How Does Terminal Velocity Arise in Fluid Dynamics?

[Gear300]

heheh...I need some help understanding something. If there was an object, such as a sphere, in water, gravity would be pushing it downwards, while a resistive force R = -bv would be pushing it upward (b as a constant).

that would imply that the net Force Fy would be
Fy = mg - bv = ma = m(dv/dt)

dv/dt = g - (b/m)v

How do I come up with

v = (mg/b)(1 - e^(-bt/m))

 

[Shooting Star]

Have you learned integration? That's how you get the final eqn.

The force R is a resistive force, acting against the direction of motion, and not necessarily pushing it up. But if the body is falling vertically downward, of course it is acting upward.

 

[Gear300]

I see...I've learned integration, but I'm still just tipping it.

I was just thinking that using dv/dt = g - (b/m)v could be rearranged for v = (mg/b)(1-(a/g)), in which dv/dt = acceleration = a. After that, I would somehow have to state that (a/g) = e^(-bt/m)...which I apparently didn't do.

Actually...nevermind...I get what's being said. The terminal velocity (when the net force is 0N) is only approached, not touched, in which the terminal velocity = (mg/b). So, the equation changes a bit.

 

[Shooting Star]

I'm glad you got the essence of it. In practice, the actual velocity gets indistinguishably close to the terminal velocity within a very short time, depending, of course on b/m. The higher this ratio is, the faster it happens.