galactus said:
Subhotosh Khan said:Did you figure out how much v[sub:rdvwvcd4]t[/sub:rdvwvcd4] was - and how does that get into your equation?
For direct time integral - you'll set up your ODE as
\(\displaystyle m\frac{dv}{dt} \ = \ Force\)
warwick said:Subhotosh Khan said:Did you figure out how much v[sub:n3ghs7r8]t[/sub:n3ghs7r8] was - and how does that get into your equation?
For direct time integral - you'll set up your ODE as
\(\displaystyle m\frac{dv}{dt} \ = \ Force\)
Well, this is what I got as this sum of the forces. If I time integrate, there's no v on the left to make for easy u-substitution.
http://i111.photobucket.com/albums/n149 ... 1284338513
Subhotosh Khan said:warwick said:http://i111.photobucket.com/albums/n149 ... 1284338513[/url]
How is it that you have (dv/dt)[sup:qqirac8g]2[/sup:qqirac8g] term in the force?
It should be very similar to the work in your other post:
viewtopic.php?f=3&t=41421[/quote:qqirac8g]
Well, it says the resistive force is proportional to the square of the instantaneous speed, i.e. dv/dt. Even if I just stick in v^2, I still don't get a nice, easy u-substitution because I don't have an extra v term. So from the other thread, I need to convert it to a hyperbolic form and then integrate, correct? Any good write-ups online for doing that?