When motion begins, do objects go through an infinite number of position derivatives? This might be a very vague and unclear question, but let me explain. When an object at rest moves, or moves from point $A$ to point $B$, we know the object must have had some velocity (1st derivative of position) during that trip. It's also true that the object had to have accelerated to gain that velocity (2nd derivative)... if we extend this argument, is it logical to say that the object must have experienced jerk (3rd deriv.) to gain that acceleration? And so on and so forth... 3rd, 4th, 5th position derivatives? Or is my logic flawed somewhere?
 A: In the classical mechanics picture, where an object's position is a real-valued function of the real parameter time, then yes, you are correct.
If the object is at rest over any extended interval, then it's position as a function of time over that interval is simply $x(t) \equiv 0$, which of course means $x^{(k)}(t) = 0$ for any $k$-th derivative at any time $t$ in the interval.
Suppose $x^{(n)}(t) = 0$ for all times $t$ in the interval $[t_0, t_1]$, where the object is at rest in an interval around $t_0$. Then for $t \in [t_0, t_1]$ it's position obeys
$$ x^{(k-1)}(t) = x^{(k-1)}(t_0) + \int_{t_0}^t x^{(k)}(t') \, \mathrm{d}t' = x^{(k-1)}(t_0) = 0 $$
for all $k \in \{2, \ldots, n\}$. This implies
$$ x(t_1) = x(t_0) + \int_{t_0}^{t_1} x'(t') \, \mathrm{d}t' = x(t_0). $$
The contrapositive of this conclusion tells us that if $x(t_1) \neq x(t_0)$, all time-derivatives of position must have been nonvanishing at least somewhere in that interval.1

On the side: There is more to be pondered in relation to this infinite regress of derivatives. For example, you may wonder how anything at rest could get moving at all in this formal system of mechanics. Certainly if $x$ were an analytic2 function of time that vanished over some interval, then the object could never move. Fortunately there are infinitely differentiable functions that are not analytic. However, loosening our restrictions on "valid" trajectories in this way makes for some worrisome issues regarding determinism in Newtonian mechanics, illustrated for example by Norton's Dome.

1 The wording here could be improved if you allow for discontinuities and delta functions, but the same argument holds.
2 The word "analytic" has taken on a multitude of different meanings in physics. Here I mean "converges to its Taylor series." That is, a function $f$ is analytic at a point $x_0$ if for all points $x \neq x_0$ in a neighborhood of $x_0$, the series
$$ \sum_{k=0}^\infty \frac{1}{k!} f^{(k)}(x_0) (x-x_0)^k $$
is defined and converges to $f(x)$.
