Uma partícula em movimento retilíneo tem a função horária dada por s(t) = 2t3 – 21t2 + 60t + 3. Considere o espaço medido em metros e o tempo em segundos. A aceleração da partícula no instante t = 4,5s é:
a.
5 m/s2
b.
2 m/s2
c.
12 m/s2
d.
4,5 m/s2
e.
1 m/s2
Respostas
respondido por:
4
Boa Tarde Leticia!
Solução!
Para resolver o exercício vamos derivar a função duas vezes.
![t=4,5s\\\\\\\
s(t)=2t^{3}-21t^{2} +60t+3\\\\\\
s'(t)=6t^{2}-42t+60\\\\\\
A ~~segunda~~derivada ~~e~~a~~equac\~ao~~da~~acelerac\~ao.\\\\\\\
s''(t)=12t-42\\\\\\\
s''(4,5)=12.(4,5)-21\\\\\\\
s''(4,5)=54-42\\\\\\
\boxed{s''(4,5)=12}](https://tex.z-dn.net/?f=t%3D4%2C5s%5C%5C%5C%5C%5C%5C%5C%0A%0As%28t%29%3D2t%5E%7B3%7D-21t%5E%7B2%7D+%2B60t%2B3%5C%5C%5C%5C%5C%5C%0As%27%28t%29%3D6t%5E%7B2%7D-42t%2B60%5C%5C%5C%5C%5C%5C%0AA+%7E%7Esegunda%7E%7Ederivada+%7E%7Ee%7E%7Ea%7E%7Eequac%5C%7Eao%7E%7Eda%7E%7Eacelerac%5C%7Eao.%5C%5C%5C%5C%5C%5C%5C%0As%27%27%28t%29%3D12t-42%5C%5C%5C%5C%5C%5C%5C%0As%27%27%284%2C5%29%3D12.%284%2C5%29-21%5C%5C%5C%5C%5C%5C%5C%0As%27%27%284%2C5%29%3D54-42%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bs%27%27%284%2C5%29%3D12%7D%0A+ "t=4,5s\\\\\\\
s(t)=2t^{3}-21t^{2} +60t+3\\\\\\
s'(t)=6t^{2}-42t+60\\\\\\
A ~~segunda~~derivada ~~e~~a~~equac\~ao~~da~~acelerac\~ao.\\\\\\\
s''(t)=12t-42\\\\\\\
s''(4,5)=12.(4,5)-21\\\\\\\
s''(4,5)=54-42\\\\\\
\boxed{s''(4,5)=12}")
Boa tarde!
Bons estudos!
Solução!
Para resolver o exercício vamos derivar a função duas vezes.
Boa tarde!
Bons estudos!
leticiasatori:
Muito obrigada !!!!
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