i am trying to calculate velocity of a free falling body with drag(b)
i want to integrate boat sides by t so:
$$\ddot{x}=\frac{mg-bv}{m}$$
$$\int\ddot{x}dt=\int\frac{mg-bv}{m}dt$$
$$\dot{x}=\int\frac{mg}{m}-\frac{bv}{m}dt$$
$$\dot{x}=\int g dt-\frac{b}{m}\int v dt$$
$$\dot{x}=gt-\frac{b}{m}x$$
but that's wrong and i don't know where my mistake is i tried to check my math but it seems ok?
from wolframalpha
Because $\int \ddot{x} dt = \int d\dot{x}$. With $\dot{x}=v$, you're not suppose to integrate $(g-b/m v)$ with respect to $t$. You have to move it to the other side and then integrate it. That is $$ \int \frac{dv}{dt} dt = \int \Big(g-\frac{b}{m} v\Big)dt \implies \int \frac{dv}{g-\frac{b}{m}v} = \int dt $$ So you'll get $$ -\frac{m}{b} \ln \Big(g-\frac{b}{m} v\Big) = t + c $$