * *":" ------*-------*-------*"E5TOBLEM. TYPE 2 TO CONTINUE: ";A:A1146023{19303:(7):(7):10:"OUTSTANDING WORK, ";Z$;"!"::"YOU GET HIGH MARKS FOR THAT.":22:100503:8:"THAT'S ABOUT ALL FOR NOW, ";Z$;".":::"SEE YOUR INSTRUCTOR IF YOU'RE STILL":4"HAVING PROBL"::" Y = 39.2^2/19.6":U2X" AND, "::15::"Y = 78.4 METERS"::`2b100502l:8:"DON'T FORGET TO ADD THE HEIGHT OF"::"THE BUILDING.":(3v6::"Y = 78.4 + 44.1 = 122.5 METERS"::20:"TYPE 1 IF YOU WANT TO REVIEW THIS"::"PRW SOLVE AND ENTER YOUR ANSWER: ";A:A122.51920A1&A78.41900s10:2:(7):"SORRY, ";Z$;". HERE IT IS--"::1:" EQUATION #3--"::" V^2 = V(0)^2 - 2*G*Y":1D" SUBSTITUTE--"::" 0 = 39.2^2 - 2*9.8*Y":'2N" THEN--EED TO SEE THE PROBLEM"::"AGAIN, 2 TO CONTINUE: ";A::A11460v0:"WHAT EQUATION WILL YOU USE? ";A::A318100"YES. THAT'S THE BEST ONE.":18200"NOTICE THAT YOU CAN'T DIVIDE THE TIME"::"BY 2 FOR THIS ONE. TRY EQUATION #3."01::"NOTLY. YOU MIGHT HAVE TO EMPLOY THE":^/"QUADRATIC FORMULA. BE SURE YOU KNOW IT!":::10050/:3:"ONE FINAL PROBLEM:"::"WHAT IS THE MAXIMUM HEIGHT ABOVE THE":/"GROUND REACHED BY THE ARROW IN THE"::"LAST PROBLEM?"::A0"TYPE 1 IF YOU NBY 4.9--"::" T^2 - 8*T - 9 = 0":\." FACTOR--"::" (T - 9)(T + 1) = 0":." FINALLY--"::9::"T = 9 OR -1 SECONDS"::1740.9:"TERRIFIC,";Z$;"! YOU'VE GOT"::"THE IDEA."$/16:"DON'T EXPECT ALL PROBLEMS TO FACTOR"::"NEA:"SORRY! HERE'S THE CORRECT SOLUTION:"::g-" EQUATION #1--"::" Y = V(0)*T - 1/2*G*T^2":-" SUBSTITUTE VALUES--"::" -44.1 = 39.2*T - .5*9.8*T^2":-" REARRANGE--"::" 4.9*T^2 - 39.2*T - 44.1 = 0"::10050*.:3:" DIVIDE SHOULD YOU USE TO FIND"::"THE TIME OF FLIGHT? (ENTER THE NUMBER):";A::A11640n,^"VERY GOOD, ";Z$;"!";,h25::"USE EQUATION #1":::10050,r:3:"OK, GO AHEAD AND FIND THE TIME."::"ENTER YOUR ANSWER WITHOUT UNITS:";A::A91730+-|(7)$+,T11200::Y06:13:31,Y2:+622:"TIME MARKERS SHOW ACCELERATION."::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE:";A:A11520+@::2:"BEGIN BY FILLING IN YOUR CHART:"::10050+J8000:8:25:"-44.1":12:17:"+39.2"S,T17:"WHICH EQUATION "LET'S WATCH. ";:10050*::9000:X515:12:17,36X::9:15,16:22:" PRESS TO SHOOT THE ARROW.";A$*:T1700::Y40.5*9:15,0Y2:0:15,0Y2:*Y06.5*9:16,Y2:0:16,Y2:*"(7):9:16,36AN STAND IT, WE'LL DO ONE MORE."::~)"A MAN SHOOTS AN ARROW UPWARDS WITH A"::"VELOCITY OF 39.2 M/S FROM THE TOP OF A":)"BUILDING 44.1 METERS TALL. IF THE ARROW"::"MISSES THE BUILDING AND HITS THE GROUND,"'*"HOW LONG IS IT IN THE AIR?"::5750X,Y3::CF#4:12:"10":2:19:"20";:26:"30";:33:"40"XP#3:35:"time":m#2:32,4932,103#1:Y06:33,499Y35,499Y:#7:3:"20":9:"V";:4:"0":11:2:"-20":#2:32,132,37#1:Y04:33,19Y35,19Y:(#176:10:"PLEASE WAIT A MOMENT WHILE PART IV IS"::"BEING LOADED FOR YOU."w>22:15:"-LOADING-":D12500::13376H(4);"BLOAD CHAIN,A520":520"GRAPHING 4"'#(#2:32,Y232,Y2#X14:3250X,Y13250X,Y3:<#1:X03:5750X,Y1IT LOOKS LIKE:";:D11000:982,49132,49:31:10050 34,0:13376:8:"WE'RE GETTING LOW ON MEMORY, ";Z$;"."::"SO, LET'S GO ON TO PART IV."::*"IF YOU WOULD LIKE TO REVIEW THIS PROGRAM":"TYPE 1. TYPE 2 TO CONTINUE: ";A:A1240J4133 t = 20 IS THE VELOCITY"::"AT t = 10 + CHANGE IN VELOCITY.":g" WHAT IS V AT TIME = 20? ";A::13376:"THE NEW VELOCITY IS 20. THAT'S NOT"::"SURPRIZING--VELOCITY IS CONSTANT WHEN": "ACCELERATION IS ZERO.";:31:10050::"HERE'S WHAT ;:31:10050f13376:"WHAT IS THE AREA UNDER THE ACCELERATION"::"CURVE FROM TIME = 10 TO 20?":" INPUT YOUR ANSWER: ";A:"THERE IS NO AREA. THUS, THERE IS NO"::"CHANGE IN VELOCITY (dV = 0).";:31:10050?13376:"THE VELOCITY AT****************"08'255:D11200:::(7):EB'"";A$: 5:%207,63232,24U%1:57,1057,16:107,0107,6:157,59157,67:207,59207,67%2:82,082,6:132,20132,28:182,80182,86:':100:8'" *************************":$'" GRAPHING MOTION - PART II".':" *********#19:4:"4":21:"A";:4:"0":23:3:"-4";:e%6:1:"X":3:23:"Position":4:23:"--------"%3:32,3357,13:X5782:X,310(82X)2625:%82,3107,3:X107132:X,320(X107)2625:%132,23157,63:X157207:X,8320(182X)2621:2:"100":6:4:"0":11:1:"-100";,#i#2:32,9732,133:14:13:"Velocity":14:14:"--------"#1:Y04:33,979Y35,979Y:#13:3:"20":15:"V";:4:"0":17:2:"-20":#2:32,14532,181#1:Y04:33,1459Y35,1459Y:03::9F#8:12:"10";:19:"20";:26:"30";:33:"40";NP#6:35:"time":Z#16:12:"10";:19:"20";:26:"30";:33:"40";d#n#23:12:"10";:19:"20";:26:"30";:33:"40";x##2:32,332,83#1:Y08:33,310Y35,310Y:&#GRAPHING MOTION - PART III'":20:1:"BYE FOR NOW, ";Z$;"!":24:10050l13376:D13:(7)::19:"END"{ D$;"PR#0"*D11500:::D$;"RUN MENU"'#(#2:32,Y232,Y2#X14:3250X,Y13250X,Y3:<#1:X03:5750X,Y15750X,Y---------":24:1:"THAT'S IT!";:31:10050?34,0:13376:10:"IF YOU WOULD LIKE TO REVIEW THIS METHOD,":"TYPE 1. TYPE 2 TO CONTINUE:";A:A143013376:10:"THAT'S ALL FOR NOW. YOU REALLY SHOULD"::"GO AHEAD AND WATCH:"::F " '-13376:16302,0:Y163:9000:9070:9170u5:33,16357,163:83,163107,163:133,163157,163:207,163233,16357,16357,17281,17281,163:107,163107,181131,181131,163:157,163157,145207,145207,163+19:7:"Acceleration":20:7:"---$:6">H THE":" SUPPORT OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">D16000::13376?H(4);"BLOAD CHAIN,A520":520"GRAPHING 2"E'#S':100:8}'" ************************":$'" GRAPHING MOTION - PART I".':" ************************"8'255:D11200:::(7):B'"";A370l*:13376:7:"LET'S LOOK AT A POSITION GRAPH AND FIND"::"BOTH THE VELOCITY AND ACCELERATION...":4:"WE'LL NEED SOME MORE MEMORY SPACE, SO"::"PLEASE WAIT A MOMENT WHILE WE LOAD IN"::"PART II OF THIS PROGRAM.">22:15:"-LOADING-":D082,6:132,20132,28:182,80182,86:4':100:8^'" *************************":$'" GRAPHING MOTION - PART IV".':" *************************"8'255:D11000:::(7):B'"";A$:---"8%3:32,3357,13:X5782:X,310(82X)2625:k%82,3107,3:X107132:X,320(X107)2625:%132,23157,63:X157207:X,8320(182X)2625:%207,63232,24%1:57,1057,16:107,0107,6:157,59157,67:207,59207,67&%2:82,#2:32,4932,1039#1:Y06:33,499Y35,499Y:j#7:3:"20":9:"V";:4:"0":11:2:"-20":}#2:32,132,37#1:Y04:33,19Y35,19Y:#1:4:"4":3:"A";:4:"0":5:3:"-4";:%6:1:"X":3:23:"Position":4:23:"-----4:3250X,Y13250X,Y3:G<#1:X03:5750X,Y15750X,Y3::{F#4:12:"10":2:19:"20";:26:"30";:33:"40"P#3:35:"time":#2:32,11532,155#1:Y04:33,11510Y35,11510Y:#15:2:"400":20:4:"0":18:1:"X"#"REVIEW. TYPE 2 TO CONTINUE: ";A:A1220v\34,0:13376:10:"THAT'S ALL THERE IS, ";Z$;".":::"SEE YOU LATER!"f20:10050p13376:D13:(7)::19:"END"zD$;"PR#0":D11500:::D$;"RUN MENU"'#(#2:32,Y232,Y2#X118207,122:31:10050E93:X207230:X,68X,84(X207)1823::5>13376:"FINALLY, THE LAST AREA IS -150, AND THE"::"FINAL POSITION IS +200.";:31:10050:13376HX182232:X,115(X182)2125:*R"THERE IT IS! TYPE 1 IF YOU WANT TO"::7178:X,59(X157)923X,66::5j X132182:X,115(182X)2250::6:182,113182,117:5:31:10050%3:X183207:X,68X,66(X182)1823::5*"FROM t = 30 TO 35, THE AREA IS -50, SO"::"THE POSITION IS +350.";4205,120210,120:207,16:3:X132157:X,50(X132)923X,66::5i"OF COURSE IT'S +75, SO THE NEW POSITION"::"IS 375."; 155,117159,117:157,115157,119:31:1005013376:"FROM t = 25 TO 30 WE ADD ANOTHER +25,"::"AND THE POSITION BECOMES 400.";"3:X15*********")8'255:D11000:::(7):>B'"";A$:D207,63232,24M%1:57,1057,16:107,0107,6:157,59157,67:207,59207,67%2:82,082,6:132,20132,28:182,80182,86:':100:8'" ************************":$'" GRAPHING MOTION - PART III" .':" ***************:4:"4":3:"A";:4:"0":5:3:"-4";:]%6:1:"X":3:23:"Position":4:23:"--------"%3:32,3357,13:X5782:X,310(82X)2625:%82,3107,3:X107132:X,320(X107)2625:%132,23157,63:X157207:X,8320(182X)2625:%z"#Xó    !Ǡ"ՠ ҠҠӠ7Ӡ0ǠǠ ǠϠΠҠҠҠҠҠ F(4);"CATALOG" A6 :5:115:" GAMES FOR THE APPLE II COMPUTER"g10:" COMPLIED FOR YOUR ENJOYMENT BY:":10:"STEVE KAMM & OTHERS":(12:"JANUARY, 1980"220:" 32K MASTER DISKETTE, UPDATED TO DOS"::13:"VERSION 3.3"<255:A11500:: J .'"****************************************":255:D1900:::(7):D VECTOR RESOLUTION"h ':" ************************"::" BY STEVEN D. KAMM": $'" 1979"::" THIS MATERIAL WAS PREPARED WITH THE":" SUPPORT OF NATIONAL SCIENCE FOUNDATION":" GRANT NO. 7900773":T NO. 7900773":k F"****************************************";:D12000:::255:(7):D$;"RUN MENU":U********"::" SOUTH OKLAHOMA CITY JUNIOR COLLEGE"::" PHYSICS TUTORIAL PROJECT": '" ************************": '" E"::" MASTER UPDATED TO DOS VERSION 3.3"{ 2:" ************************"::" BY STEVEN D. KAMM": <" COPYRIGHT 1979"::" THIS MATERIAL WAS PREPARED WITH THE":" SUPPORT OF NATIONAL SCIENCE FOUNDATION":12:"GRANl! D$(4): CHR$(4)IS CTRL-D :100:"****************************************":" SOUTH OKLAHOMA CITY JUNIOR COLLEGE"::" PHYSICS TUTORIAL PROJECT":" ************************":, (" TUTORIALS, VOL. 1, 32K MASTER DISKETT    膢 ȱ " <67 L  8ij v vo>21:"IN FACT, YOU DON'T NEED THE AXES, ONCE":"YOU HAVE THE COMPONENTS."f1:5,2732:10,31524:10050:2:910021:"THAT'S BECAUSE THE VECTOR AND ITS":"COMPONENTS REPRESENT THE SAME THI Q22570 1:9600:9700z 21:"STEP 4 - USE GRAPHICAL METHODS OR TRIG-":"ONOMETRY TO FIND THE TWO COMPONENTS." 24:10050: 0:9100:9200:V117:V225:9060 21:"ONCE YOU HAVE THE COMPONENTS, YOU DON'T":"NEED THE ORIGINAL VECTOR A4:10050:$ V117:V225:15:9060v 21:"STEP 2 - MEASURE THE ANGLE BETWEEN THE":"VECTOR AND ONE AXIS (THETA)." F24:10050: x1:9200 21:"STEP 3 - DRAW PERPENDICULAR LINES FROM":"THE HEAD OF THE VECTOR TO THE AXES." 24:10050: 0E :21:"CONSIDER ANY VECTOR , POINTING":"IN ANY DIRECTION."V 24:10050:d L12:9300q ~3:9400~ 0:5,32 21:"STEP 1 IS TO ADD A SET OF AXES. YOU DO":"NOT HAVE TO USE X & Y, BUT THE AXES" "SHOULD BE AT RIGHT-ANGLES." 2ST"::"ALSO BE POSSIBLE TO REPLACE ONE VECTOR": "BY TWO EQUIVALENT VECTORS. THIS TECH-"::"NIQUE, CALLED RESOLVING A VECTOR INTO": "COMPONENTS, IS VERY USEFUL IN PHYSICS."::"LET'S LOOK AT THE PROCESS:" "24:10050  2 :910Z "TRIG BEFORE YOU START THIS LESSON. BE"::"SURE YOU HAVE YOUR CALCULATOR HANDY!": :"WHEN YOU'RE READY TO BEGIN,"::10050 :4:Z$;", THE BASIC IDEA IS"::"VERY SIMPLE. SINCE TWO VECTORS CAN BE":1 "ADDED TOGETHER TO MAKE ONE, IT MUXd - VECTOR RESOLUTION&n10000N:3:"HELLO! WHAT'S YOUR NAME?";Z$6:"THIS PROGRAM IS FOR YOU, ";Z$;"."::"IT SHOULD HELP YOU UNDERSTAND HOW TO": "RESOLVE A VECTOR INTO TWO COMPONENTS."::"YOU SHOULD BE ABLE TO DO RIGHT-TRIANGLE":           < A+B(X)^2 + A+B(Y)^2 >"[" = SQRT < 10.1^2 + -5.9^2 >":" = SQRT < 136.82 > :" = 11.7 NT":::" THETA = ARCTAN < A+B(Y) / A+B(X) >"" = ARCTAN < -5.9 / 10.1 >":" = ARCTAN < -0.584 >"::" = -30.3 DEGR" A+B(Y) = A(Y) + B(Y)":" = 2.9 - 8.8 = -5.9"::"NOW WE HAVE TWO RIGHT-ANGLE VECTORS"::"THAT MUST BE COMBINED USING THE ":"PYTHAGOREAN RULE.":24:10050:16304,0:9900:24:10050::3:"HERE IT IS:"::" A+B = SQRTY76:9100:X104:9500:X110:9200:X118:9600:X138:Y151:9100:X146:9500:X152:9300:X160:9600z24:10050:::3"THE NEXT STEP IS TO COMBINE THE X- &"::"Y-COMPONENTS:":" A+B(X) = A(X) + B(X)":" = 13.7 - 3.6 = 10.1":;35,81251,81:247,79247,83:134,80134,60:130,63136,638200:2:133,81102,81:104,79104,83:133,82133,146:131,143137,1433:X138:Y63:9000:X146:9500:X152:9300:X160:9600:X242:Y93:9000:X250:9500:X256:9200:X264:9600dX96:(68) = -8.8":jN"AGAIN, USING THIS METHOD, YOU MUST ADD"::"NEGATIVE SIGNS WHEN THE COMPONENTS ARE":X"IN A NEGATIVE DIRECTION!":::"ARE YOU FILLING IN YOUR CHART? GOOD!"::b"NOW LET'S LOOK AT IT.":24:10050l::8000:81009v1:1X&"JUST BE SURE TO USE NEGATIVE SIGNS"::"WHEN THE COMPONENTS ARE IN A NEGATIVE":s0"DIRECTION!"::10050::3:" A(X) = 14 * COS(12) = 13.7"::" A(Y) = 14 * SIN(12) = 2.9":D:" B(X) = 9.5 * COS(68) = -3.6"::" B(Y) = 9.5 * SIND"::"THE X- & Y-COMPONENTS OF & .":v"BY USING THE ANGLE WITH THE X-AXIS, "::"THIS PROCESS IS EASY:":" A(X) = A * COS(THETA-A)"::" A(Y) = A * SIN(THETA-A), AND":" B(X) = B * COS(THETA-B)"::" B(Y) = B * SIN(THETA-B)":X3490X4510'24:10050O:5:"NOW WE'LL FILL IN THE DATA."c385:24:1005012:16:"14. * 9.5":14:16:"12 * 68"21:"ANGLES ARE MEASURED FROM THE X-AXIS!":24:10050-::"SO FAR, SO GOOD! NOW WE NEED TO FIN--------"Z"DIRECTION * * *":"---------------------------------------""X-COMPONENT * * *":"---------------------------------------""Y-COMPONENT * * *":"---------------------------------------": IS TO USE A CHART. IT WILL HELP":R |"KEEP THINGS STRAIGHT. COPY THIS ONE:"::\ XX1 10:" A B A + B":"---------------------------------------" "MAGNITUDE * * *":"-------------------------------THIS METHOD &":h T"YOU WON'T GO WRONG LATER! NO MATTER"::"HOW THE PROBLEM IS STATED, CHANGE ALL": ^"ANGLES SO YOU MEASURE FROM THE X-AXIS.":19:10050 h:16304,0:24:10050 mX0" r::3:"NOW TO SOLVE THE PROBLEM. AN EASY"::"METHOD 0:X111:9100:X119:9600R 6X205:Y77:9400:X213:9500:X219:9000:X227:9600 @:21:" HERE ARE THE ANGLES. YOU SHOULD":" MAKE A SKETCH FOR YOUR NOTES."::10050 J::7:"NOTE THAT WE HAVE MEASURED THE ANGLES"::"FROM THE X-AXIS. ADOPT AL PROBLEM:":::"ADD: = 14 NT AT 12 DEG. N OF E": " = 9.5 NT AT 22 DEG. W OF S":24:"WRITE DOWN THE PROBLEM & ";A$ ::8000:8100:8200 "22:" HERE ARE THE VECTORS.":24:10050 ,X97:Y96:9400:X105:950GNED TO HELP YOU"::"LEARN HOW TO ADD VECTORS GIVEN IN": "THE MAGNITUDE/DIRECTION FORM, I.E."::5::" = 23 METERS AT 30 N OF E":: "YOU SHOULD KNOW HOW TO DO RIGHT-ANGLE"::"TRIGONOMETRY BEFORE YOU BEGIN."::: "LET'S LOOK AT A TYPICEd- VECTOR ADDITION - I'n10000h:5:"HELLO, PHYSICS STUDENT!"::"WHAT'S YOUR NAME? ";Z$:10:Z$;", YOU WILL NEED PAPER, PENCIL,"::"AND YOUR CALCULATOR."::"IF YOU'RE READY, LET'S GO!":20:10050: :3:"THIS PROGRAM IS DESI       YOU STILL CAN'T FIND THE"::"PROBLEM, REVIEW THE METHOD (RERUN THIS":"PROGRAM). THEN, SEE YOUR INSTRUCTOR.":24:10050:8:"IF YOU WANT TO REVIEW THIS PROGRAM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A1230(:9:"THAT'S THE END FOR NOW, " S(Y) = 26.0":24:10050T:3:"5. R+S(X) = -42.4":" R+S(Y) = -5.9"::"6. R+S = 42.8 NT":::"7. THETA = 7.9 DEGREES S OF W"::"YOU SHOULD HAVE THE CORRECT ANSWERS."::"IF YOU DON'T, CHECK EACH STEP FOR MATH":H"ERRORS. IFMl" = 52 NT AT 30 DEGREES N OF W"::"3. R(X) = 34 * COS(70) = 11.6"v" R(Y) = 34 * SIN(70) = 31.9"::" S(X) = 52 * COS(30) = 54.0"" S(Y) = 52 * SIN(30) = 26.0":"4. R(X) = 11.6":" R(Y) = -31.9":" S(X) = -54.0":0 DEGREES E OF S"::" = 52 NT AT 30 DEGREES N OF W"::N"WORK OUT THE ANSWER COMPLETELY. THEN"::"CHECK YOUR WORK. THE ANSWERS ARE ON":X"THE NEXT PAGE.":20:10050b:3:"1. MAKE A CHART."::"2. = 34 NT AT 70 DEGREES S OF E"= 13.1 DEGREES S OF W"::s&"IF YOU USE ARCTAN < A+B(Y)/A+B(X) >,"::"THE ANGLE THETA WILL BE MEASURED FROM":0"THE X-AXIS.":24:10050::3:"THAT'S ALL THERE IS TO IT!"::"NOW HERE'S ONE TO DO ON YOUR OWN:"::AD"ADD-- = 34 NT AT 224:10050Z::"6. ADD A+B(X) & A+B(Y):"::" A+B = SQRT <(-37.90)^2 + (-8.84)^2>":" = SQRT <1436.4 + 78.1>"::" = 38.9 CM"::"7. THETA(A+B) = ARCTAN < -8.84/-37.90 >"::" = ARCTAN < .233 >"::" , IT WILL HELP.":J" A(X) = -5.13"::" A(Y) = 14.10"::" B(X) = -32.77"::" B(Y) = -22.94"::"5. ADD X- & Y-COMPONENTS:"::" A+B(X) = -37.90"::" A+B(Y) = -8.84": "ARE YOU DOING O.K., ";Z$;"?": MEASURED FROM THE X-AXIS."::m"3. A(X) = 15 * COS(70) = 5.13"::" A(Y) = 15 * SIN(70) = 14.10"::" B(X) = 40 * COS(35) = 32.77"::" B(Y) = 40 * SIN(35) = 22.94":24:10050::"4. ADD MINUS SIGNS. IF YOU'VE MADE A"::" SKETCHWRITE DOWN THE PROBLEM. THEN WE'LL"::"SOLVE IT TOGETHER.":24:10050z::"1. WE'LL ASSUME YOU'VE MADE A CHART."::"2. DATA: = 15 CM AT 70 DEGREES"::" = 40 CM AT 35 DEGREES":" CHANGE THE ANGLE OF SO IT IS"::" AL VECTOR USING THE"::" PYTHAGOREAN THEOREM."::n |"7. FIND THE ANGLE USING TRIG (ARCTAN).":24:10050 :6:"NOW WE'RE READY. HERE'S A PROBLEM:":::"ADD-- = 15 CM AT 20 DEG. W OF N": " = 40 CM AT 35 DEG. S OF W"::F"RK ONLY IF ANGLES ":? T" ARE FROM THE X-AXIS!":24:10050 ^:3:"4. ADD MINUS SIGNS TO COMPONENTS IN"::" THE NEGATIVE DIRECTION.": h:"5. ADD THE X-COMPONENTS TOGETHER. DO"::" THE SAME FOR THE Y-COMPONENTS."::5 r"6. FIND THE TOTTA. REMEMBER TO"::" CHANGE ALL ANGLES SO THEY ARE": 6" MEASURED FROM THE X-AXIS.":::"3. FIND THE X- & Y-COMPONENTS OF EACH": @" VECTOR:"::" A(X) = A * COS(THETA-A)": J" A(Y) = A * SIN(THETA-A)"::" THESE FORMULAS WO"---------------------------------------"{ "X-COMPONENT * * *":"---------------------------------------" "Y-COMPONENT * * *":"---------------------------------------": "24:100508 ,:3:"2. FILL IN THE GIVEN DAA CHART TO KEEP YOU ON"::" THE RIGHT TRACK:": " A B A+B":"---------------------------------------" "MAGNITUDE * * *":"---------------------------------------"* "DIRECTION * * *":d- VECTOR ADDITION - II(n10000d:5:"HELLO AGAIN!"::"REMIND ME OF YOUR NAME? ";Z$:"GOOD. NOW, ";Z$;", WE'LL TRY TO SOLVE"::"A FEW PROBLEMS. FIRST, A QUICK REVIEW":"OF THE BASIC METHOD.":20:100502 :3:"1. START WITH         OUR":+"ANSWER: (DON'T TYPE UNITS) ";L@L921L925680:(7):"INCORRECT. REMEMBER THAT:"::" . = A * B * COS(THETA)":"TRY AGAIN. ENTER YOUR NEW ANSWER: ";L::" . = 923.5 NT-M"::"SO MUCH FOR THAT ONE. LEZb(7):"** THAT'S NOT RIGHT. MAKE A SKETCH"::"OF THE X-Y PLANE AND TRY AGAIN."::l"WHAT IS THE ANGLE BETWEEN & ?"::"ENTER THE NUMBER OF DEGREES: ";Lv:3:"** THE ANGLE IS 55 DEGREES. **":::"NOW FINISH THE SOLUTION AND ENTER Y10050I::3:"NOW SOLVE THIS ONE YOURSELF:":::" FIND . IF":D" = 70 METERS, 10 DEGREES E OF N"::" = 23 NT, 45 DEGREES N OF W"::N"WHAT IS THE ANGLE BETWEEN & ?"::"ENTER THE NUMBER OF DEGREES: ";LXL556300^!::5:"NOW PROCEED WITH THE SOLUTION:":::" . = A * B * COS(THETA)":#" = 12 * 11 * COS(65)":&" = 12 * 11 * 0.423"::" = 55.8 NT-M":::0"EASY, ISN'T IT, ";Z$;"?":24:N THE X-Y PLANE."::10050y::9:"DRAW A SKETCH. THEN FIND THETA. YOU"::"SHOULD SEE THAT THE ANGLE BETWEEN THE":"VECTORS IS 65 DEGREES. RIGHT?":20:1005016304,0:3:X160:Y71:9500:21:"HERE'S THE ANGLE BETWEEN & .":24:1005NE OF THE ANGLE":N"BETWEEN THEM. FOR EXAMPLE:":::"FIND . IF:":" = 12 METERS, 10 DEGREES N 0F E"::" = 11 NT, 15 DEGREES E OF N":24:10050 :16304,0:8100:8200:21:"HERE'S WHAT IT LOOKS LIKE. BOTH":"VECTORS ARE IRDINATE SYSTEM.":24:10050j::8000:21:"REMEMBER:":" X = EAST Y = NORTH Z = UP":24:10050::3:"FOR DOT PRODUCTS,":::" . = A * B * COS(THETA)"::"SO, SIMPLY MULTIPLY THE MAGNITUDES OF"::"THE VECTORS AND THE COSI"YOU MIGHT NOT BE READY TO CONTINUE. BE":"SURE YOU UNDERSTAND THE MEANING OF"::"VECTOR PRODUCTS BEFORE YOU GO ON."::15:"IF YOU'RE READY, LET'S DO DOT PRODUCTS."::"THEY'RE VERY EASY SINCE THE ANSWER IS":"A SCALAR. HERE'S THE COOX IS ALWAYS PERPENDICULAR":" TO THE PLANE MADE BY & ."::390:" * EXCELLENT! THE VECTOR X IS":" PERPENDICULAR TO THE PLANE FORMED"" BY VECTORS & .":::10050,:3:"IF YOU HAD TROUBLE WITH THE QUESTIONS,"::YOU ARE SLEEPING, ";Z$;"!"::350w |:" * OF COURSE! THE VECTOR PRODUCT IS THE":" SAME AS THE CROSS PRODUCT.":: "3. THE VECTOR X LIES IN THE SAME":" PLANE AS & ."::" ANSWER T OR F?";T$ T$"F"420J(7):" * NO. * SORRY. THE DOT PRODUCT IS ANOTHER"::" NAME FOR THE SCALAR PRODUCT."::310 T:" * GOOD! A DOT PRODUCT GIVES A SCALAR.":: ^"2. A CROSS PRODUCT IS THE SAME AS A":" VECTOR PRODUCT."::" ANSWER T OR F? ";T$ hT$"T"380! r(7):" * MEMBER TO ALWAYS USE THE SMALLEST"::"ANGLE BETWEEN THE VECTORS!":24:10050 ,:3:"JUST TO BE SURE YOU'RE AWAKE, ANSWER"::"THESE QUESTIONS:":: 6"1. A DOT PRODUCT GIVES A SCALAR ANSWER."::" ANSWER T OR F? ";T$ @T$"T"340P J(7):" E'S THE NOTATION WE'LL USE:":::" = VECTOR A": " = VECTOR B"::" . = DOT PRODUCT OF & ": " X = CROSS PRODUCT OF & "::" THETA = SMALLEST ANGLE BETWEEN THE": " VECTORS"::M ""REO WATCH:":::" 'CROSS PRODUCTS - M/D FORM'":::"LET'S WORK ONLY IN THE MAGNITUDE/": "DIRECTION (M/D) FORM. IN OTHER WORDS,"::"WE'LL LOOK AT VECTORS IN THE FORM:":: (2)::" = 15 KM AT 30 DEGREES N OF E"::24:100509 :3:"HERr d- DOT PRODUCTS - M/D FORM+n10000:9:"HELLO, PHYSICS STUDENT!"::"WHAT'S YOUR NAME? ";Z$::::"THANK YOU!":24:10050:3:Z$;", THIS PROGRAM IS DESIGNED TO"::"HELP YOU FIND THE DOT PRODUCT OF TWO":\ "VECTORS. YOU SHOULD ALS     "TYPE 1 TO WATCH AGAIN.":"TYPE 2 TO CONTINUE. ";TC4T2840L>740H::3:"THUS, THE DIRECTION OF X IS DOWN.":::"HERE'S ONE FINAL PROBLEM. FIND X:":R" = 62 NT, 25 DEGREES ABOVE EAST"::" = 15 M, DUE EAST"::21:"HERE'S X.":24:10050CX120:3:74,87173X,54X]5:61,90X61,903XxX4Ē1:21,110181,30D150:D 0:74,87173X,54X:5:59,14863,148:57,14665,146 3:X70:Y150:9100:X78:9600:X86:90006*:21:8000:8100:8200:570M::3:"THUS, THE DIRECTION OF X IS UP."::"WHAT IS THE DIRECTION OF X?"::"IF YOU SAID DOWN, YOU'RE RIGHT!"::" X = - X":::"LET'S LOOK AT IT.":24:10050"::8000:8100:8200޾.-~u瓈v翑ӽ-,~vu翂ɽ>o~=?-,~v瓈v6>)/-~,~!-ooooT+󓟇9Emy, ST BE A VECTOR IN THE +Z":v&"OR -Z DIRECTION. REMEMBER, THE CROSS"::"PRODUCT ANSWER MUST BE PERPENDICULAR TO":0"THE PLANE OF & . WATCH!":24:10050::16304,0:21:"ROTATE TO WITH YOUR RIGHT HAND.":24:10050DX12055)"::" X = 12 * 11 * 0.819"::x" X = 108.1 FT-LBS":::"THAT'S NOT SO BAD, IS IT? ";:10050:3:"NOW LET'S SEE HOW TO FIND THE DIRECTION"::"OF X."::"SINCE & LIE IN THE X-Y PLANE,"::"THE ANSWER MU480I:CHRS$(7):"OPPS! THAT'S INCORRECT. TRY AGAIN.":24:10050:390::"GOOD! NOW CALCULATE X."::"ENTER YOUR ANSWER WITHOUT UNITS: ";T::T107T110520(7):"ERROR! HOW DID YOU MESS UP?":*" X = 12 * 11 * SIN(>.":24:10050\:8200:21:"HERE'S . BOTH VECTORS ARE IN THE":"X-Y PLANE."::10050:X160:Y71:9500:21:"HERE'S THE ANGLE BETWEEN & .":24:10050::3:"WHAT IS THE ANGLE THETA?"::"ENTER THE NUMBER OF DEGREES? ";TT55ĺ::" = 12 FEET, 20 DEGREES N OF E":" = 11 POUNDS, 15 DEGREES E OF N":::"WRITE DOWN THE PROBLEM. THEN WE'LL"::"LOOK AT IT.":24:10050::8000:21:"HERE'S OUR COORDINATE SYSTEM.":24:10050:8100:21:"HERE'S VECTOR TO":l r" THROUGH THE SMALLEST ANGLE. YOUR"::" THUMB SHOWS THE RESULT.":: |"THIS RULE IS FOR X. TWIST TO"::" FOR X. MAKE A NOTE OF THIS!":24:10050,:3:"HERE'S A PROBLEM. FIND X IF:":B * SIN(THETA)":::"WHERE THETA IS THE SMALLEST ANGLE": T"BETWEEN & .":::"THE HARD PART IS FINDING THE DIRECTION": ^"OF THE RESULTING VECTOR. WE USE THE"::"RIGHT HAND RULE.":24:10050 h:3:"RIGHT HAND RULE:"::" USE YOUR RIG"NO"320 6:3:"NO? THEN YOU HAD BETTER WATCH THE"::"PROGRAM: 'DOT PRODUCTS - M/D FORM'."::"LOAD THAT PROGRAM BEFORE YOU GO ON.": ;:10050:1000 @:3:"GREAT! THE MATHEMATICS OF CROSS "::"PRODUCTS ARE EASY:"::: J" X = A * N DOT PRODUCTS, YOU KNOW":w "THE NOTATION WE USE & YOU KNOW THAT THE"::"CROSS PRODUCT OF TWO VECTORS PRODUCES": "A THIRD VECTOR AT RIGHT ANGLES TO THE"::"PLANE FORMED BY THE FIRST TWO.": ""RIGHT, ";Z$;:"? (TYPE YES OR NO): ";T$ ,T$, " 'DOT PRODUCTS - M/D FORM'": :"DON'T START THIS PROGRAM WITHOUT SEEING"::"THE OTHER ONE FIRST!":24:10050 :3:"WHAT'S YOUR NAME? ";Z$:::"THANK YOU, ";Z$;"." ::"LET'S GET STARTED. SINCE YOU'VE DONE"::"THE PROGRAM O!d-CROSS PRODUCTS - M/D FORM,n10000k:4:"HELLO!"::"THIS PROGRAM IS DESIGNED TO HELP YOU":"FIND THE CROSS PRODUCT OF TWO VECTORS"::"EXPRESSED IN MAGNITUDE/DIRECTION FORM.": "YOU SHOULD HAVE ALREADY WATCHED THE"::"PROGRAM:":          AND CROSS PRODUCT OF UNIT VECTORS TO"::"SOLVE THESE PROBLEMS!"::O10050:"STEP 4 - COLLECT LIKE TERMS:"::"*DOT PRODUCTS - ADD I.I, J.J, K.K TERMS"::"AS A SCALAR.":::"*CROSS PRODUCTS - ADD , , "::"TERMS AS A VECTOR."::= = = 1"::"*CROSS PRODUCTS - USE RIGHT-HAND RULE:"::" = 0 = <-K> = "::" = = 0 = <-I>"::" = <-J> = = 0"::D"YOU MUST BE ABLE TO DETERMINE THE DOT"::" EA + EB + EC"::" FA + FB + FC":b"DO NOT REVERSE ANY I,J,K TERMS! ":"IS NOT NECESSARILY EQUAL TO .":l10050v:"STEP 3 - EVALUATE TERMS:"::"*DOT PRODUCTS - ALL TERMS EQUAL ZERO"::"EXCEPT: M 2, TOP 1 BY BOTTOM 3, TOP 2 BY"::"BOTTOM 1, TOP 2 BY BOTTOM 2, ETC.":N"THE SEQUENCE IS: S1R1, S1R2, S1R3":" S2R1, S2R2, S2R3":" S3R1, S3R2, S3R3":CX"FOR THIS PROBLEM:"::" DA + DB + DC"::"BOVE R."::j&"EXAMPLE: = D + E + F":" TIMES = A + B + C":0"BE CAREFUL! REVERSING GIVES ERRORS.":::::10050JD:"STEP 2 - CROSS MULTIPLYING"::"MULTIPLY TOP 1 BY BOTTOM 1, TOP 1 BY"::"BOTTOBINE LIKE TERMS"::"ASSUME 2 VECTORS OF THE FORM:"::" = A + B + C"::" = D + E + F":::10050 :"STEP 1 - SET UP VECTORS:"::"TO FIND TIMES , WRITE R ABOVE S."::"FOR TIMES , WRITE S A. IF YOU MISSED"::"SOME, STUDY YOUR TEXT BEFORE CONTINUING."K:10050:"NOW WE'RE READY TO BEGIN MULTIPLYING."::"THERE ARE FOUR STEPS TO THE PROCESS:":" 1. SET UP VECTORS"::" 2. CROSS MULTIPLY"::" 3. EVALUATE TERMS"::" 4. COMER 1,2OR 3";::A:&A430,440,450:"NO. SINCE YOU HAVE & , THE"::"ANSWER MUST BE EITHER OR <-J>."::390:"WRONG. DID YOU USE THE RIGHT HAND RULE?"::390>:"YES! IF YOU GOT ALL THE ABOVE RIGHT,"::"YOU'RE READY TO GO ON KNOW THAT A DOT PRODUCT"::"CANNOT GIVE A VECTOR ANSWER. TRY AGAIN."::320|:"YES! A DOT PRODUCT OF A UNIT VECTOR"::"WITH ITSELF IS ALWAYS ONE."::"4. WHAT IS THE RESULT OF X?":" (1) <-I> (2) <-J> (3) "::"ANSW) J" (1) 0 (2) 1 (3) ":H T:"ANSWER 1,2 OR 3";::A[ ^A360,380,370 h:"INCORRECT. THE DOT PRODUCT USES THE"::"COSINE TERM. SINCE THE ANGLE BETWEEN A"::"VETOR AND ITSELF IS 0, THE COSINE IS 1."::320Mr:"SORRY. YOU MUST"2. ANOTHER NAME FOR THE CROSS PRODUCT"::"IS THE VECTOR PRODUCT.":d :"TRUE OR FALSE";::B${ ":(B$,1)"T"310 ,:"INCORRECT. TRY IT AGAIN."::270 6:"VERY GOOD, ";Z$;". LET'S CONTINUE.": @:"3. WHAT IS THE RESULT OF .?":":6 :"1. THE DOT PRODUCT . IS A SCALAR.":T :"TRUE OR FALSE";::B$k :(B$,1)"T"260 :"WRONG. YOU SHOULD KNOW A DOT PRODUCT"::"ALWAYS GIVES A SCALAR. TRY AGAIN."::220 :"EXCELLENT! DOT PRODUCTS YIELD SCALARS.":F :X::2,529:28,6:27,6:26,5:6,28:7,28o 13:3,3519:3,712:15,3:14,4:13,5:14,6:15,7:16,34:17,34 "THIS IS THE AXIS SYSTEM WE'LL USE.":"COPY IT DOWN. NOTE <-J> IS TOWARDS YOU.": 10050 ::"NOW WE NEED TO CHECK YOUR KNOWLEDGE.:: " MEANS THE VECTOR A"::" MEANS THE UNIT VECTOR I"::" . MEANS THE DOT PRODUCT OF A&B"::" X MEANS THE CROSS PRODUCT OF A&B": 10050 :1:7,3019:33,3516:16,2034:33,3520:4,17:5,17* 12:X928:X,38@ d100003n"HELLO! WHAT'S YOUR NAME? ";Z$:x"WELL, ";Z$;", THIS PROGRAM IS"::"DESIGNED TO HELP YOU FIND THE"::"DOT & CROSS PRODUCT OF TWO VECTORS.": "LET'S WORK WITH VECTORS IN UNIT VECTOR"::"FORM ONLY. WE'LL USE THESE NOTATIONS:"        "DO YOU WANT TO CONTINUE (Y/N)?";B$:(B$,1)"Y"580A 9000q :"DON'T YOU THINK YOU SHOULD STUDY MORE?" D12000::9000X" V OR S?";C$Y(C$,1)"?d <*f'qq   ynnsn<*g'<*^'jsno<5+_' 00T0 "8. ELECTRIC FIELD INTENSITY (FORCE":" DIVIDED BY ELECTRIC CHARGE)":7010k: "9. ENERGY":7000D "10.ANGULAR VELOCITY":7010N D1300:X b 7:Z$;", YOU HAD ";M;" WRONG ANSWERS.":l M1ī3220v "TERRIFIC. DO ANOTHER.":7":" ELECT & MAG FIELD INTENSITIES":7010D "3. VELOCITY":7010 "4. ANGULAR MOMUNTUM (MOMENT OF INERTIA":" TIMES ANGULAR VELOCITY VECTOR)":7010 "5. DISPLACEMENT":7010 "6. MASS":7000& "7. WORK (DOT PROD.-FORCE & DISAPLM'T)":70000/:"YOU CAN DO BETTER. READ YOUR TEXT!"ED12000::9000K R 2} " VECTORS AND SCALARS, QUIZ 3": "****************************************": M0 :"1. DISTANCE":7000+ "2. INTENSITY OF E/M WAVES (CROSS PROD.-7000U\"10.ANGULAR MOMENTUM (CROSS PROD.-RADIUS":" AND LINEAR MOMENTUM)":7010dfD1300:jpz7:Z$;", YOU GOOFED ON ";M;".":M1ī2220"EXCELLENT. DO ANOTHER.":"DO YOU WANT TO CONTINUE (Y/N)?";B$:(B$,1)"Y"5809 "2. DENSITY":7000."3. VOLUME":7000M "4. KINETIC ENERGY":7000m*"5. ELECTRIC CHARGE":70004"6. ANGULAR VELOCITY (CROSS PROD.-RADIUS":" AND LINEAR VELOCITY)":7010>"7. SPEED":7000H"8. DISPLACEMENT":7010R"9. TIME": (Y/N)?";B$:(B$,1)"Y"580&9000W:"NOT SO GOOD. I THINK YOU NEED TO STUDY!"mD12000::9000sz2" VECTORS AND SCALARS, QUIZ 2":"****************************************":M0:"1. ACCELEREATION":7010ORCE & TIME)":7010Gj"9. MOMENTUM (PROD. OF MASS & VELOCITY)":7010zt"10.POWER (DOT PROD.-FORCE & VELOCITY)":7000~D1300::7:Z$;", YOU MISSED ";M;".":M1ī1220"VERY GOOD! DO ANOTHER.":"DO YOU WANT TO CONTINUE0:"1. AREA":70002$"2. VELOCITY":7010G."3. TIME":7000|8"4. WORK (DOT PROD.-FORCE & DISPLACEM'T)":7000B"5. TEMPERATURE":7000L"6. TORQUE (CROSS PROD.-RADIUS & FORCE)":7010V"7. MASS":7000`"8. IMPULSE (PRODUCT OF FS FOR YOU TO USE.":5 X:"CHOOSE QUIZ 1, 2, OR 3."S b:"WHICH DO YOU WANT?";X] g630s lX1000,2000,3000 v:"INPUT ERROR!"::600  2 " VECTORS AND SCALARS, QUIZ 1": "****************************************":MIF YOU CAN":2 "IDENTIFY VECTORS AND SCALARS."d &9:"FOR EACH QUANTITY LISTED, INDICATE IF": 0"IT IS A VECTOR OR SCALAR BY CHOOSING ": :"EITHER 'V' OR 'S'.":::"WHEN YOU ARE READY, PRESS : ";A$ D N:::"THERE ARE 3 QUIZE1)"F"460( "YOU HIT THE WRONG KEY."= D11500::390t :8:(7):"SORRY, ";Z$;", YOU NEED TO REVIEW.": "STUDY YOUR TEXT SOME MORE!": :10050:9000 "VERY GOOD, ";Z$;"!" D11500:  5 "HERE'S A CHANCE TO SEE $ ^"COME BACK WHEN YOU'RE READY."1 h:10050; r9000m |:"GREAT! NOW HERE'S ANOTHER."::D11000: :::"TRUE OR FALSE? A DOT PRODUCT OF TWO" :"VECTORS IS A SCALAR.": :"ANSWER T OR F";::B$:: (B$,1)"T"490 (B$,GNITUDE & DIRECTION. *":8 :"ENTER T OR F";::B$:M "(B$,1)"F"380b ,(B$,1)"T"330 6"YOU HIT THE WRONG KEY.":::(7) @D11200::260 J:8:(7):"YOU'RE NOT READY FOR THIS QUIZ!": T"STUDY THE DEFINITIONS IN YOUR TEXT.":qd- VECTOR/SCALAR QUIZ&n10000L:5:"HI! WHAT'S YOUR NAME? ";Z$S8"WELL, ";Z$;", I'M HERE TO HELP YOU.":"SO, LET'S GET STARTED!"::D12000:13"IS THIS STATEMENT TRUE OR FALSE?": ::"* A SCALAR HAS MA       "CHART AND ";:10050~:3:"DID YOU NOTICE THAT THERE ARE FIVE"::"EMPTY PLACES IN THE CHART? GOOD!"::10050::"IN ANY PROBLEM, YOU WILL BE GIVEN THREE"::"OF THE VALUES. THEN YOU MUST SOLVE FOR":8"THE OTHER TWO. SO, BEFORE YOU HELPFUL TO ORGANIZE YOUR DATA.":m"A CHART IS A HANDY WAY TO KEEP TRACK"::"OF THE INFORMATION.":::"HERE'S ONE YOU CAN USE..."::10050:2:800017:"NOTICE THAT TWO VALUES ARE FILLED IN."::"THIS WILL ALWAYS BE TRUE. COPY THE":SIMPLIFIES THE EQUATIONS":"WE USE. JUST REMEMBER THAT IF THE FINAL":l"POSITION WORKS OUT TO HAVE A NEGATIVE"::"VALUE, IT IS LOCATED TO THE LEFT OF THE":v"INITIAL POSITION.":::10050":6:"IN ORDER TO ORGANIZE YOUR THINKING,"::"IT ISTHUS, VELOCITY TO THE RIGHT IS POSITIVE,":"WHILE ACCELERATION TO THE LEFT IS":lD"NEGATIVE.":31:10050N:3:"** INITIAL CONDITIONS:":::" THE INITIAL TIME, T(0), AND THE":X" INITIAL POSITION, X(0), ARE ZERO."::Fb"THIS CONVENTION 050:2:8300m:3:"REMEMBER THAT THE FIRST STEP IN OUR"::"METHOD IS TO ESTABLISH CONVENTIONS.":&"HERE ARE TWO USEFUL ONES:":::"** DIRECTIONS:"::" AND ARE POSITIVE. ":0" AND ARE NEGATIVE."::P:"$;"!";:31:10050r:7:"SINCE THE METHOD WE USE WORKS FOR BOTH"::"SYSTEMS, WE'LL EXPRESS MOTION IN TERMS":"OF 'X', 'V', AND 'A'. JUST REMEMBER THAT":"YOU CAN SUBSTITUTE 'THETA', 'OMEGA',":"AND 'ALPHA' FOR ROTATIONAL PROBLEMS.":22:10"ACCELERATION. ";:10050::"IN THE ROTATIONAL SYSTEM, WE MUST USE:"::" OMEGA - ANGULAR VELOCITY (RAD/SEC)":" ALPHA - ANGULAR ACCELERATION (RAD/S/S)"::10050::"BE SURE TO WRITE DOWN ANYTHING YOU DON'T":"KNOW, ";Z"::" Z - FOR MOTION ON THE Z-AXIS":j:"** ROTATION"::" THETA - FOR MOTION AROUND AN":" AXLE (IN RADIANS)"::10050:3:"IN THE TRANSLATIONAL SYSTEM, WE USE THE"::"FAMILIAR 'V' FOR VELOCITY AND 'A' FOR":-INE PATH":A"CAN SPEED UP, SLOW DOWN, AND REVERSE."::10050:4:"SO, IF WE WANT TO REPRESENT THE POSITION":"OF AN OBJECT, WE CAN USE:"::"** TRANSLATION"::" X - FOR MOTION ON THE X-AXIS":+" Y - FOR MOTION ON THE Y-AXIS:"FOR ROTATIONAL ONE DIMENSIONAL MOTION,"::"THE OBJECT ROTATES AROUND A FIXED AXLE": r"WITH A CONSTANT RADIUS--FOR EXAMPLE,"::"A WHEEL. IT CAN SPEED UP, SLOW DOWN, ": |"AND REVERSE ROTATION DIRECTION JUST"::"LIKE AN OBJECT ON A STRAIGHT LON MEANS AN OBJECT"::"CAN MOVE IN ONE DIMENSION ONLY.": T"FOR TRANSLATIONAL MOTION, WE SAY THAT"::"THE OBJECT MAY MOVE BACK AND FORTH ALONG" ^"A STRAIGHT LINE--USUALLY THE X-AXIS."::"(WE COULD ALSO CHOOSE THE Y-AXIS, ETC.)"::10050X h:6"HERE'S THE METHOD:"::8100 6:"WRITE DOWN THESE STEPS. BE SURE TO NOTE"::"THAT YOU MUST LEARN THE EQUATIONS!"::10050:: @:5:"BEFORE WE LOOK AT THE EQUATIONS, LET'S"::"DISCUSS THE VARIABLES. ";:10050::: J"ONE DIMENSIONAL MOTIMETHOD FOR SOLVING PROBLEMS."x "BE SURE TO COPY DOWN THIS METHOD AS WE"::"GO ALONG. YOU'LL WANT TO REFER TO THE": "METHOD WHEN WE SOLVE PROBLEMS."::10050: :"NOW LET'S LOOK AT A FEW EXAMPLES."::10050 "9000:9100:9200 ,::3:L ONLY LOOK AT THE CASE":r "WHERE ACCELERATION IS CONSTANT. YOU"::"WILL NEED TO KNOW A FEW EQUATIONS AND": "SOME BASIC ALGEBRA.":::"GET YOUR CALCULATOR READY AND ";:10050 :4:"THE MOST IMPORTANT PART OF THIS PROGRAM"::"DESCRIBES A 5#d- ONE DIMENSIONAL KINEMATICS.n10000f:3:"WELCOME!"::"PLEASE GIVE ME YOUR NAME: ";Z$8:"THANK YOU, ";Z$;". IN THIS PROGRAM"::"WE'LL SPEND SOME TIME LEARNING A METHOD": "OF SOLVING ONE-DIMENSIONAL MOTION "::"PROBLEMS. WE'L               1:17,36T11200:5Y05:11:30,11Y2:22:"TIME MARKERS SHOW THE ACCELERATION."::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1630::3:"HERE'S THE PROBLEM:":::"A STUDENT DROPS A ROCK FROM THE TOP OF":H"A CLIFF AND OB:Ul"THE INITIAL VELOCITY IS ZERO. LET'S"::"LOOK AT IT. ";:10050sv::9000:9010:1:15,1122:"PRESS TO DROP THE ROCK.";A$17,11:0:15,11:T1700::Y05.501:17,11Y2:0:17,11Y2: (7):"SOLVE PROBLEMS, TYPE 1. IF YOU'RE READY"::"TO CONTINUE, TYPE 2: ";A:A1310X12:"OK, ";Z$;". LET'S LOOK AT AN EASY"::"PROBLEM. WE'LL TAKE THE CASE OF A ROCK":b"WHICH IS DROPPED FROM THE TOP OF A "::"CLIFF. RIGHT AWAY YOU SHOULD SEE THAT"ONLY TO MOTION"::"IN THE Y DIRECTION WHERE IS POS-"::"ITIVE & GRAVITATIONAL ACCELERATION IS"::"NEGATIVE! COPY THE EQUATIONS AND"::10050D:3:"IF YOU WOULD LIKE TO REVIEW THE FIRST"::"PART OF THIS PROGRAM BEFORE WE BEGIN TO":ONHEN YOU SOLVE FOR THE OTHER":"TWO. ";:10050:3:"HERE ARE THE EQUATIONS YOU'LL NEED:":::" 1. Y = V(0)*T - 1/2*G*T^2"::&" 2. V = V(0) - G*T":::" 3. V^2 = V(0)^2 - 2*G*Y"::<0"NOTICE THAT THESE APPLY :"TO MAKE A CHART. LET'S SEE WHAT IT":Q"LOOKS LIKE: ";:10050_::8000" COPY THIS CHART ";:10050::"NOTICE THAT THERE ARE FOUR UNKNOWNS. IN"::"ANY PROBLEM YOU WILL BE GIVEN AT LEAST":D"TWO VALUES--TLL IS DROPPED":k"FROM THE TOP OF A CLIFF, THE TOP OF THE"::"CLIFF IS Y(0) = 0, AND THE BOTTOM OF":"THE CLIFF WILL HAVE A NEGATIVE POSITION"::"VALUE: FOR EXAMPLE Y = -20 METERS.":(10050:::"A GOOD WAY TO ORGANIZE THE VARIABLES IS": NEGATIVE"::10050t:6:"NOW HERE'S ANOTHER CONVENTION WHICH"::"SIMPLIFIES THE EQUATIONS. WE ALWAYS":"SET THE INITIAL TIME AND POSITION EQUAL"::"TO ZERO."::10050:4:"DON'T BE CONFUSED, ";Z$;"!"::"IT JUST MEANS THAT IF A BA - INITIAL":U" Y - FINAL":::" TIME T(0) - INITIAL":" T - FINAL":::" VELOCITY V(0) - INITIAL":" V - FINAL":::" ACCELERATION -G - ACCELERATION IS"" WN> AS NEGATIVE. THUS,":s h"WHEN A ROCK IS THROWN UPWARD, IT HAS A"::"POSITIVE INITIAL VELOCITY, WHILE THE": r"ACCELERATION DUE TO GRAVITY IS NEGATIVE."::"OK, ";Z$;"?":24:10050 |:3:"NOW HERE ARE THE VARIABLES:":::" POSITION Y(0)::" 3. ORGANIZE DATA":p J" 4. APPLY EQUATIONS--YOU MUST KNOW THEM!":" 5. SOLVE FOR UNKNOWNS"::10050:: T"NOW LET'S LOOK AT A VERY IMPORTANT SIGN"::"CONVENTION."::10050 ^:5:"WE WILL ESTABLISH THE DIRECTION AS"::"POSITIVE & STUDY IT A MINUTE.";:31:10050::|&5:149,6191,6:"WHAT IS THE SLOPE OF THE GRAPH FROM 'C'"0:"TO 'D'? INPUT YOUR ANSWER: ";A::A0580:"NO! SINCE THE LINE IS HORIZONTAL, THE"::"SLOPE IS ZERO.";:31:10050:59%3:X3272:X,50(X32)2160:W72,40112,16:X112152:X,6(152X)2160:h152,6192,62:74,3674,42:112,13112,19:148,3148,9:192,3192,95:10:"A":2:15:"B":3:22:"C";:28:"D"9:34,9=9:"HERE'S A POSITION F":U"THE VELOCITY. PLOT THAT VALUE ON THE"::"VELOCITY GRAPH.";:31:10050::"REMEMBER:"::" THE SLOPE OF X IS THE VALUE OF V."::"HERE'S WHAT WE MEAN..."::31:1005013376::2:32,532,50200,504:3:"X":7:31:"TIME""THE DERIVATIVE OF A GRAPH MEANS THE"::"SLOPE OF THE GRAPH.";:31:1005013376:3:"SO TO FIND THE VELOCITY AT ANY TIME,"::"DRAW A LINE TANGENT TO THE POSITION":"GRAPH AT THE TIME IN QUESTION. THE"::"SLOPE OF THE LINE WILL BE THE VALUE OMENT. SO IF YOU HAVE A GRAPH":w|"OF POSITION VS. TIME, YOU SIMPLY TAKE"::"ITS DERIVATIVE TO FIND THE GRAPH OF":"VELOCITY.";:31:10050::"REMEMBER YOUR CALCULUS. WHAT IS THE"::"GRAPHICAL MEANING OF THE DERIVATIVE?"::10050::JadngcdT1Y UnvutttxvvvvtlltzvvvttyvnTYx:16302,0\- TURN ON ROMPLUS+ BOARDhD$(4)#\TpovwavT\Tp|eg}T\TpT\TpvIvnytttth{{jtT\TpovevTR]T؝eggcbngnvxtuvnnvst tktvo         THE"::"AXES, THEN THE ZERO SLOPES. FINALLY, WE"::"CALCULATE THE SLOPES.";:31:1005013376:20:"WHAT IS THE ACCELERATION (SLOPE) FROM"::"TIME 5 TO 10?";A:13376:A270020:"EXCELLENT!";22:"A = (0-10)/(10-5) = -2"::31:10050bbbڷP`NhLTzb`^u؀FTzbOTHER CONSTANT"::"VELOCITIES.";:31:10050Pb133,133157,133:207,97231,97l13376:20:"PLOT THESE VALUES. BEWARE OF NEGATIVE"::"SLOPES!";:31:10050v57,10681,115:107,115131,133:157,133207,97>13376:20:"SINCE ACCELERATION IS C357,13?013376:20:"SLOPE = (75-25)/(5-0) = +10"::10050R:33,10657,106D13376:20:"#6. PLOT THIS VALUE ON THE VELOCITY"::" GRAPH.";:31:10050N131,25131,63155,63:209,63233,63233,25,X13376:20:"FIND THE SLOPES FOR THE 17:183,113183,117s13376:20:"#3. HERE WE SHOW THE ZERO VELOCITIES ON"::"THE VELOCITY GRAPH.";:31:1005013376:20:"#4. THE SLOPE (VELOCITY) IS CONSTANT"::" FROM TIME 0 TO 5."::"#5. CALCULATE THE SLOPE.";:31:10050&33,3357,3POSITIONS WHERE THE"::"SLOPE IS ZERO."::31:10050x0:182,79182,87:5:83,3107,3:181,78181,88:183,78183,8813376:20:"THE SLOPE IS ZERO IN TWO PLACES."::100500:181,115183,115:182,115182,117:5:83,115107,115:181,113181,19030:9100:960013:34,13t14:"HERE'S THE POSITION GRAPH."::"NOW WE'LL ADD AXES FOR VELOCITY."::1005013376:Y115:9000:9050:914019:34,1920:"NOTICE THAT THE TIMES CORRESPOND."::10050513376:20:"STEP #2. LOCATE 50::\" 8. FINALLY, FIND THE ACCELERATION"::" GRAPH FROM THE VELOCITY GRAPH BY":" REPEATING STEPS #1 TO #6 ON YOUR"::" NEW VELOCITY GRAPH.";:31:10050:::"NOW LET'S USE THE METHOD...";:31:1005013376::Y43:9000:䎖ޭү#f偎が*肌鎖邎:tlinesand H-t!calculatetheslopestofindwievaluesofvelocity -~!heresatrickyoucanuse iguid`bbdmds`uhnohrbnoruantafteryouhave 8.H!foundthezerovelocitiesaonstantvelocities simULATE THE SLOPE IN THESE PLACES."::" 6. INDICATE THESE PLACES ON THE": @" VELOCITY GRAPH WITH A VELOCITY EQUAL":" TO THE CALCULATED SLOPE.":::31:10050 J13376:3:" 7. FOR REGIONS WHERE THE POSITION GRAPH":" IS CURVED, DRAW TANGEN VELOCITY ON THE VELOCITY GRAPH.":5 31:10050: "13376:3:" 4. ON THE POSITION GRAPH, LOCATE THE"::" PLACES WHERE VELOCITY IS CONSTANT.": ," (THAT IS, THE POSITION GRAPH IS A"::" STRAIGHT LINE.)";:31:10050I 6::" 5. CALCDRAW AXES FOR THE VELOCITY GRAPH":t " DIRECTLY BELOW THE POSITION GRAPH"::" SO THE TIMES CORRESPOND.": " 2. ON THE POSITION GRAPH, LOCATE THE"::" PLACES WHERE THE SLOPE IS ZERO.":$ " 3. INDICATE THOSE PLACES WITH A ZERO"::" :U "IN PART II WE'LL APPLY THE IDEAS OF"::"PART I TO A COMPLETE PROBLEM.":: "FIRST, LET'S WRITE OUT A METHOD WHICH": "MAY BE OF HELP TO YOU. YOU SHOULD COPY"::"THIS METHOD FOR REFERENCE.":::10050# 13376:5:"METHOD:"::" 1. G2)d - GRAPHING MOTION - PART II6n:10000Fx:16302,0d- TURN ON ROMPLUS+ BOARDpD$(4)D$;"PR#5"R$(13)R$M$"":- THERE IS A HEREM$;"1A" 13376:5:"THAN