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PostPosted: Fri, 5 Sep 2003 03:05:53 UTC 
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Joined: Fri, 5 Sep 2003 02:59:44 UTC
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Consider a body that moves horizontally through a medium whose resistance is proportional to the square of the velocity V, so that dv/dt = -kV^2. Show that
V(t) = Vo/ (1 + Vokt)
and taht
X(t) = Xo + 1/k ln(1 + Vokt).

NOte that, in contrast with the results of Problem 2, X(t) --> + infinity as t --> +infinity

Please help i don't get application problems real well[/code]


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PostPosted: Fri, 5 Sep 2003 04:43:27 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Hello, beaversjoe!

Quote:
Consider a body that moves horizontally through a medium
whose resistance is proportional to the square of the velocity V,
so that dv/dt = -kV^2.
Show that: V(t) = Vo/ (1 + Vokt)
and that: X(t) = Xo + 1/k ln(1 + Vokt)


We have: \frac{dV}{dt} = -kV^2\;\;\Rightarrow\;\;\frac{dV}{V^2} = -k\ dt

Integrating: -\frac{1}{V} = -kt + C\;\;\Rightarrow\;\;V = \frac{1}{kt - C}

When t = 0,\ V = V_0:\;\; V_0 = \frac{1}{-C}\;\;\;\Rightarrow\;\;C = -\frac{1}{V_0}

Then: V = \frac{1}{kt + \frac{1}{V_0}}\;\;\Rightarrow\;\;\therefore V(t) = \displaystyle{\frac{V_0}{1 + V_0kt}}


To find X(t), integrate V:

X(t) = \int \frac{V_0}{1 + V_0kt}dt = \frac{1}{k}\ln|1 + V_0kt| + C


When t = 0,\ X = X_0:

X_0 = \frac{1}{k}\ln|1| + C\;\;\Rightarow\;\;C = X_0


\therefore X(t) = X_0 + \frac{1}{k}\ln|1 + V_0kt|


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