So here is the question: A particle moves in a straight line. When the particle’s displacement from a fixed origin is $x$ m, its velocity is $v$ m/s and its acceleration is $a$ m/$s^2$. Given that $a = 2v^3$ and that $v = 2$ when $x = 0$, find the relationship between $x$ and $v$.
Because:
- $a = 2v^3$
$v \frac{dv}{dx} = 2v^3 $
$\frac{1}{2} v^2 \frac{dv}{dx}$= $\frac{1}{2} v^4$
$\frac{dv}{dx} = v^2$
I am stuck here.
We have a separable ODE:
$$v\frac{dv}{dx}=2v^3\implies v^{-2}dv=2dx\implies -v^{-1}=2x+c,$$
and using the initial condition $v=2$ when $x=0$ implies $c=-1/2$, so
$$v=\frac{2}{1-4x}.$$