{"id":28474,"date":"2021-09-08T13:00:45","date_gmt":"2021-09-08T13:00:45","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28474"},"modified":"2024-09-16T11:43:09","modified_gmt":"2024-09-16T11:43:09","slug":"mouvement-rectiligne-uniformement-accelere-mrua","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/mouvement-rectiligne-uniformement-accelere-mrua\/","title":{"rendered":"Mouvement Rectiligne Uniform\u00e9ment Acc\u00e9l\u00e9r\u00e9 (MRUA)"},"content":{"rendered":"<p><center><\/p>\n<h1>Mouvement Rectiligne Uniform\u00e9ment Acc\u00e9l\u00e9r\u00e9 (MRUA)<\/h1>\n<p>    <em><strong>R\u00e9sum\u00e9 :<\/strong><br \/>\n    Dans cette le\u00e7on, nous passerons en revue le concept de mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 (MRUA). Nous expliquerons comment ce type de mouvement implique une acc\u00e9l\u00e9ration constante en ligne droite et est mod\u00e9lis\u00e9 \u00e0 l&#8217;aide d&#8217;\u00e9quations obtenues par int\u00e9gration.<\/em><\/p>\n<p>    <strong>Objectifs d&#8217;apprentissage :<\/strong><br \/>\n    \u00c0 la fin de cette le\u00e7on, l&#8217;\u00e9tudiant sera capable de :<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Comprendre<\/strong> le concept de mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 (MRUA) et ses caract\u00e9ristiques.<\/li>\n<li><strong>D\u00e9duire<\/strong> les \u00e9quations de trajet pour le MRUA \u00e0 partir de l&#8217;acc\u00e9l\u00e9ration constante.<\/li>\n<li><strong>Appliquer<\/strong> les \u00e9quations du MRUA pour analyser et r\u00e9soudre des probl\u00e8mes de mouvement unidimensionnel.<\/li>\n<li><strong>Interpr\u00e9ter<\/strong> les conditions initiales et les constantes dans les \u00e9quations du MRUA.<\/li>\n<\/ol>\n<p>    <strong>INDEX DES CONTENUS<\/strong><br \/>\n    <a href=\"#1\">Qu&#8217;est-ce que le mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 ?<\/a><br \/>\n    <a href=\"#2\">Le MRUA et le cas de la chute libre<\/a><br \/>\n    <a href=\"#3\">Exercices relatifs au mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9<\/a><br \/>\n<\/center><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/P21bsNFF9Fw\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Qu&#8217;est-ce que le mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 ?<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw&amp;t=143s\" target=\"_blank\" rel=\"noopener\"><strong>Le mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9<\/strong><\/a>, abr\u00e9g\u00e9 en MRUA, est un type de mouvement que nous avons d\u00e9j\u00e0 \u00e9tudi\u00e9 de mani\u00e8re implicite. Vous le verrez lorsque nous examinerons comment il est mod\u00e9lis\u00e9 \u00e0 partir des \u00e9quations de trajet. Mais si nous voulons une description rapide, le MRUA est un type de mouvement o\u00f9 l&#8217;acc\u00e9l\u00e9ration est constante, tant en magnitude qu&#8217;en direction, et qui se d\u00e9veloppe sur une ligne droite, c&#8217;est-\u00e0-dire dans une dimension.<\/p>\n<h3>Mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 \u00e0 partir des \u00e9quations de trajet<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw&amp;t=395s\" target=\"_blank\" rel=\"noopener\"><strong>La d\u00e9duction du MRUA<\/strong><\/a> est une copie directe du travail que nous avons r\u00e9alis\u00e9 pour obtenir les \u00e9quations de trajet par int\u00e9gration dans les le\u00e7ons pr\u00e9c\u00e9dentes. Comme le MRUA est un mouvement avec une acc\u00e9l\u00e9ration constante et unidimensionnelle, il suffit de faire les d\u00e9ductions sur un seul axe de coordonn\u00e9es ; si nous raisonnons sur l&#8217;axe <span class=\"katex-eq\" data-katex-display=\"false\">\\hat{x}<\/span>, nous obtenons ce qui suit :<\/p>\n<p style=\"text-align: justify;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\na_x(t) &amp; =&amp; a_{0x} \\\\ \\\\\n\nv_x(t) &amp; =&amp;  \\int a_{0x}dt = a_{0x}t + v_{0x} \\\\ \\\\\n\nx(t) &amp; =&amp; \\displaystyle \\int v_{x}(t)dt = \\frac{1}{2}a_{0x}t^2 + v_{0x}t + x_0\n\n\\end{array}\n\n<\/span>\n<p style=\"text-align: justify;\">Ici, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a_{0x}, v_{0x}<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span> sont toutes des constantes, et les deux derni\u00e8res sont des constantes d&#8217;int\u00e9gration. Avec cela, nous avons un mod\u00e8le complet du mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 dans la direction de l&#8217;axe <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\hat{x}<\/span><\/span>. Le raisonnement pour tout autre axe est enti\u00e8rement analogue.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Le MRUA et le cas de la chute libre<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw&amp;t=512s\" target=\"_blank\" rel=\"noopener\"><strong>L&#8217;un des cas les plus repr\u00e9sentatifs<\/strong><\/a> du MRUA est la chute libre. C&#8217;est un mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 qui se d\u00e9veloppe verticalement et qui est provoqu\u00e9 par l&#8217;acc\u00e9l\u00e9ration de la gravit\u00e9. Son mod\u00e8le \u00e0 travers les \u00e9quations de trajet est le suivant :<\/p>\n<p style=\"text-align: justify;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\na_y(t) &amp; =&amp;  -g \\\\ \\\\\n\nv_y(t) &amp; =&amp;  -gt + v_{0y} \\\\ \\\\\n\ny(t) &amp; =&amp; \\displaystyle - \\frac{1}{2}gt^2 + v_{0y}t+ y_0\n\n\\end{array}\n\n<\/span>\n<p style=\"text-align: justify;\">Ici, l&#8217;acc\u00e9l\u00e9ration de la gravit\u00e9 est <span class=\"katex-eq\" data-katex-display=\"false\">g=9,81[m\/s^2]<\/span>. Dans la chute libre, il est typique que cela commence initialement au repos (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v_{0y}=0<\/span><\/span>) et \u00e0 une hauteur initiale <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y_0=h<\/span><\/span>, de sorte que les \u00e9quations se r\u00e9duisent \u00e0<\/p>\n<p style=\"text-align: justify;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\na_y(t) &amp; =&amp; -g \\\\ \\\\\n\nv_y(t) &amp; =&amp; -gt \\\\ \\\\\n\ny(t) &amp; =&amp; \\displaystyle - \\frac{1}{2}gt^2 + h\n\n\\end{array}\n\n<\/span>\n<p style=\"text-align: justify;\">Quel que soit l&#8217;ensemble d&#8217;\u00e9quations que vous ayez, il est possible d&#8217;extraire des informations en \u00abposant les bonnes questions\u00bb aux \u00e9quations.<\/p>\n<h4>Si un corps part du repos \u00e0 une hauteur <span class=\"katex-eq\" data-katex-display=\"false\">h<\/span><\/h4>\n<h5>Combien de temps met-il pour tomber ?<\/h5>\n<p style=\"text-align: justify;\">Si nous posons cette question aux \u00e9quations, elles nous diront que \u00able corps touche le sol lorsque la hauteur est z\u00e9ro\u00bb, c&#8217;est-\u00e0-dire <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(t)=0.<\/span><\/span> Si cela se produit, alors nous devons r\u00e9soudre le temps dans l&#8217;\u00e9quation <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{1}{2}gt^2 + h = 0.<\/span><\/span> \u00c0 partir de cela, deux r\u00e9sultats possibles sont obtenus :<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle t=\\pm\\sqrt{\\frac{2h}{g}}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Le temps n\u00e9gatif regarde vers le pass\u00e9, et le positif vers le futur. Comme la chute se produit dans le futur, nous pouvons d\u00e9finir le temps de chute comme<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle t_{chute}=+\\sqrt{\\frac{2h}{g}}<\/span><\/span><\/p>\n<h6>\u00c0 quelle vitesse arrive-t-il au sol ?<\/h6>\n<p style=\"text-align: justify;\">Nous pouvons r\u00e9pondre \u00e0 cette question simplement en rempla\u00e7ant le temps de chute dans l&#8217;\u00e9quation de la vitesse. Si nous faisons cela, nous obtenons la vitesse de chute :<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle v_{chute} = v_y(t_{chute})=-g\\sqrt{\\frac{2h}{g}}=-\\sqrt{\\frac{2g^2h}{g}} = -\\sqrt{2gh}<\/span><\/span><\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Exercices relatifs au mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9<\/h2>\n<ol style=\"text-align: justify;\">\n<li>Un mobile passe par l&#8217;origine avec une vitesse initiale <span class=\"katex-eq\" data-katex-display=\"false\">v_0=10[km\/h]<\/span> et avec une acc\u00e9l\u00e9ration de <span class=\"katex-eq\" data-katex-display=\"false\">a_0=\\displaystyle \\frac{20[km\/h]}{5[s]}.<\/span>\nCalculez la position et la vitesse du mobile aux instants a) <span class=\"katex-eq\" data-katex-display=\"false\">t=5[s],<\/span> b) <span class=\"katex-eq\" data-katex-display=\"false\">t=10[s],<\/span> c) <span class=\"katex-eq\" data-katex-display=\"false\">t=15[s]<\/span> et d) <span class=\"katex-eq\" data-katex-display=\"false\">t=1[min].<\/span> <a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw&amp;t=1329s\" target=\"_blank\" rel=\"noopener\"><strong>[SOLUTION]<\/strong><\/a><\/li>\n<li>Une personne laisse tomber une boule d&#8217;acier et une pierre d&#8217;une hauteur de 20[m] en m\u00eame temps depuis le repos. Les deux objets ont les m\u00eames dimensions, mais des poids diff\u00e9rents. Combien de temps mettent-ils pour tomber et \u00e0 quelle vitesse se d\u00e9placent-ils au moment de l&#8217;impact avec le sol ? Un de ces corps peut-il tomber plus vite que l&#8217;autre ou arriver avec une plus grande vitesse ? <a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw&amp;t=1873s\" target=\"_blank\" rel=\"noopener\"><strong>[SOLUTION]<\/strong><\/a><\/li>\n<li>Une pi\u00e8ce est jet\u00e9e au fond d&#8217;un puits. Le bruit indiquant que la pi\u00e8ce est arriv\u00e9e au fond est entendu apr\u00e8s 10 [s]. Quelle est la profondeur du puits ? <a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw&amp;t=1972s\" target=\"_blank\" rel=\"noopener\"><strong>[SOLUTION]<\/strong><\/a><\/li>\n<li>Une personne crache verticalement dans le ciel et en 1,2[s], elle retombe sur son visage. a) \u00c0 quelle vitesse a-t-elle crach\u00e9 ? b) Quelle hauteur le crachat a-t-il atteint ? <a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw&amp;t=2090s\" target=\"_blank\" rel=\"noopener\"><strong>[SOLUTION]<\/strong><\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Mouvement Rectiligne Uniform\u00e9ment Acc\u00e9l\u00e9r\u00e9 (MRUA) R\u00e9sum\u00e9 : Dans cette le\u00e7on, nous passerons en revue le concept de mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 (MRUA). Nous expliquerons comment ce type de mouvement implique une acc\u00e9l\u00e9ration constante en ligne droite et est mod\u00e9lis\u00e9 \u00e0 l&#8217;aide d&#8217;\u00e9quations obtenues par int\u00e9gration. Objectifs d&#8217;apprentissage : \u00c0 la fin de cette le\u00e7on, l&#8217;\u00e9tudiant [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28458,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":66,"footnotes":""},"categories":[663,647],"tags":[],"class_list":["post-28474","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fondements-de-la-mecanique","category-physique"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Mouvement Rectiligne Uniform\u00e9ment Acc\u00e9l\u00e9r\u00e9 (MRUA) - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Le mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9 (MRUA) est un mouvement en ligne droite o\u00f9 un objet subit une acc\u00e9l\u00e9ration constante au fil du temps.\" \/>\n<meta name=\"robots\" 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