{"id":26259,"date":"2023-06-06T17:04:09","date_gmt":"2023-06-06T16:04:09","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=26259"},"modified":"2023-06-07T22:52:12","modified_gmt":"2023-06-07T21:52:12","slug":"caminhando","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=26259","title":{"rendered":"Caminhando"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_26259' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_26259' class='GTTabs_curr'><a  id=\"26259_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_26259' ><a  id=\"26259_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_26259'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"26261\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=26261\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13.png\" data-orig-size=\"731,555\" 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;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"8_Pag210-T13\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13.png\" class=\"alignright wp-image-26261\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13-300x228.png\" alt=\"\" width=\"320\" height=\"243\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13-300x228.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13.png 731w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><\/a>A figura mostra as pegadas de um homem. O comprimento do passo, <em>P<\/em>, \u00e9 a dist\u00e2ncia entre a parte de tr\u00e1s de duas pegadas consecutivas.<\/p>\n<p>Para os homens, a f\u00f3rmula estabelece uma rela\u00e7\u00e3o aproximada entre <em>n<\/em> e <em>P<\/em>, em que<\/p>\n<ul>\n<li><em>n<\/em> = n\u00famero de passos por minuto, e<\/li>\n<li><em>P<\/em> = comprimento do passo em metros.<\/li>\n<\/ul>\n<ol>\n<li>Se esta f\u00f3rmula se aplicar ao caminhar do Pedro e ele der 70 passos por minuto, qual \u00e9 o comprimento, em cent\u00edmetros, do passo do Pedro? Apresenta os c\u00e1lculos que efetuares.<\/li>\n<li>O Bernardo sabe que o comprimento do seu passo \u00e9 de 0,80 metros. A f\u00f3rmula aplica-se ao caminhar do Bernardo. Calcula, em metros por minuto e em quil\u00f3metros por hora, a velocidade a que o Bernardo caminha.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_26259' onClick='GTTabs_show(1,26259)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_26259'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><\/p>\n<blockquote>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"26261\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=26261\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13.png\" data-orig-size=\"731,555\" 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;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"8_Pag210-T13\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13.png\" class=\"alignright wp-image-26261\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13-300x228.png\" alt=\"\" width=\"320\" height=\"243\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13-300x228.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13.png 731w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><\/a>A figura mostra as pegadas de um homem. O comprimento do passo, <em>P<\/em>, \u00e9 a dist\u00e2ncia entre a parte de tr\u00e1s de duas pegadas consecutivas.<\/p>\n<p>Para os homens, a f\u00f3rmula estabelece uma rela\u00e7\u00e3o aproximada entre <em>n<\/em> e <em>P<\/em>, em que<\/p>\n<\/blockquote>\n<ul>\n<li>\n<blockquote><em>n<\/em> = n\u00famero de passos por minuto, e<\/blockquote>\n<\/li>\n<li>\n<blockquote><em>P<\/em> = comprimento do passo em metros.<\/blockquote>\n<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ol>\n<li>Substituindo <em>n<\/em> por 70, vem: \\(\\begin{array}{*{20}{c}}{\\frac{{70}}{P} = 140}&amp; \\Leftrightarrow &amp;{P = \\frac{{70}}{{140}}}&amp; \\Leftrightarrow &amp;{P = 0,5}\\end{array}\\).<br>Se o Pedro der 70 passos por minuto, o comprimento do passo do Pedro \u00e9 50 cent\u00edmetros.<br><br><\/li>\n<li>Substituindo P por 0,80, vem: \\(\\begin{array}{*{20}{c}}{\\frac{n}{{0,8}} = 140}&amp; \\Leftrightarrow &amp;{n = 0,8 \\times 140}&amp; \\Leftrightarrow &amp;{n = }\\end{array}112\\).<br>Portanto, o Bernardo d\u00e1 112 passos por minuto.<br>Ou seja, o Bernardo caminha \u00e0 velocidade de \\(\\left( {0,8 \\times 112} \\right)\\) m\/min = \\(89,6\\) m\/min, que \u00e9 equivalente a \\(\\frac{{89,6\\,{\\rm{m}}}}{{1\\,{\\rm{min}}}} = \\frac{{89,6 \\times 60\\,{\\rm{m}}}}{{1 \\times 60\\,{\\rm{min}}}} = \\frac{{5376\\,{\\rm{m}}}}{{60\\,{\\rm{min}}}} = 5,376\\;{\\rm{km\/h}}\\).<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_26259' onClick='GTTabs_show(0,26259)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado A figura mostra as pegadas de um homem. O comprimento do passo, P, \u00e9 a dist\u00e2ncia entre a parte de tr\u00e1s de duas pegadas consecutivas. Para os homens, a f\u00f3rmula estabelece&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":26262,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,714],"tags":[424,192,345],"series":[],"class_list":["post-26259","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-equacoes-literais-e-sistemas","tag-8-o-ano","tag-equacoes-literais","tag-funcao-afim"],"views":134,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2023\/06\/8_Pag210-T13_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/26259","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=26259"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/26259\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/26262"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=26259"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=26259"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=26259"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=26259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}