If $|a(t)|\ll b$ is it alright to take $a\left({a\cdot \dot{a} \over b^2}\right)$ as $0$?
Would the following argument make sense?
I know that we can take $\left({a\cdot a \over b^2}\right)$ as $0$ and $0={d\over dt}\left({a\cdot a \over b^2}\right)=2\left({a\cdot \dot{a} \over b^2}\right)$.
So since $a$ is bounded by finite number (namely $b$) therefore $a\left({a\cdot \dot{a} \over b^2}\right)$ can be taken to be $0$.