{"id":6654,"date":"2011-03-22T22:44:29","date_gmt":"2011-03-22T22:44:29","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6654"},"modified":"2022-01-25T00:21:36","modified_gmt":"2022-01-25T00:21:36","slug":"um-objecto-move-se-ao-longo-de-uma-recta","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6654","title":{"rendered":"Um objecto move-se ao longo de uma reta"},"content":{"rendered":"<p><ul id='GTTabs_ul_6654' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6654' class='GTTabs_curr'><a  id=\"6654_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6654' ><a  id=\"6654_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<li id='GTTabs_li_2_6654' ><a  id=\"6654_2\" onMouseOver=\"GTTabsShowLinks('C\u00e1lculo de $d&#8217;(t)$'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>C\u00e1lculo de $d&#8217;(t)$<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_6654'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Um objeto move-se ao longo de uma reta e a sua dist\u00e2ncia, em cent\u00edmetros, a um ponto de refer\u00eancia fixo \u00e9 dada em fun\u00e7\u00e3o do tempo t, em segundos, por \\[\\begin{matrix}<br \/>\nd\\,(t)=2\\,t+\\frac{8}{t+1} &amp; (t\\ge 0)\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<p>Recorrendo exclusivamente a processos anal\u00edticos, resolva as tr\u00eas al\u00edneas seguintes.<\/p>\n<ol>\n<li>Determine o per\u00edodo de tempo durante o qual o objeto distou do ponto de refer\u00eancia 15 cm ou menos.<\/li>\n<li>Prove que a taxa m\u00e9dia de varia\u00e7\u00e3o de d no intervalo $\\left[ 1,\\ 3 \\right]$ \u00e9 1.<br \/>\nInterprete este valor no contexto da situa\u00e7\u00e3o descrita.<\/li>\n<li>Sabe-se que \\[d&#8217;(t)=2-\\frac{8}{{{(t+1)}^{2}}}=\\frac{2\\,(t-1)(t+3)}{{{(t+1)}^{2}}}\\] (d&#8217; designa a derivada de d)<br \/>\nVerifique se a fun\u00e7\u00e3o d tem um m\u00ednimo absoluto e, em caso afirmativo, determine-o.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6654' onClick='GTTabs_show(1,6654)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6654'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Ora, \\[\\begin{array}{*{35}{l}}<br \/>\n2\\,t+\\frac{8}{t+1}\\le 15 &amp; \\Leftrightarrow\u00a0 &amp; 2\\,t+\\frac{8}{t+1}-15\\le 0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\frac{2\\,{{t}^{2}}+2\\,t+8-15\\,t-15}{t+1}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\frac{2\\,{{t}^{2}}-13\\,t-7}{t+1}\\le 0\u00a0 \\\\<br \/>\n\\end{array}\\le 0\\]Como \\[\\begin{array}{*{35}{l}}<br \/>\n2{{t}^{2}}-13t-7=0 &amp; \\Leftrightarrow\u00a0 &amp; t=\\frac{13\\mp \\sqrt{169+56}}{4}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; t=\\frac{13\\mp 15}{4}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nt=7 &amp; \\vee\u00a0 &amp; t=-0,5\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n\\end{array}\\]vem:<\/p>\n<table class=\"aligncenter\" style=\"width: 70%;\" border=\"2\" cellspacing=\"4\" cellpadding=\"4\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\">$t$<\/td>\n<td style=\"text-align: center;\">$0$<\/td>\n<td><\/td>\n<td style=\"text-align: center;\">$7$<\/td>\n<td style=\"text-align: right;\">$+\\infty $<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">$2\\,{{t}^{2}}-13\\,t-7$<\/td>\n<td style=\"text-align: center;\">&#8211;<\/td>\n<td style=\"text-align: center;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0&#8211;<\/td>\n<td style=\"text-align: center;\">$0$<\/td>\n<td style=\"text-align: center;\">\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 +<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">$t+1$<\/td>\n<td style=\"text-align: center;\">+<\/td>\n<td style=\"text-align: center;\">+<\/td>\n<td style=\"text-align: center;\">+<\/td>\n<td style=\"text-align: center;\">+<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">$\\tfrac{2{{t}^{2}}-13t-7}{t+1}$<\/td>\n<td style=\"text-align: center;\">&#8211;<\/td>\n<td style=\"text-align: center;\">&#8211;<\/td>\n<td style=\"text-align: center;\">$0$<\/td>\n<td style=\"text-align: center;\">+<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Portanto, durante os primeiros 7 segundos, o objeto distou do ponto de refer\u00eancia 15 cm ou menos.<\/p>\n<\/li>\n<li>Ora, \\[tm{{v}_{\\left[ 1,\\ 3 \\right]}}=\\frac{d(3)-d(1)}{3-1}=\\frac{(6+\\tfrac{8}{4})-(2+\\tfrac{8}{2})}{2}=1\\]<br \/>\nA dist\u00e2ncia do objeto ao ponto de refer\u00eancia aumentou, em m\u00e9dia, 1 cent\u00edmetro por segundo, entre os instantes $t=1$ e $t=3$ (entre o primeiro e o terceiro segundos de observa\u00e7\u00e3o).<br \/>\n\u00ad<\/li>\n<li>Tendo em considera\u00e7\u00e3o as propriedades da fun\u00e7\u00e3o quadr\u00e1tica e reparando que ${{(t+1)}^{2}}&gt;0\\,,\\ \\forall t\\in \\mathbb{R}_{0}^{+}$, temos:<br \/>\n<table class=\"aligncenter\" style=\"width: 70%;\" border=\"2\" cellspacing=\"4\" cellpadding=\"4\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\">$t$<\/td>\n<td style=\"text-align: center;\">$0$<\/td>\n<td><\/td>\n<td style=\"text-align: center;\">$1$<\/td>\n<td style=\"text-align: right;\">$+\\infty $<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">$d&#8217;(t)=\\tfrac{2\\,(t-1)(t+3)}{{{(t+1)}^{2}}}$<\/td>\n<td style=\"text-align: center;\">&#8211;<\/td>\n<td style=\"text-align: center;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0&#8211;<\/td>\n<td style=\"text-align: center;\">$0$<\/td>\n<td style=\"text-align: center;\">\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 +<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">$d(t)$<\/td>\n<td style=\"text-align: center;\">$8$<\/td>\n<td style=\"text-align: center;\">$\\searrow $<\/td>\n<td style=\"text-align: center;\">$6$<\/td>\n<td style=\"text-align: center;\">$\\nearrow $<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Confirma-se que a fun\u00e7\u00e3o d tem um m\u00ednimo absoluto, que \u00e9 igual a $d(1)=2\\times 1+\\frac{8}{1+1}=6$.<\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/23-03-2011Ecra001.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6655\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6655\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/23-03-2011Ecra001.png\" data-orig-size=\"690,468\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Gr\u00e1fico\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/23-03-2011Ecra001.png\" class=\"aligncenter wp-image-6655\" title=\"Gr\u00e1fico\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/23-03-2011Ecra001.png\" alt=\"\" width=\"552\" height=\"374\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/23-03-2011Ecra001.png 690w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/23-03-2011Ecra001-300x203.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/23-03-2011Ecra001-150x101.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/23-03-2011Ecra001-400x271.png 400w\" sizes=\"auto, (max-width: 552px) 100vw, 552px\" \/><\/a><\/p>\n<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6654' onClick='GTTabs_show(0,6654)'>&lt;&lt; Enunciado<\/a><\/span><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6654' onClick='GTTabs_show(2,6654)'>C\u00e1lculo de $d&#8217;(t)$ &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_2_6654'>\n<span class='GTTabs_titles'><b>C\u00e1lculo de $d&#8217;(t)$<\/b><\/span><\/p>\n<p>Seja ${{t}_{0}}\\in \\mathbb{R}_{0}^{+}$ e $h\\ne 0$.<\/p>\n<p>\\[\\begin{array}{*{35}{l}}<br \/>\n\\frac{d({{t}_{0}}+h)-d({{t}_{0}})}{h} &amp; = &amp; \\frac{2({{t}_{0}}+h)+\\frac{8}{{{t}_{0}}+h+1}-2{{t}_{0}}-\\frac{8}{{{t}_{0}}+1}}{h}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{2h+\\frac{8}{{{t}_{0}}+h+1}-\\frac{8}{{{t}_{0}}+1}}{h}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 2+\\frac{\\frac{8{{t}_{0}}+8-8{{t}_{0}}-8h-8}{({{t}_{0}}+h+1)({{t}_{0}}+1)}}{h}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 2-\\frac{8h}{h({{t}_{0}}+h+1)({{t}_{0}}+1)}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 2-\\frac{8}{({{t}_{0}}+h+1)({{t}_{0}}+1)}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Quando $h\\to 0$, $\\frac{d({{t}_{0}}+h)-d({{t}_{0}})}{h}\\to 2-\\frac{8}{{{({{t}_{0}}+1)}^{2}}}$.<\/p>\n<p>Logo, \\[d&#8217;(t)=2-\\frac{8}{{{(t+1)}^{2}}}=\\frac{2{{(t+1)}^{2}}-8}{{{(t+1)}^{2}}}=2\\times \\frac{{{(t+1)}^{2}}-{{2}^{2}}}{{{(t+1)}^{2}}}=2\\times \\frac{(t+1-2)(t+1+2)}{{{(t+1)}^{2}}}=\\frac{2(t-1)(t+3)}{{{(t+1)}^{2}}}\\]<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6654' onClick='GTTabs_show(1,6654)'>&lt;&lt; Resolu\u00e7\u00e3o<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Um objeto move-se ao longo de uma reta e a sua dist\u00e2ncia, em cent\u00edmetros, a um ponto de refer\u00eancia fixo \u00e9 dada em fun\u00e7\u00e3o do tempo t, em segundos, por \\[\\begin{matrix}&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20938,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,134],"tags":[145,144],"series":[],"class_list":["post-6654","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-derivadas","tag-derivadas-2","tag-extremos-relativos"],"views":2941,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/11_ao-longo-de-uma-reta.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6654","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=6654"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6654\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20938"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6654"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6654"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6654"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6654"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}