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Le saut s'effectue depuis le \ t\[EAcute]l\[EAcute]ph\[EAcute]rique des Deux Alpes \[AGrave] 2200 m \ d'altitude." }], "Body", CellMargins->{{2.625, Inherited}, {Inherited, Inherited}}], Cell[BoxData[ FormBox[GridBox[{ {Cell["", "CellLabel"], RowBox[{ StyleBox[\(Donn\[EAcute]es\ techniques\), FontWeight->"Bold"], "\[IndentingNewLine]", "\[IndentingNewLine]", Cell[TextData[{ StyleBox[ "- Hauteur sol - t\[EAcute]l\[EAcute]ph\[EAcute]rique : 140 m\ \[EGrave]tres. \n- Longueur des \[EAcute]lastiques : 28 m\[EGrave]tres. \n- \ Nombres d'\[EAcute]lastiques : 3 suivant les diff\[EAcute]rentes gammes de \ poids.\n- Nombre de fibres latex par \[EAcute]lastique : environ 1000.\n- \ Poids de l'\[EAcute]lastique : 40 kg. \n- Hauteur du saut : 120 \ m\[EGrave]tres (arr\[EHat]t au ", FontFamily->"Arial"], Cell[BoxData[ \(TextForm\`1\^er\)]], StyleBox[ " rebond \[AGrave] 20 m\[EGrave]tres du sol). \n- Remont\ \[EAcute] au 1er rebond : 85%. \n- Vitesse maximale de chute : environ 180 \ km/h (\[AGrave] v\[EAcute]rifier avec le mod\[EGrave]le!)\n- Temps de chute \ au 1er rebond : 6 secondes.", FontFamily->"Arial"] }], "Body", CellMargins->{{2.625, Inherited}, {Inherited, Inherited}}]}]} }, ColumnWidths->Automatic, RowAlignments->Baseline, ColumnAlignments->{Left}], InlineCell]], "BilateralCell", TextAlignment->Left, TextJustification->0], Cell["\<\ Plus pr\[EGrave]s de Gen\[EGrave]ve, en Haute Savoie, dans la r\ \[EAcute]gion Morzine-Avoriaz, la soci\[EAcute]t\[EAcute] Snorange offre un \ saut depuis le t\[EAcute]l\[EAcute]ph\[EAcute]rique (85 m) ou depuis la \ passerelle (50 m) de Nyon.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Probl\[EGrave]me", "Subsubsection", CellMargins->{{2.625, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, FontFamily->"Lucida Grande"], Cell["\<\ a) Construisez un mod\[EGrave]le permettant de simuler le mouvement \ de chute du sauteur en faisant les hypoth\[EGrave]ses suivantes : - le saut s'effectue selon une verticale et l'origine de l'axe est situ\ \[EAcute]e sur le sol; - la force de frottement due \[AGrave] l'air est proportionnelle \[AGrave] la \ vitesse du sauteur; - la force de rappel due \[AGrave] l'\[EAcute]lastique est proportionnelle \ \[AGrave] son \[EAcute]longation. b) Donnez l'\[EAcute]quation du mouvement et sa solution lorsque la distance \ parcourue : - est inf\[EAcute]rieure ou \[EAcute]gale \[AGrave] la longueur de \ l'\[EAcute]lastique (l'\[EAcute]lastique est d\[EAcute]tendu); - est sup\[EAcute]rieure \[AGrave] la longueur de l'\[EAcute]lastique \ (mouvement lorsque l'\[EAcute]lastique est tendu). c) Comparez les r\[EAcute]sultats de la simulation aux donn\[EAcute]es \ techniques de la pr\[EAcute]sentation. d) Que doit valoir la longueur de l'\[EAcute]lastique pour effectuer les deux \ autres sauts (depuis le t\[EAcute]l\[EAcute]ph\[EAcute]rique et depuis la \ passerelle de Nyon) en toute s\[EAcute]curit\[EAcute] si votre masse vaut 80 \ kg ? e) Comment les r\[EAcute]sultats sont-ils modifi\[EAcute]s si la force de \ frottement due \[AGrave] l'air est proportionnelle au carr\[EAcute] de la \ vitesse du sauteur ?\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Solution", "Subsubsection", CellMargins->{{2.625, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, FontFamily->"Lucida Grande"], Cell[TextData[{ "Cherchons la solution g\[EAcute]n\[EAcute]rale ", Cell[BoxData[ \(TextForm\`z\_1\)]], "(", StyleBox["t", FontSlant->"Italic"], ") lorsque le sauteur n'est soumis qu'\[AGrave] une force de frottement \ proportionnelle \[AGrave] sa vitesse et \[AGrave] son poids (premi\[EGrave]re \ \[EAcute]galit\[EAcute]) pour une vitesse initiale nulle (deuxi\[EGrave]me \ \[EAcute]galit\[EAcute]) et une position initiale \[EAcute]gale \[AGrave] ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["h", FontSlant->"Italic"], "0"], TextForm]]], " (troisi\[EGrave]me \[EAcute]galit\[EAcute]) :" }], "Body", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(sol1 = DSolve[{\(h''\)[t] \[Equal] \(-mu\)*\(h'\)[t]/m - g, \(h'\)[0] \[Equal] 0, h[0] \[Equal] h0}, h[t], t]\)], "Input"], Cell[BoxData[ \({{h[ t] \[Rule] \(\[ExponentialE]\^\(-\(\(mu\ t\)\/m\)\)\ \((\(-g\)\ m\ \^2 + \[ExponentialE]\^\(\(mu\ t\)\/m\)\ g\ m\^2 + \[ExponentialE]\^\(\(mu\ t\ \)\/m\)\ h0\ mu\^2 - \[ExponentialE]\^\(\(mu\ t\)\/m\)\ g\ m\ mu\ \ t)\)\)\/mu\^2}}\)], "Output"] }, Open ]], Cell["\<\ Introduisons les donn\[EAcute]es num\[EAcute]riques dans la \ solution et calculons le temps n\[EAcute]cessaire pour tendre \ l'\[EAcute]lastique :\ \>", "Body", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ StyleBox[\( (*\ mu\ : \ coeff . \ de\ prop . \ entre\ la\ vitesse\ et\ la\ force\ de\ frottement\ *) \), FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ g\ : \ grandeur\ de\ l' acc\[EAcute]l\[EAcute]ration\ terrestre\ *) \), FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ m\ : \ masse\ du\ sauteur\ *) \), FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ h0\ : \ hauteur\ initiale\ *) \), FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ le\ : \ longueur\ de\ l' \[EAcute]lastique\ *) \), FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ tmax\ : \ temps\ d' observation\ *) \), FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ dn\ : \ donn\[EAcute]es\ num\[EAcute]riques\ *) \), FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], RowBox[{ RowBox[{ StyleBox[\(tmax = 30;\), FontColor->RGBColor[0, 0, 1]], StyleBox["\[IndentingNewLine]", FontColor->RGBColor[0, 0, 1]], StyleBox[\(dn = {mu -> 10, k -> 17, g \[Rule] 9.8, m \[Rule] 80, h0 \[Rule] 140, \ le \[Rule] 28};\), FontColor->RGBColor[0, 0, 1]], "\[IndentingNewLine]", \(temps = Solve[\((sol1[\([1, 1, 2]\)] \[Equal] h0 - le)\) /. dn, t]\)}], ";"}]}]], "Input"], Cell[BoxData[ RowBox[{\(InverseFunction::"ifun"\), \(\(:\)\(\ \)\), "\<\"Inverse \ functions are being used. Values may be lost for multivalued inverses. \\!\\(\ \\*ButtonBox[\\\"More\[Ellipsis]\\\", ButtonStyle->\\\"RefGuideLinkText\\\", \ ButtonFrame->None, ButtonData:>\\\"InverseFunction::ifun\\\"]\\)\"\>"}]], \ "Message"], Cell[BoxData[ RowBox[{\(Solve::"ifun"\), \(\(:\)\(\ \)\), "\<\"Inverse functions are \ being used by \\!\\(Solve\\), so some solutions may not be found; use Reduce \ for complete solution information. \ \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", ButtonStyle->\\\"RefGuideLinkText\ \\\", ButtonFrame->None, ButtonData:>\\\"Solve::ifun\\\"]\\)\"\>"}]], \ "Message"] }, Open ]], Cell["Repr\[EAcute]sentons l'horaire du sauteur :", "Text", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}, FormatType->TextForm], Cell[BoxData[{ \(\(fs = 10;\)\), "\[IndentingNewLine]", \(\(h1 = Plot[sol1[\([1, 1, 2]\)] /. dn, {t, 0, temps[\([2, 1, 2]\)]}, PlotStyle \[Rule] {Hue[2/3]}, AxesLabel \[Rule] {"\", "\"}, TextStyle \[Rule] {FontFamily \[Rule] Verdana, FontSize \[Rule] fs, FontSlant \[Rule] Italic}];\)\)}], "Input"], Cell["\<\ Calculons la vitesse du sauteur durant cette premi\[EGrave]re phase \ et repr\[EAcute]sentons-la en fonction du temps :\ \>", "Body", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(\(D[sol1[\([1, 1, 2]\)], t] /. dn\) /. t -> temps[\([2, 1, 2]\)]\)], "Input"], Cell[BoxData[ \(\(-21.153603114249723`\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[{\(D[sol1[\([1, 1, 2]\)], t] /. dn\), "\[IndentingNewLine]", RowBox[{ RowBox[{"v1", "=", RowBox[{"Plot", "[", RowBox[{ "%", ",", \({t, 0, temps[\([2, 1, 2]\)]}\), ",", \(PlotStyle \[Rule] {Hue[1/3]}\), ",", RowBox[{"AxesLabel", "\[Rule]", RowBox[{"{", RowBox[{ "\"\\"", ",", "\"\\""}], "}"}]}], ",", \(TextStyle \[Rule] {FontFamily \[Rule] Verdana, FontSize \[Rule] fs, FontSlant \[Rule] Italic}\)}], "]"}]}], ";"}]}], "Input"], Cell[BoxData[ \(\(-\(1\/800\)\)\ \[ExponentialE]\^\(\(-t\)/8\)\ \ \((\(-62720.00000000001`\) + 76720.`\ \[ExponentialE]\^\(t/8\) - 7840.`\ \[ExponentialE]\^\(t/8\)\ t)\) + 1\/100\ \[ExponentialE]\^\(\(-t\)/8\)\ \((1750\ \ \[ExponentialE]\^\(t/8\) - 980.0000000000001`\ \[ExponentialE]\^\(t/8\)\ t)\)\)], "Output"] }, Open ]], Cell[TextData[{ "La premi\[EGrave]re solution nous permet de calculer la vitesse du sauteur \ au moment o\[UGrave] l'\[EAcute]lastique commence \[AGrave] se tendre. A ce \ moment sa position est \[EAcute]gale \[AGrave] ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["h", FontSlant->"Italic"], "0"], TextForm]]], "-", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["l", FontSlant->"Italic"], "e"], TextForm]]], ". Cherchons la solution g\[EAcute]n\[EAcute]rale ", Cell[BoxData[ \(TextForm\`z\_2\)]], "(", StyleBox["t", FontSlant->"Italic"], ") lorsque le sauteur est soumis, en plus des deux forces pr\[EAcute]c\ \[EAcute]dentes, \[AGrave] une force de rappel proportionnelle \[AGrave] l'\ \[EAcute]longation de l'\[EAcute]lastique :" }], "Body", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(sol2 = DSolve[{\(h''\)[t] \[Equal] \(-mu\)/m*\(h'\)[t] - g - k/m \((h[t] - \((h0 - le)\))\), \(\(h'\)[0] \[Equal] D[sol1[\([1, 1, 2]\)], t] /. dn\) /. t -> temps[\([2, 1, 2]\)], h[0] == h0 - le}, h[t], t]\)], "Input"], Cell[BoxData[ \({{h[ t] \[Rule] \(\(1\/\(k\ \@\(\(-4.`\)\ k\ m + 1.`\ \ mu\^2\)\)\)\((21.153603114249723`\ \[ExponentialE]\^\(\(\((\(-mu\) - \@\(\(-4\ \)\ k\ m + mu\^2\))\)\ t\)\/\(2\ m\)\)\ k\ m - 21.153603114249723`\ \[ExponentialE]\^\(\(\((\(-mu\) + \@\(\(-4\ \)\ k\ m + mu\^2\))\)\ t\)\/\(2\ m\)\)\ k\ m - 0.5`\ \[ExponentialE]\^\(\(\((\(-mu\) - \@\(\(-4\)\ k\ m + \ mu\^2\))\)\ t\)\/\(2\ m\)\)\ g\ m\ mu + 0.5`\ \[ExponentialE]\^\(\(\((\(-mu\) + \@\(\(-4\)\ k\ m + \ mu\^2\))\)\ t\)\/\(2\ m\)\)\ g\ m\ mu + h0\ k\ \@\(\(-4.`\)\ k\ m + 1.`\ mu\^2\) - 1.`\ k\ le\ \@\(\(-4.`\)\ k\ m + 1.`\ mu\^2\) - 1.`\ g\ m\ \@\(\(-4.`\)\ k\ m + 1.`\ mu\^2\) + 0.5`\ \[ExponentialE]\^\(\(\((\(-mu\) - \@\(\(-4\)\ k\ m + \ mu\^2\))\)\ t\)\/\(2\ m\)\)\ g\ m\ \@\(\(-4.`\)\ k\ m + 1.`\ mu\^2\) + 0.5`\ \[ExponentialE]\^\(\(\((\(-mu\) + \@\(\(-4\)\ k\ m + \ mu\^2\))\)\ t\)\/\(2\ m\)\)\ g\ m\ \@\(\(-4.`\)\ k\ m + 1.`\ \ mu\^2\))\)\)}}\)], "Output"] }, Open ]], Cell["Repr\[EAcute]sentons l'horaire du sauteur :", "Body", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[ \(sol2[\([1, 1, 2]\)] /. dn\)], "Input"], Cell[BoxData[ \(\((\(\(0.`\)\(\[InvisibleSpace]\)\) - 0.000804971399270549`\ \[ImaginaryI])\)\ \((\((\(\(0.`\)\(\ \[InvisibleSpace]\)\) + 81844.3400608741`\ \[ImaginaryI])\) + \ \((\(\(24848.900235379624`\)\(\[InvisibleSpace]\)\) + 28645.519021305932`\ \[ImaginaryI])\)\ \ \[ExponentialE]\^\(1\/160\ \((\(-10\) - 2\ \[ImaginaryI]\ \@1335)\)\ t\) - \ \((\(\(24848.900235379624`\)\(\[InvisibleSpace]\)\) - 28645.519021305932`\ \[ImaginaryI])\)\ \ \[ExponentialE]\^\(1\/160\ \((\(-10\) + 2\ \[ImaginaryI]\ \@1335)\)\ \ t\))\)\)], "Output"] }, Open ]], Cell[BoxData[{ \(\(sol2[\([1, 1, 2]\)] /. dn;\)\), "\[IndentingNewLine]", \(\(h2 = Plot[%, {t, 0, tmax}, TextStyle \[Rule] {FontFamily \[Rule] Verdana, FontSize \[Rule] fs, FontSlant \[Rule] Italic}];\)\)}], "Input"], Cell["\<\ Cette courbe doit \[EHat]tre d\[EAcute]cal\[EAcute]e vers la droite \ car la force de rappel n'agit qu'apr\[EGrave]s un certains laps de temps. \ Translatons la courbe vers la droite et redessinons-la:\ \>", "Body", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}], Cell[BoxData[{ \(\(\(sol2[\([1, 1, 2]\)] /. dn\) /. t \[Rule] t - temps[\([2, 1, 2]\)];\)\), "\[IndentingNewLine]", \(\(h2 = Plot[%, {t, temps[\([2, 1, 2]\)], tmax}, AxesOrigin \[Rule] {0, 0}, PlotStyle \[Rule] {Hue[1]}, TextStyle \[Rule] {FontFamily \[Rule] Verdana, FontSize \[Rule] fs, FontSlant \[Rule] Italic}];\)\)}], "Input"], Cell["\<\ Calculons et repr\[EAcute]sentons la vitesse du sauteur en fonction \ du temps :\ \>", "Text", FormatType->TextForm], Cell[BoxData[{\(\(D[sol2[\([1, 1, 2]\)], t] /. dn\) /. t \[Rule] t - temps[\([2, 1, 2]\)];\), "\[IndentingNewLine]", RowBox[{ RowBox[{"v2", "=", RowBox[{"Plot", "[", RowBox[{ "%", ",", \({t, temps[\([2, 1, 2]\)], tmax}\), ",", \(PlotStyle \[Rule] {Hue[1/3]}\), ",", RowBox[{"AxesLabel", "\[Rule]", RowBox[{"{", RowBox[{ "\"\\"", ",", "\"\\""}], "}"}]}], ",", \(TextStyle \[Rule] {FontFamily \[Rule] Verdana, FontSize \[Rule] fs, FontSlant \[Rule] Italic}\)}], "]"}]}], ";"}]}], "Input"], Cell["Cherchons la vitesse minimale (maximale en valeur absolue) :", "Text", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell[BoxData[{ \(\(FindRoot[\(D[sol2[\([1, 1, 2]\)], {t, 2}] \[Equal] 0 /. dn\) /. t \[Rule] t - temps[\([2, 1, 2]\)], {t, 3}];\)\), "\[IndentingNewLine]", \(\(\(D[sol2[\([1, 1, 2]\)], t] /. dn\) /. t \[Rule] t - temps[\([2, 1, 2]\)]\) /. t \[Rule] %[\([1, 2]\)] // Chop\)}], "Input"], Cell[BoxData[ \(\(-25.740853770498898`\)\)], "Output"] }, Open ]], Cell[TextData[StyleBox["La valeur indiqu\[EAcute]e pour la grandeur de la \ vitesse maximale (180 km/h) est incompatible avec une longueur \ d'\[EAcute]lastique de 28 m !", FontColor->GrayLevel[0]]], "Text", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}, FontColor->RGBColor[1, 0, 0]], Cell["Assemblons les deux horaires sur un m\[EHat]me graphique:", "Body", CellMargins->{{2.375, Inherited}, {Inherited, Inherited}}], Cell[BoxData[ \(\(Show[h1, h2, PlotRange \[Rule] All, AxesOrigin \[Rule] {0, 0}, TextStyle \[Rule] {FontFamily \[Rule] Verdana, FontSize \[Rule] fs, FontSlant \[Rule] Italic}];\)\)], "Input"], Cell["\<\ Horaire : en bleu premi\[EGrave]re phase du saut (\[EAcute]lastique \ d\[EAcute]tendu); en rouge deuxi\[EGrave]me phase (\[EAcute]lastique \ tendu).\ \>", "Text"] }, Open ]] }, Open ]] }, FrontEndVersion->"5.2 for Macintosh", ScreenRectangle->{{0, 992}, {0, 746}}, ScreenStyleEnvironment->"Printout", PrintingStyleEnvironment->"Printout", WindowToolbars->{"RulerBar", "EditBar"}, WindowSize->{885, 719}, WindowMargins->{{4, Automatic}, {Automatic, 1}}, PrintingCopies->2, PrintingPageRange->{2, 2}, PageHeaders->{{ "Club Math\[EAcute]matiques Appliqu\[EAcute]es & Sciences", Inherited, Inherited}, { "http://p7app.geneve.ch:8007/math", Inherited, "SEM-prospective. Raymond Morel"}}, PageFooters->{{ "Document disponible sur http://Hypatie.ge.ch", Inherited, Cell[ TextData[ {"bernard.vuilleumier@edu.ge.ch", " ", ValueBox[ "Date"]}], "Footer"]}, {Cell[ TextData[ {"Club Math Appl & Sciences.", " ", ValueBox[ "FileName"]}], "Footer"], Inherited, Cell[ TextData[ {"bernard.vuilleumier@edu.ge.ch", " ", ValueBox[ "Date"]}], "Footer"]}}, PageFooterLines->{True, True}, PrintingOptions->{"PrintingMargins"->{{85, 56.6875}, {48.1875, 48.1875}}, "PrintCellBrackets"->False, "PrintRegistrationMarks"->True, "PrintMultipleHorizontalPages"->False, "FacingPages"->True}, ShowSelection->True, ShowCellLabel->False, FontFamily->"Lucida Grande", Magnification->1.5 ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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