A particle of mass $m$ moves under a retarding force that is proportional to the cube of the speed. Find how long it takes to travel a distance $d$ from the instant speed was $u$.
I've been stuck on this question for an hour now. Can someone please point me in the right direction? The answer is
$$ \dfrac{d}{u} + \dfrac{1}{2}\dfrac{k}{m}d^2 $$
My initial equation is:
$ a = - \dfrac{kv^3}{m} $ then trying $\dfrac{dv}{dt}$ or $\dfrac{dv}{dx}v$ for $a$, without to much success.
Here is my thought process:
$$ v \dfrac{dv}{dx} = - \dfrac{kv^3}{m} $$ $$ \dfrac{dv}{dx} = - \dfrac{kv^2}{m} $$ $$ \dfrac{dx}{dv} = - \dfrac{m}{kv^2} $$ $$ x = - \dfrac{m}{k} \int v^{-2} dv $$ $$ x = \dfrac{m}{kv} + c $$
When $x = 0$ , $v = u$, therefor $c = - \dfrac{m}{ku} $
So I have
$$ x = \dfrac{m}{kv} - \dfrac{m}{ku} $$
I've solved it......
$$ x + \dfrac{m}{ku} = \dfrac{m}{kv} $$ $$ \dfrac{m}{kv} = \dfrac{kux + m}{ku} $$
Cancel the $k$ on the numerator.
$$ \dfrac{m}{v} = \dfrac{kux + m}{u} $$
Take reciprocal of both sides.
$$ \dfrac{v}{m} = \dfrac{u}{kux + m} $$
$$ v= \dfrac{mu}{kux + m} $$ $$ \dfrac{dx}{dt} = \dfrac{mu}{kux + m} $$
Take reciprocal of both sides.
$$ \dfrac{dt}{dx} = \dfrac{kux + m}{mu} $$ $$ \dfrac{dt}{dx} = \dfrac{m + kux}{mu} $$ $$ \dfrac{dt}{dx} = \dfrac{1}{u} + \dfrac{kx}{m}$$
$$ t = \int \Bigg[\dfrac{1}{u} + \dfrac{kx}{m} \Bigg] dx $$
Using initial conditions $x = 0$ to $x = d$.
$$ t = \int_{x=0}^{x=d} \Bigg[\dfrac{1}{u} + \dfrac{kx}{m} \Bigg] dx $$ $$ t = \Bigg[\dfrac{x}{u} + \dfrac{kx^2}{2m} \Bigg]_{x=0}^{x=d} dx $$
$$ t= \dfrac{d}{u} + \dfrac{1}{2}\dfrac{k}{m}d^2 $$
Your solution is probably the more elegant one, but here is an alternative with tedious algebraic manipulations omitted for brevity.
From $$ a = {\mathrm{d}v \over \mathrm{d}t} = -{kv^3 \over m} $$ solve for $v(t)$ $$ \int{\mathrm{d}v \over v^3} = -\int{k \over m}\mathrm{d}t\\ {1 \over v^2} = {2kt \over m} + c. $$ Applying the initial condition $v(0) = u$, we find $$c = {1 \over u^2}$$ and so $$v(t) = \sqrt{mu^2 \over 2ku^2t + m}.$$
Integrating $v(t)$ from $0$ to the unknown time $t_d$ should result in the desired distance $d$, so $$ d = \int_0^{t_d}\sqrt{mu^2 \over 2ku^2t + m}\mathrm{d}t = {\sqrt{m} \over 2ku}\int_m^{2ku^2t_d + m}s^{-1/2}\mathrm{d}s = {\sqrt{2ku^2mt_d + m^2} - m\over ku} $$ using the substitution $s = 2ku^2t+m$. Solving for $t_d$ yields $$ t_d = {k \over 2m}d^2 + {d \over u}. $$