My Terminal Velocity is Zero? (DE Question)

[Bazzinga]

[PLAIN]http://img204.imageshack.us/img204/11/60272414.png [Broken]

Hey guys, I'm stuck on this question, part b)

I figured out a) and v(t) for V(0) = 0 ends up being

mg - mge^-kt/m
k

(Sorry, latex was being difficult)

But then when I try to figure out the limit as t->infinity i get 0. I'm pretty sure the terminal velocity of jumping out of an airplane wouldn't be 0

I'm hoping I just made a stupid mistake somewhere!

 

[Char. Limit]

What you're getting, I assume, is...

$$\frac{mg\left(1-e^{-\frac{kt}{m}}\right)}{k}$$

I must say that as [itex]t\rightarrow\infty[/tex], that doesn't approach zero.

 

[Bazzinga]

Ha, see? stupid mistake :P

 

[Bazzinga]

Hold up, you still get 0... e^0 = 1, 1 - 1 = 0, mg0 = 0, 0/k = 0

 

[Char. Limit]

But as t approaches infinity, you don't get e^0. You get e^(-infinity), which is 0.

 

[Bazzinga]

Ohhh! I get it, thanks! About part c)... distance = velocity x time, so would I just multiply my equation by t? I can't seem to wrap my head around this stuff

 

[Char. Limit]

No, x=vt only for constant v. If the v is not constant (as this v isn't), you have to integrate v with respect to t to get the distance.