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06/02/2020

Uma fórmula simples e útil no m.r.u.v.

Num movimento rectilíneo uniformemente variado (ou seja, $a$ constante) é válida a fórmula: \[v^2-v_0^2=2a\Delta x\] Consegue deduzi-la?

\[ \left\{ {\begin{array}{l} {x = x_0 + v_0 t + \displaystyle\frac{1}{2}at^2 } \\ {v = v_0 + at} \end{array}} \right. \] \[ \Leftrightarrow \left\{ {\begin{array}{l} {x - x_0 = \left( {v_0 + \displaystyle\frac{1}{2}at} \right)t} \\ {v - v_0 = at} \end{array}} \right. \] \[ \Leftrightarrow \left\{ {\begin{array}{l} {\Delta x = \left( {v_0 + \displaystyle\frac{{v - v_0 }}{2}} \right)\left( {\displaystyle\frac{{v - v_0 }}{a}} \right)} \\ {v - v_0 = at} \end{array}} \right. \] \[ \Leftrightarrow \left\{ {\begin{array}{l} {\Delta x = \left( {\displaystyle\frac{{v + v_0 }}{2}} \right)\left( {\displaystyle\frac{{v - v_0 }}{a}} \right)} \\ {v - v_0 = at} \end{array}} \right. \] \[ \Leftrightarrow \left\{ {\begin{array}{l} {\Delta x = \displaystyle\frac{{\left( {v + v_0 } \right)\left( {v - v_0 } \right)}}{2a}} \\ {v - v_0 = at} \end{array}} \right. \] \[ \Leftrightarrow \left\{ {\begin{array}{l} {2a\Delta x = v^2 - v_0 ^2 } \\ {v - v_0 = at} \end{array}} \right. \]
$\blacksquare$

PS: Não faço ideia se esta fórmula é permitida no exame de FQ do ensino secundário. Como manda o bom senso, não utilizem fórmulas sem antes confirmar se as podem usar com o vosso professor.

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