{"id":7800,"date":"2012-04-01T16:31:26","date_gmt":"2012-04-01T15:31:26","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7800"},"modified":"2022-01-30T18:57:42","modified_gmt":"2022-01-30T18:57:42","slug":"uma-esfera-metalica","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7800","title":{"rendered":"Uma esfera met\u00e1lica"},"content":{"rendered":"<p><ul id='GTTabs_ul_7800' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7800' class='GTTabs_curr'><a  id=\"7800_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7800' ><a  id=\"7800_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_7800'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Uma esfera met\u00e1lica M move-se sobre uma reta r durante 12 segundos.<\/p>\n<p>A sua posi\u00e7\u00e3o em rela\u00e7\u00e3o ao ponto O, em fun\u00e7\u00e3o do tempo, \u00e9 dada pela equa\u00e7\u00e3o $$d(t) = {t^3} &#8211; 16{t^2} + 50t + 40$$ com $d$ em cent\u00edmetros.<\/p>\n<p>Uma posi\u00e7\u00e3o $-1$ significa que a esfera se encontra 1 cent\u00edmetro \u00e0 esquerda de O e $+1$ significa que se encontra 1 cent\u00edmetro \u00e0 direita de O.<\/p>\n<ol>\n<li>No instante inicial, em que posi\u00e7\u00e3o se encontra a esfera? E no instante final?<\/li>\n<li>Qual foi a dist\u00e2ncia m\u00e1xima da esfera a O? Em que instante se verificou?<\/li>\n<li>Indique os intervalos de tempo em que a esfera se desloca para a esquerda e para a direita.<\/li>\n<li>Em que instante \u00e9 zero a velocidade da esfera? E quando \u00e9 m\u00ednima a velocidade?<br \/>\nComo interpreta, no gr\u00e1fico, estes resultados?<\/li>\n<li>Represente, graficamente, a posi\u00e7\u00e3o e a velocidade no intervalo considerado.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7800' onClick='GTTabs_show(1,7800)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7800'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Como $d(0) = 40$ e $d(12) = {12^3} &#8211; 16 \\times {12^2} + 50 \\times 12 + 40 = 64$, a esfera encontra-se a 40 cent\u00edmetros \u00e0 direita de O, no instante inicial, e a 64 cent\u00edmetros \u00e0 direita de O, no instante final.<br \/>\n\u00ad<\/li>\n<li>Ora, $v(t) = d&#8217;(t) = 3{t^2} &#8211; 32t + 50$, cujos zeros s\u00e3o:$$\\begin{array}{*{20}{l}}<br \/>\n{v(t) = 0}&amp; \\Leftrightarrow &amp;{t = \\frac{{32 \\pm \\sqrt {{{32}^2} &#8211; 600} }}{6}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{t = \\frac{{32 \\pm \\sqrt {424} }}{6}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{t = \\frac{{16 &#8211; \\sqrt {106} }}{3} \\vee t = \\frac{{16 + \\sqrt {106} }}{3}}<br \/>\n\\end{array}$$<\/p>\n<table class=\" aligncenter\" style=\"width: 80%;\" border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$t$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$0$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\"><\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">${\\frac{{16 &#8211; \\sqrt {106} }}{3}}$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\"><\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">${\\frac{{16 + \\sqrt {106} }}{3}}$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\"><\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$12$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$d(t)$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$40$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$ \\nearrow $<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$\\frac{{88 + 212\\sqrt {106} }}{{27}}$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$ \\searrow $<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$\\frac{{88 &#8211; 212\\sqrt {106} }}{{27}}$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$ \\nearrow $<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$64$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$v(t)$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$+$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$+$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$0$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$-$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$0$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$+$<\/td>\n<td style=\"text-align: center; border: #2f4f4f 1px solid;\">$+$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>$${t_1} = \\frac{{16 &#8211; \\sqrt {106} }}{3} \\approx 1,9$$<br \/>\n$${t_2} = \\frac{{16 + \\sqrt {106} }}{3} \\approx 8,8$$<br \/>\n$$d({t_1}) = d(\\frac{{16 &#8211; \\sqrt {106} }}{3}) = \\frac{{88 + 212\\sqrt {106} }}{{27}} \\approx 84,1$$<br \/>\n$$d({t_2}) = d(\\frac{{16 + \\sqrt {106} }}{3}) = \\frac{{88 &#8211; 212\\sqrt {106} }}{{27}} \\approx\u00a0 &#8211; 77,6$$<br \/>\nA dist\u00e2ncia m\u00e1xima da esfera ao ponto O foi de 84 metros, aproximadamente, cerca de 2 segundos ap\u00f3s o in\u00edcio do movimento.<br \/>\n\u00ad<\/li>\n<li>A esfera desloca-se para a direita quando a velocidade \u00e9 positiva e desloca-se para a esquerda quando a velocidade \u00e9 negativa.\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12p215-51G.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"7859\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=7859\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12p215-51G.png\" data-orig-size=\"682,462\" 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\/2012\/04\/12p215-51G.png\" class=\"aligncenter wp-image-7859 size-full\" title=\"Gr\u00e1fico\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12p215-51G.png\" alt=\"\" width=\"682\" height=\"462\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12p215-51G.png 682w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12p215-51G-300x203.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12p215-51G-150x101.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12p215-51G-400x270.png 400w\" sizes=\"auto, (max-width: 682px) 100vw, 682px\" \/><\/a><\/p>\n<p>Logo, a esfera desloca-se para a direita nos intervalos $$\\left] {0,\\frac{{16 &#8211; \\sqrt {106} }}{3}} \\right[ \\approx \\left] {0;1,9} \\right[$$ e $$\\left] {\\frac{{16 + \\sqrt {106} }}{3},12} \\right[ \\approx \\left] {8,8;12} \\right[$$ Desloca-se para a esquerda no intervalo $$\\left] {\\frac{{16 &#8211; \\sqrt {106} }}{3},\\frac{{16 + \\sqrt {106} }}{3}} \\right[ \\approx \\left] {1,9;8,8} \\right[$$<br \/>\n\u00ad<\/li>\n<li>A velocidade \u00e9 nula para $${t_1} = \\frac{{16 &#8211; \\sqrt {106} }}{3} \\approx 1,9$$ e $${t_2} = \\frac{{16 + \\sqrt {106} }}{3} \\approx 8,8$$ em segundos.<br \/>\nEstes instantes correspondem \u00e0queles em o movimento da esfera muda de sentido.<\/p>\n<p>Ora, $a(t) = v'(t) = 6t &#8211; 32$, cujo zero \u00e9: $a(t) = 0 \\Leftrightarrow t = \\frac{{16}}{3}$.<br \/>\nPortanto, a velocidade \u00e9 m\u00ednima para ${t_3} = \\frac{{16}}{3} \\approx 5,3$, em segundos.<br \/>\n\u00ad<\/li>\n<li><\/li>\n<\/ol>\n<p style=\"text-align: center;\"><script src=\"https:\/\/cdn.geogebra.org\/apps\/deployggb.js\"><\/script>\r\n<div id=\"ggbApplet\" style=\"margin: 0 auto;\"><\/div>\r\n<script>\r\nvar parameters = {\r\n\"id\": 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