If $a=s^3$, how do I find $v$? 
Let $a=s^3$. Given that $v=3$ when $s=4$, find $v$.

This question really threw me off, because the acceleration is defined in terms of the displacement, and not in terms of time like most kinematics questions are that I have met. I didn't know how to solve it, and the book's worked answer didn't make much sense to me. My question is: if you were to see a question like this, how would you approach it?
 A: There is a standard "trick" involving chain rule that's used to relate $a$ and $s$ and $v$ independent of $t$. It's useful in deriving the classical kinetic energy formula, for example. It goes like this:
$$\frac{ds}{dt} = v$$
$$\frac{dv}{dt} = a$$
Dividing and applying chain rule, you get:
$$\frac{ds}{dv} = \frac va$$
Separating variables,
$$vdv = ads$$
Integrating,
$$\int_3^{v_f}vdv=\int_4^{s_f}ads $$
$$\int_3^{v_f}vdv=\int_4^{s_f}s^3ds$$
Note that I specified the lower bounds as the initial conditions to avoid the constant of integration. The upper bounds are time-variable (the $f$ subscript refers to "final"). In a physics context, they would just name them $v$ and $s$ (identical to the variable names), but this is an abuse of notation.
So you get:
$$\frac 12 v_f^2 - \frac {3^2}{2} = \frac 14 s_f^4 - \frac{4^4}{4}$$
And I'm sure you can proceed from here.
A: Velocity is the antiderivative of acceleration. Then use $v=3$ and $s=4$ so as to find the value of the constant of integration.
