I have a fluids problem where I wind up with the following equation.

mx''+(.3)*(x')^2-mg=0

where x'' is acceleration, x' is velocity. I've calculated terminal velocity to be 44.294 m/s, where a~0. I need to find the distance it takes to reach 90% of terminal velocity, so v=39.865 m/s. Any advice on how to approach this? Mass=60 kg, and g=9.81 . Any help at all would be appreciated.

[Edited on August 31, 2006 at 6:02 PM. Reason : clarification]

8/31/2006 5:42:11 PM

Is that all you're given?

8/31/2006 6:03:24 PM

What I am given is that drag force =k*v^2, where k=.3 . Skydiver has mass of 60kg. He falls straight down. I needed to find the terminal velocity, the vertical distance to reach 90% of the terminal speed, and distance to reach that speed neglecting air resistance. I found the first and last, I just need help with the middle one. If it helps, I know that the answer for the distance is 166.1 m, I just need to know how to get to that point

8/31/2006 6:06:13 PM

I mean I think I have to integrate to find an equation for x, but it has been a while since I have done that and I can't find all my calc and Diff. Eq. notes. Also, the (x')^2 is throwing me off I think.

8/31/2006 6:10:40 PM

v * dv = a * ds

8/31/2006 9:14:11 PM