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My Terminal Velocity is Zero? (DE Question)

  • Thread starter Bazzinga
  • Start date
  • #1
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[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 :confused:

I'm hoping I just made a stupid mistake somewhere!
 
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Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
14
What you're getting, I assume, is...

[tex]\frac{mg\left(1-e^{-\frac{kt}{m}}\right)}{k}[/tex]

I must say that as [itex]t\rightarrow\infty[/tex], that doesn't approach zero.
 
  • #3
45
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Ha, see? stupid mistake :P
 
  • #4
45
0
Hold up, you still get 0... e^0 = 1, 1 - 1 = 0, mg0 = 0, 0/k = 0
 
  • #5
Char. Limit
Gold Member
1,204
14
But as t approaches infinity, you don't get e^0. You get e^(-infinity), which is 0.
 
  • #6
45
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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
 
  • #7
Char. Limit
Gold Member
1,204
14
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.
 

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