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Time it takes to achieve terminal velocity?

  • Thread starter mcovalt
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1. Homework Statement
A 100kg skydiver has a terminal velocity of 58 m/s. Assuming air resistance of the skydiver is Fair=.3(v)2, how long will it take the skydiver to reach 58 m/s?

gravity is 10 m/s2 in our class btw.

2. Homework Equations
Fair=.3(v)2
Fg=mg


3. The Attempt at a Solution
I know the air resistance force will end up equaling 1000N at 58 m/s. I just don't know how long that will take.
 

ehild

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Write up Newton's second law, solve it for v(t) and see if 58 m/s will be ever reached.

ehild
 
Last edited:
57
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I integrated to find the answer, which lead to about 17.55 seconds.
 
28
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Write up Newton's second law, solve it for v(t) and see if 58 m/s will be ever reached.

ehild

I'm sorry, I am not in calculus based physics so excuse the ignorance :) Sound's like a differential equation. Newton's second is F=m * dv/dt. If this is a differential, how do I set this up to calculate via Wolframalpha?
 
28
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I integrated to find the answer, which lead to about 17.55 seconds.
How would one go about integrating it?
 
57
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Set up the sum of forces:
[tex]\Sigma F = F_{g} - F_{A}[/tex]
[tex]\Sigma F = (100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2[/tex]
[tex] ma= (100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2[/tex]
[tex] a= \frac{(100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2}{m}[/tex]
[tex] \frac{dv}{dt}= 10 \frac{m}{s^{2}} - 0.003v^2[/tex]
[tex] \frac{dv}{10 \frac{m}{s^{2}} - 0.003v^2}= dt[/tex]
[tex] \int\frac{dv}{10 \frac{m}{s^{2}} - 0.003v^2}= \int dt[/tex]
 
28
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Thank you so much! this helps greatly!
 
57
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I assume you are able to solve separable differential equations right?
 
28
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Well, no. But you got me to a place where I can start learning.
 
28
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Alright. I completely understand the math thanks to http://www.youtube.com/watch?v=nNHlSB6b1HU".

But I have one more q before I call it a day. How exactly did integrating give us the answer in seconds?

Let me guess and tell me how wrong I am: by getting F=ma down to an acceleration equation we found the integral of the acceleration equation to get the velocity equation. We were able to do this by getting the v's on one side and the t's on the other. Once we integrated both sides we found the velocity equation. Solving the equation for v=(58) we found that t=17.5498 seconds.

Correct?
 
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Ray Vickson

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Set up the sum of forces:
[tex]\Sigma F = F_{g} - F_{A}[/tex]
[tex]\Sigma F = (100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2[/tex]
[tex] ma= (100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2[/tex]
[tex] a= \frac{(100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2}{m}[/tex]
[tex] \frac{dv}{dt}= 10 \frac{m}{s^{2}} - 0.003v^2[/tex]
[tex] \frac{dv}{10 \frac{m}{s^{2}} - 0.003v^2}= dt[/tex]
[tex] \int\frac{dv}{10 \frac{m}{s^{2}} - 0.003v^2}= \int dt[/tex]
It does not make sense to write an integral such as
[tex]
\displaystyle
\int\frac{dv}{10 \frac{m}{s^{2}} - 0.003v^2}= \int dt[/tex].
If you gave this to a computer algebra package, it would assume that "m" and "s" are some constants, and would give you an answer involving them in an incorrect way. For example, Maple gives the integral as
[tex]
\displaystyle
\frac{5.773502692\, s \, \mbox{ arctanh}( \frac{.01732050808 s v}{\sqrt{m}} )}{\sqrt{m}}
[/tex]

You need to remove the "units" inside the integral---after all, you don't have any units in the "0.003 v^2" or in the "dv" parts.

RGV
 
57
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Whops my bad, Vickson is right about removing the units, although I only placed them there initially so you would know where those numbers came from. The solution of the left integral would just actually come from partial fractions method though.
 
28
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I understood about the values, but thanks for the heads up!

I finished my paper and I was wondering, if you all had the time, to look at page 3 and the bottom of page 5 to make sure I properly explained the integration aspect.

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B4NW_8WLoKvwMmViMzFkOWYtNTM3NS00ZmU0LWI5OTgtYzk5ZDIwYWQyYTM5&hl=en_US&authkey=CNLHqaMP [Broken]
 
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rude man

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I integrated to find the answer, which lead to about 17.55 seconds.
If the terminal velocity is 58 m/s then it takes infinite time to reach it. And of course infinite distance. But of course the diver would hit the ground first.

If you do the integrations in post 6 you will get tterminal = tanh-1(1)/√(cg) = ∞
where c = 0.003 according to post 6.
 

rude man

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I understood about the values, but thanks for the heads up!

I finished my paper and I was wondering, if you all had the time, to look at page 3 and the bottom of page 5 to make sure I properly explained the integration aspect.

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B4NW_8WLoKvwMmViMzFkOWYtNTM3NS00ZmU0LWI5OTgtYzk5ZDIwYWQyYTM5&hl=en_US&authkey=CNLHqaMP [Broken]
Link won't open.
 
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Redbelly98

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. . . we found that t=17.5498 seconds.

Correct?
Well, one person got that answer, and rude man has posted a quite different answer.
 

ehild

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Terminal velocity means the velocity at the end of time and time never ends... More precisely terminal velocity is the limit of velocity when the time goes to infinity. It is a constant, when the acceleration is zero, that is, when the forces - gravity and air drag balance each other.

In this problem, ∑F=mg-kv2, and it is zero when v2=mg/k=1000/0.3, that is [itex]v_∞=\sqrt{mg/k}=57.74 m/s[/itex], not 58 m/s. If the skydriver's initial speed is zero, its speed will increase and tends to v. If the initial speed is greater than that, it will slow down and the speed tends to vagain. If it happens to travel with the initial speed it will be unchanged.

Solving the differential equation, you get [tex]\ \ln|\frac{v_∞+v}{v_∞-v}|-\ln|\frac{v_∞+v_0}{v_∞-v_0}|=\frac{2gt}{v_∞}[/tex], (v0 is the initial velocity), that is


[tex]\frac{(v_∞+v)(v_∞-v_0)}{(v_∞-v)(v_∞+v_0)}=e^{2gt/v_∞}[/tex]

As the exponential function on the right side is never negative, v_∞-v has the same sign as v_∞-v_0. If the initial speed is less than the terminal speed, the speed can never exceed the terminal speed.



ehild
 

rude man

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Or simply

t = (v/g)tanh-1(v/v) and

v = (v)tanh(gt/v).

Assuming initial v = 0.
 

Redbelly98

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It's possible the teacher wanted to see if students know, conceptually, that the velocity approaches the terminal value without ever reaching it.
 

ehild

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Sure, the intention of the teacher was to see if the students knew what terminal velocity was... He gave the terminal velocity (58 m/s) and asked when would be the speed equal to 58 m/s. But the thread went aside by performing the integration (although it was useful for the OP). As the terminal velocity v(terminal)= sqrt(mg/k) was not the given value (58 m/s), integration gave a finite time by substituting v=58 m/s but keeping v(terminal)= sqrt(mg/k) although v was greater than that. That confused a few people posting to the thread.

ehild
 
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