Admitindo que a dist\u00e2ncia (em dec\u00edmetros) do ponto A ao ponto C, $t$ segundos ap\u00f3s o instante em que o carrinho foi solto, \u00e9 dada em fun\u00e7\u00e3o do tempo $t$ (em segundos) por $$\\begin{array}{*{20}{c}}
\n{d(t) = 4{e^{ – 0,2t}}\\cos \\left( {\\frac{\\pi }{3}t} \\right) + 8}&{({\\text{com }}t \\geqslant 0)}
\n\\end{array}$$<\/p><\/blockquote>\n
\u00ad<\/p>\n
\n- No instante em que o carrinho \u00e9 solto, o ponto C encontra-se a $d(0) = 4{e^0}\\cos \\left( {\\frac{\\pi }{3} \\times 0} \\right) + 8 = 4 \\times 1 + 8 = 12$ dec\u00edmetros do ponto A.
\n\u00ad<\/li>\n - No contexto da situa\u00e7\u00e3o, a equa\u00e7\u00e3o permite determinar os instantes\u00a0em que o carrinho passa pelo ponto de equil\u00edbrio.
\n$$\\begin{array}{*{20}{l}}
\n{d(t) = 8}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{4{e^{ – 0,2t}}\\cos \\left( {\\frac{\\pi }{3}t} \\right) + 8 = 8}& \\wedge &{t \\geqslant 0}
\n\\end{array}} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{4{e^{ – 0,2t}}\\cos \\left( {\\frac{\\pi }{3}t} \\right) = 0}& \\wedge &{t \\geqslant 0}
\n\\end{array}} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{\\cos \\left( {\\frac{\\pi }{3}t} \\right) = 0}& \\wedge &{t \\geqslant 0}
\n\\end{array}} \\\\
\n{}& \\Leftrightarrow &{\\frac{\\pi }{3}t = \\frac{\\pi }{2} + k\\pi ,k \\in \\mathbb{Z}_0^ + } \\\\
\n{}& \\Leftrightarrow &{t = \\frac{3}{2} + 3k,k \\in \\mathbb{Z}_0^ + }
\n\\end{array}$$
\n\u00ad<\/li>\n - A fun\u00e7\u00e3o $d$ \u00e9 cont\u00ednua no seu dom\u00ednio, logo \u00e9 tamb\u00e9m cont\u00ednua no intervalo $\\left[ {0,\\frac{3}{2}} \\right]$.
\nComo $d(0) = 12$ e $d(\\frac{3}{2}) = 0$, ent\u00e3o $d(\\frac{3}{2}) < 10 < d(0)$.
\nLogo, de acordo com o teorema de Bolzano, $\\exists t \\in \\left] {0,\\frac{3}{2}} \\right[:d(t) = 10$.
\nPortanto, existe pelo menos um instante em que o ponto C esteve a 10 dec\u00edmetros do ponto A.
\n\u00ad<\/li>\n - Ora, $$\\begin{array}{*{20}{l}}
\n{\\mathop {\\lim }\\limits_{t \\to\u00a0 + \\infty } d(t)}& = &{\\mathop {\\lim }\\limits_{t \\to\u00a0 + \\infty } \\left( {4{e^{ – 0,2t}}\\cos \\left( {\\frac{\\pi }{3}t} \\right) + 8} \\right)} \\\\
\n{}& = &{\\underbrace {\\mathop {\\lim }\\limits_{t \\to\u00a0 + \\infty } \\left( {4{e^{ – 0,2t}}\\cos \\left( {\\frac{\\pi }{3}t} \\right)} \\right)}_{0\\,\\,(*)} + 8} \\\\
\n{}& = &8
\n\\end{array}$$
\n(*) Ainda que n\u00e3o exista $\\mathop {\\lim }\\limits_{t \\to\u00a0 + \\infty } \\cos \\left( {\\frac{\\pi }{3}t} \\right)$, tem-se que $ – 1 \\leqslant \\cos \\left( {\\frac{\\pi }{3}t} \\right) \\leqslant 1,\\forall t \\in \\mathbb{R}$ e $\\mathop {\\lim }\\limits_{t \\to\u00a0 + \\infty } 4{e^{ – 0,2t}} = 0$.
\nO carrinho vai oscilando em torno do ponto de equil\u00edbrio e, com o decorrer do tempo, tende a parar a 8 dm do ponto A.
\n\u00ad<\/li>\n<\/ol>\n
![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag126-3.jpg\" "\\"Carrinho\\"")
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