Solve for distance(and time) it takes to accelerate from v=0 to v=v_f

I think the solutions can be written as:

$v=\sqrt{\frac{k_1}{k_2}}\tanh{(\sqrt{k_1k_2}t)}\text{ and }x=\ln{(\cosh{(\sqrt{k_1k_2}t)})}/k_2$

which I obtained by guessing a few forms for the velocity and manipulating them until they matched numerical solutions for all the values of k1 and k2 that I tried!!   I've no doubt there is a more straightforward way of doing it!!

If you've been there before, so to speak, and you know what the solution is to look like, then Alan's ' trial solution ' method is quickest.

Assume that

 $\displaystyle v=A\tanh(Bt), \text{ then}\\\frac{dv}{dt}=AB\text{ sech} ^{2}(Bt)=AB(1-\tanh^{2}(Bt))=AB -\frac{B}{A}v^{2}$

Equate coefficients with the given equation and solve for A and B.

If you have to get there from scratch, with one eye on the upcoming algebra, write $\displaystyle \frac{dv}{dt}=k_{2}(K^{2}-v^{2}),\text{ where, for convenience, } K^{2}=k_{1}/k_{2}.$

then  $\displaystyle \int \frac{dv}{K^{2}-v^{2}}=\int k_{2}dt$

$\displaystyle \frac{1}{2K}\int \frac{1}{K+v}+\frac{1}{K-v}dv=\frac{1}{2K}\ln\left(\frac{K+v}{K-v}\right)=k_{2}t + C$,   $\displaystyle v=0 \text{ when } t=0, \text{ so } C=0.$

$\displaystyle \frac{K+v}{K-v}=e^{2Kk_{2}t} \text{ from which }\quad v= K\left(\frac{e^{2Kk_{2}t}-1}{e^{2Kk_{2}t}+1}\right)=K\left(\frac{e^{Kk_{2}t}-e^{-Kk_{2}t}}{e^{Kk_{2}t}+e^{Kk_{2}t}}\right)=K\left(\frac{\sinh (Kk_{2}t)}{\cosh(Kk_{2}t)}\right)\\=K\tanh(Kk_{2}t)$.

Substitution for K now gets us the earlier mentioned solution (Alan's solution).

To find x, integrate wrt t,

$\displaystyle \int v\,dt= x =K\int\tanh(Kk_{2}t)\,dt=\frac{K}{Kk_{2}}\ln(\cosh(Kk_{2}t))+D\\\text{ and since } x=0\text{ when }t=0, D=0.$

Substituting for K,

$\displaystyle x=\frac{1}{k_{2}}\ln(\cosh(\sqrt{k_{1}k_{2}}t)$

Here is a brief version of how I approached it - pretty much as Tiggsy explained, though I did a numerical solution first to get a feel for the shape of the velocity as a function of time:

Note: Tiggsy approach from scratch is the better way to do this.