I've been working on this problem a while.
I am supposed to prove that, for a particle of mass $m$, which is subject to a force $\vec{F}$ = $-av^2\vec{\hat{v}}$, (with $a$ a positive constant) the total work done in one-dimension as the particle moves a distance $b$ is:
$W = \frac{1}{2}mv_0^2(e^{-2ab/m} - 1)$
where $v_0$ is the initial velocity.
I have obtained an equation for $v$ in terms of the direction $x$, which I think is right.
$v = e^{-ax/m}$
so $F = -ae^{-2ax/m}\vec{\hat{v}}$ and $d\vec{s} = dx$. Using all this, I get a similar answer but without the $v_0^2$. Can someone tell me where I went wrong?
Thank you