Gauss's blog

Gauss's blog

I’d go back to December turn around and make it alright.

记一个物理问题

posted on 2021-04-16 13:38:56 | under 物理 |

汽车以恒定功率 $P$ 驱动,所受的阻力恒为 $f$,$v(0)=0$,求 $v(t)$.

不难列出微分方程$$\frac{P}{v}-f=m\frac{\text{d}v}{\text{d}t}$$ 分离变量得$$\frac{mv}{P-fv}\text{d}v=\text{d}t$$ 两边积分得$$\int\frac{mv}{P-fv}\text{d}v=t+C_1$$ 接下来考虑求左端积分$$\begin{aligned} \int\frac{mv\text{d}v}{P-fv}&=\frac{m}{P}\int\frac{v\text{d}v}{1-\dfrac{f}{P}v}\\ &=-\frac{m}{P}\int\frac{\dfrac{P}{f}\left(1-\dfrac{f}{p}v\right)-\dfrac{P}{f}}{1-\dfrac{f}{P}v}\text{d}v \\&=-\frac{m}{P}\left(\int\frac{P}{f}\text{d}v-\frac{P}{f}\int\frac{\text{d}v}{1-\dfrac{f}{P}v}\right) \\&=-\frac{m}{P}\left[\frac{Pv}{f}+\frac{P^2}{f^2}\ln\left(1-\frac{fv}{P}\right)\right]+C_2 \\&=-\frac{mv}{f}-\frac{mP}{f^2}\ln\left(1-\frac{fv}{P}\right)+C_2 \end{aligned}$$ 即得$$t=-\frac{mv}{f}-\frac{mP}{f^2}\ln\left(1-\frac{fv}{P}\right)+C$$ 代入初始条件,可知 $C=0$.

两边 exp 可得$$e^t=e^{-mv/f}\left(1-\frac{fv}{P}\right)^{-mP/f^2}$$ 两边取 $-f^2/mP$ 次幂$$e^{-f^2t/mp}=e^{fv/P}\left(1-\frac{fv}{P}\right)$$ 作代换$u=1-\dfrac{fv}{P}$,则有$v=\dfrac{P}{f}(1-u)$,代入得$$e^{-f^2t/mP}=ue^{1-u}$$ 两边同除 $e$

$$e^{-f^2t/mP-1}=ue^{-u}$$ 引入朗博函数$$u=-W(-e^{-f^2t/mP-1})$$ 回代得$$v(t)=\frac{P\left(1+W(-e^{-f^2t/mP-1})\right)}{f}$$ 极限速度 $v=\dfrac{P}{f}$.