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Distance and time integrals

  • Thread starter armolinasf
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  • #1
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Homework Statement



If you jump out of an airplane and your parachute fails to open our downward velocity t seconds after jumping is approximated for g=9.8m/sec^2 and k=.2 sec, by

v(t)=(g/k)(1-e^(-kt))


So, if you jump from 5000 meters above the ground write an equation whose solution is the number of seconds you fall before hitting the ground.



The Attempt at a Solution



This is coming from the section in my book on definite integrals but I'm not sure where to start. Thanks for any help.
 

Answers and Replies

  • #2
Dick
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v(t)=x'(t) where x(t) is the vertical displacement. So to find x(t) given v(t) you would integrate v(t), right?
 
  • #3
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Integrating v(t) would give me the total meters travelled but if I'm looking for the number of seconds it took to fall 5000 meters I would need to integrate something else.

Or could I use the integral that gives me distance and say that t=d/v where d is the integral and v is v(t)? This was my first thought but I'm uncertain since v(t) is a function of time.
 
  • #4
Dick
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Integrating v(t) would give me the total meters travelled but if I'm looking for the number of seconds it took to fall 5000 meters I would need to integrate something else.

Or could I use the integral that gives me distance and say that t=d/v where d is the integral and v is v(t)? This was my first thought but I'm uncertain since v(t) is a function of time.
If you can find x(t) you can certainly try and solve x(t)=5000 for t, right? Using d/v is wrong. The speed isn't constand.
 
  • #5
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So then finding x(t) would just be finding the antiderivative of v(t) correct?
 
  • #6
Dick
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Sure.
 

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