Now I'm doing a research on the matter similar to this thread : Terminal Velocity of identical shape/size objects
which is very self explanatory and very helpful. However in my case, the objects will be sliding down an inclined plane instead of falling perpendicularly to the ground. In that case, would the equation for the terminal velocity be
$$
v_{t} = \sqrt{\frac{2mg\sin{\theta}}{{\rho}AC_{d}}} ?
$$
Then I'd have to differentiate
$$
\dot{v}(t) = g\sin{\theta}-\frac{1}{2}\frac{{\rho}AC_{d}}{m}v(t)^2
$$
to find the time it takes to reach the terminal velocity?
Best Answer
Direct integration give you the time it takes to go between $v_1$ and $v_2$.
$$ t = \int \limits_{v_1}^{v_2} \frac{1}{ \dot{v}(t)} \, {\rm d}v $$
So if $\dot{v}(t) = a_0 - \beta v^2$ then
$$ t = \frac{1}{2 \sqrt{a_0 \beta}} \ln \left( \frac{a_0+\sqrt{a_0 \beta}(v_2-v_1)-\beta v_1 v_2 }{a_0-\sqrt{a_0 \beta}(v_2-v_1)-\beta v_1 v_2} \right) $$