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速度の二乗に比例する抵抗

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\dot{v}\:=\:aとする。

m\dot{v}\:=\:mg-kv^2

\dot{v}\:=\:g-\frac{k}{m}v^2

\dot{v}\:=\:-\frac{k}{m}(v^2-\frac{mg}{k})

\dot{v}\:=\:-\frac{k}{m}(v-\alpha)(v+\alpha)

(但し\alpha=\sqrt{\frac{mg}{k}})

\frac{\dot{v}}{(v-\alpha)(v+\alpha)}\:=\:-\frac{k}{m}

\frac{1}{2a}\cdot\left[\frac{\dot{v}}{(v-\alpha)}\:-\:\frac{\dot{v}}{(v+\alpha)}\right]\:=\:-\frac{k}{m}

\frac{\dot{v}}{(v-\alpha)}\:-\:\frac{\dot{v}}{(v+\alpha)}\:=\:-\frac{2ak}{m}

\int\frac{f^{\prime}\left(x\right)}{f\left(x\right)}\;dx\:=\:\ln(f\left(x\right))より

\int\frac{\dot{v}}{(v-\alpha)}\:-\:\frac{\dot{v}}{(v+\alpha)}\;dt\:=\:-\int\frac{2ak}{m}\;dt

\ln(v-\alpha)\:-\:\ln(v+\alpha)\:=\:-\frac{2ak}{m}t+C

\frac{v-\alpha}{v+\alpha}\:=\:e^{-\frac{2ak}{m}t+C}\:=\:Ae^{-\frac{2ak}{m}t}

f\left(t\right)\:=\:Ae^{-\frac{2ak}{m}t}として

v-\alpha\:=\:(v+\alpha)f\left(t\right)

v(1-f\left(t\right))\:=\:\alpha(1+f\left(t\right))

v\:=\:\alpha\frac{1+f\left(t\right)}{1-f\left(t\right)}

v\:=\:\alpha\left(1+\frac{2f\left(t\right)}{1-f\left(t\right)}\right)

これを積分するが、

\int\:v\;dt\:=\:\alpha\int\1+\frac{2f\left(t\right)}{1-f\left(t\right)}\;dt

ところで \frac{d(1-f\left(t\right))}{dt}\:=\:-f^{\prime}\left(t\right)\:=\:-\frac{dAe^{-\frac{2ak}{m}t}}{dt}\:=\:-\cdot-\frac{2ak}{m}Ae^{-\frac{2ak}{m}t}\:=\:\frac{2ak}{m}f\left(t\right)

\frac{2f\left(t\right)}{1-f\left(t\right)}\:=\:\frac{\frac{2}{\frac{2ak}{m}}(1-f^{\prime}\left(t\right))}{(1-f\left(t\right))}\:=\:\frac{m(1-f\left(t\right))^{\prime}}{ak(1-f\left(t\right))}

よって \int\:v\;dt\:=\:\alpha\int\:1+\frac{2f\left(t\right)}{1-f\left(t\right)}\;dt\:=\:\alpha\int\:1+\frac{m(1-f\left(t\right))^{\prime}}{ak(1-f\left(t\right))}\;dt\:=\:\alpha\left{t+\frac{m}{ak}\cdot\ln(1-f\left(t\right))\right}+B

x\:=\:\alpha\left{t+\frac{m}{ak}\ln\left(1-Ae^{-\frac{2ak}{m}t}\right)\right}+B\;\box

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