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Álvaro Antônio de Lacerda Rosário 2025-01-05 11:29:17 -03:00
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Física/1. Tópicos Fundamentais/Exercícios/Cinemática/Movimento uniforme.md
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@ -258,20 +258,23 @@ $$
\begin{gathered} \begin{gathered}
x(2) = 9,75 + 1,5 \cdot 8 = 21,75cm \\ x(2) = 9,75 + 1,5 \cdot 8 = 21,75cm \\
x(3) = 9,75 + 1,5 \cdot 8 = 50,25cm \\ x(3) = 9,75 + 1,5 \cdot 8 = 50,25cm \\
\\ \end{gathered}
\begin{align}
\dfrac{x(2) + x(3)}{2} = \dfrac{21,75 + 50,25}{2} = 36 \\ \dfrac{x(2) + x(3)}{2} = \dfrac{21,75 + 50,25}{2} = 36 \\
\\ \end{a}
36 = 9,75 + 1,50 t^3 \Rightarrow \\ 36 = 9,75 + 1,50 t^3 \Rightarrow \\
1,50t ^3 = 36 - 9,75 \Rightarrow \\ 1,50t ^3 = 36 - 9,75 \Rightarrow \\
1,50t ^3 = 26,25 \Rightarrow \\ 1,50t ^3 = 26,25 \Rightarrow \\
t^3 = 17,5 \Rightarrow \\ t^3 = 17,5 \Rightarrow \\
t = \sqrt[3]{17,5} t = \sqrt[3]{17,5} \\
\\ \\
f'(t) = 4,5 t ^2 \Rightarrow \\ f'(t) = 4,5 t ^2 \Rightarrow \\
f'(\sqrt[3]{17,5}) = 4,5 \cdot (\sqrt[3]{17,5}) ^2 = 4,5 \cdot \sqrt[3]{17,5^2} \Rightarrow \\ f'(\sqrt[3]{17,5}) = 4,5 \cdot (\sqrt[3]{17,5}) ^2 = 4,5 \cdot \sqrt[3]{17,5^2} \Rightarrow \\
f'(3) = 4,5 \cdot \sqrt[3]{17,5^3 \cdot 17,5^{-1}} \Rightarrow \\ f'(\sqrt[3]{17,5}) = 4,5 \cdot \sqrt[3]{17,5^3 \cdot 17,5^{-1}} \Rightarrow \\
f'(3) = 4,5 \cdot \sqrt[3]{17,5^3 \cdot 17,5^{-1}} \Rightarrow \\ f'(\sqrt[3]{17,5}) = 4,5 \cdot 17,5^{-1} \Rightarrow \\
\end{gathered} f'(\sqrt[3]{17,5}) = 4,5 \cdot \dfrac{1}{17,5} = \dfrac{4,5}{17,5} \Rightarrow \\
f'(\sqrt[3]{17,5}) = 4,5 \cdot \dfrac{1}{17,5} = \dfrac{4,5}{17,5} \Rightarrow \\
$$ $$
🧐 **Resposta:** $2,1m/s$ 🧐 **Resposta:** $2,1m/s$