autonomous equation with only one equilibrium 
Cell-sized organisms move differently from the way large objects do. The velocity of a cell-sized organism can be described by the differential equation $$v' = (a-v^2)v.$$ For which values of $a$ does this autonomous equation have only $1$ equilibrium?

I'm very confused because I thought we get the equilibrium values by setting $v' = 0$. Hence we have $v=0$ or $a-v^2 = 0$. The first one doesn't concern us (bc no $a$ value), but in the second one, we have $a = v^2$, and seeing as how $v$ is a real value (velocity), we must have $a \ge 0$. However, that's apparently wrong, and I don't see what other values of $a$ I would have. 
 A: You just need to take the negation of your statement :). If $a > 0$, then there are three equilibrium points: $v = 0$ (unstable) and $v = \pm \sqrt a$ (stable). If $a \leq 0$, there is only one equilibrium point $v = 0$ (stable).
A: $$v'=\frac{dv}{dt}=(a-v^2)v \tag 1$$
As you correctly pointed out, $\quad v'=0\quad$ if $\quad v=0\quad$ or if $\quad v^2=a$.
But is it possible that $v=0$ ? 
And is it possible than $v^2=a$ ?
To answer one have to solve the ODE.
$$ $$
SOLVING IN THE CASE $a\neq 0$ :
$$t=\int\frac{dv}{(a-v^2)v}$$
$$t=\frac{1}{2a}\ln\left|\frac{v^2}{a-v^2}\right|+c$$
With initial condition $v(0)=v_0$ :
$$t=\frac{1}{2a}\ln\left|\frac{v^2(a-v_0^2)}{v_0^2(a-v^2)}\right|$$
$$v^2=\frac{a\,v_0^2\,e^{2at}}{a+v_0^2\,(e^{2at}-1)} \tag 2$$
Thus $v$ cannot be nul except if $v_0=0$. But we don't know what is the value of $v_0 .$
One see that fully answering to the question is not possible if the initial condition is not specified in the wording of the question.
Then, is it possible that $v^2=a$ ?
$\quad a=\frac{a\,v_0^2\,e^{2at}}{a+v_0^2\,(e^{2at}-1)}\quad$ after simplification leading to $\quad v_0^2=a$ .
So, this is possible if $v_0^2=a$ . But we don't know what is the value of $v_0 .$ Again fully answering to the question is not possible if the initial condition is not specified in the wording of the question.
Note that if $v_0^2=a$ the equation $(1)$ gives $v'=0$. Thus the solution would be $v=$ constant : This is the trivial case $v(t)=v_0=\sqrt{a}\:$ if $a>0$.
Note : 
For $t\to\infty$ equation $(2)$ gives $v^2\to a.\quad$ So if $a>0$ there is an "equilibrium" at limit.
SOLVING IN THE CASE $a=0$ :
$$v'=\frac{dv}{dt}=-v^3 \tag 3$$
$$t=-\int \frac{dv}{v^3}\quad\implies\quad t=\frac{1}{2\,v^2}+c$$
With initial condition $v(0)=v_0$
$$t=\frac{1}{2\,v^2}-\frac{1}{2\,v_0^2}$$
$$v^2=\frac{v_0^2}{1+v_0^2t}\tag 4$$
$v'=0$ if $v_0=0$. Same comment than above.
Note : For $t\to\infty$ equation $(4)$ gives $v\to 0.\quad$ So there is an "equilibrium" at limit.
OVERALL CONCLUSION : 
With no initial condition specified in the wording of the question one cannot give a definitive answer.
