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I'm hating these variable resistance questions.

A body of mass $m$ falls from rest in a medium that produces a resistance of magnitude $m\cdot k \cdot v$. where $k$ is a constant, where the speed of the particle is $v$. Show that when the body has reached a speed $V$ it will have fallen for a time

$\frac{1}{k} \ln \left(\frac{g}{g-k\cdot V}\right) \; .$

Help much appreciated.

  • 0
    Shouldn't you ask this at [physics](http://physics.stackexchange.com/)?2011-02-16

2 Answers 2

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Well the force on the particle due to gravity is $mg,$ where $g$ is the acceleration due to gravity, so the net downward force on the particle is $m(g-kv).$ So its acceleration is $g-kv.$ So you need to solve $dv/dt = g-kv$ given the initial conditions. That is

$\int_0^V \frac{ \text{d}v}{g-kv} = \int_0^T \text{d}t.$

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Okay here is my noob way of solving this (I only learned this less than a year ago but I have forgotten everything and semester starts in a few weeks and I'll probably be required to do this stuff EEEEEEEEEEEEEEEEEEEEEEP),

The forces on the "body" are the resistance and its weight (gravity) as far as I can tell from your description

So $ \vec{F}_{net} = m\mathbf{a} =mg-mk\mathbf{v} $

which can be simplified to

$ \mathbf{a} = g-k\mathbf{v} $

which can be written in a slightly different form as because acceleration is derivative of velocity

v'(t)+k v(t)=g

This looks type of equation looks familiar to me and can be solved as it is a "linear first order differential equation"

Now here is where I may start to make errors so please forgive me

First we multiple everything an "integrating factor" equal to $e^{\int f(t)\,dt}$ where $f(t)=k,$ to make it all work so

$e^{\int f(t)\,dt}=e^{\int k\,dt}=e^{kt}$

We don't need a "plus C" (I'm not entirely clear on this but I don't think it makes a difference to anything in the long run, SOMEONE PLEASE CONFIRM)

So when we multiply everything by $e^{kt} $

v'(t)e^{kt}+k v(t)e^{kt}=ge^{kt}

which is "equivalent" to this (look at the product rule)

(v(t)e^{kt})'=ge^{kt}

Now if we integrate both sides

$ v(t)e^{kt}={\int ge^{kt}\,dt}$

$ v(t)e^{kt}=\frac{ge^{kt}}{k}+C$

$ v(t)=\frac{C}{e^{kt}}+\frac{g}{k}$

To get rid of the C, you said it starts from rest so $v(0)=0$

$ v(0)=0=\frac{C}{e^{k0}}+\frac{g}{k} = \frac{C}{1}+\frac{g}{k}$ $ 0= C+\frac{g}{k}$ $ C=-\frac{g}{k}$

So

$ v(t)=\frac{-\frac{g}{k}}{e^{kt}}+\frac{g}{k}$

and I would simplify this to

$ v(t)=g\frac{1-e^{-kt}}{k}$

Now Wolfram Alpha would have done this for you

Now to the second part of your question when v(t)=V what is t?

$ v(t)=V=g\frac{1-e^{-kt}}{k}$

$ \frac{kV}{g}={1-e^{-kt}}$

$ e^{-kt}=1-\frac{kV}{g}$

$ e^{-kt}=\frac{g-kV}{g}$

$ e^{kt}=\frac{g}{g-kV}$

$ kt=\log_e {\frac{g}{g-kV}}$

$ t=\frac{1}{k}\log_e ({\frac{g}{g-kV}})$

Wolfram Alpha would do this too but is down for me now

I hope this is correct

  • 0
    I feel dumb after seeing [this](http://math.stackexchange.com/questions/22312/a-body-falls-through-a-medium/22343#22343) TOTAL BRAIN FADE2011-02-16