A Putnam question:
A particle of unit mass moves on a straight line under the action of a force which is a function $f(v)$ of the velocity $v$ of a particle, but the form of this function is not known. A motion is observed, and the distance $x$ covered in time $t$ is found to be $x = at + bt^2 + ct^3$, where a, b and c have numerical values determined by observing the motion. Find the function $f(v)$ for the range of $v$ covered by the experiment.
I reasoned that since $F=ma$ and the force is $f(v)$, then $x'' = v' = a = {f(v) \over 1}$ and $x'' = 2b + 6ct$, so $f(v) = 2b +6ct,$ where $m=1$ is the mass of the particle. Is that correct? If not, please provide a correct solution.
Assume for simplicity that $m=1$. Then we have $x''=f(x')$, since $x''$ is the acceleration and $x'$ is the speed. You can re-write this as $$ 6ct+2b=f(a+2bt+3ct^2). $$ Let $s=a+2bt+3ct^2$, then $$ t=\frac{ -b\pm \sqrt {(b^2-3ca)+3cs}}{3c} $$ and thus $$ 6ct+2b=\pm 2\sqrt {(b^2-3ca)+3cs} $$ Assuming that $f$ is positive, we get $$ f(s)=2\sqrt {(b^2-3ca)+3cs} $$