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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.

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What have you done so far? You should at least be able to get the differential equation(s) which need(s) solving. – Peter Taylor Feb 16 '11 at 9:45
The formula for the time should include big V for the "final" speed. Not small v. – Raskolnikov Feb 16 '11 at 10:36
Shouldn't you ask this at physics? – Graviton Feb 16 '11 at 13:55
up vote 3 down vote accepted

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 $v(0)=0$

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


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

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

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I feel dumb after seeing this TOTAL BRAIN FADE – user7152 Feb 16 '11 at 13:16

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