Let $M$ be a connected manifold equipped with a connection $\nabla$. By Hopf-Rinow theorem, we know that if $M$ is complete then for any $x,y$ there exist a curve $\gamma:[0,1] \to M$ such that $\gamma(0) = x, \gamma(1) = y$ and $\nabla\_{\gamma'(t)} \gamma'(t)=0$ for all $t$. This is a way to say that $\gamma$ is a geodesic.


Suppose now that $M$ is possibly non complete. Given a threeshold $\varepsilon$, is it always possible to find a $\gamma$ between two fixed points $x,y$ such that
$$ \frac{\lVert \nabla\_{\gamma'(t)} \gamma'(t) \rVert }{\lVert \gamma' \rVert^2 } < \varepsilon $$


I kind of solved the case in which $M$ is of the form $\mathbb{R}^n \setminus \cup\_{i=1}^k N\_i $, where $N\_i$ are submanifolds of codimension at least 2. In this case you can take a segment from $x,y$ and perturb it to be transverse to each $N\_i$ in the $C\_2$ topology (see Hirsch, differential topology, transversality chapter), so that $\gamma''$ will be almost zero and $\gamma'$ almost costantly $(y-x)$.
Since $\dim \gamma + \dim N\_i < \dim M = n$, transversality means $\textrm{Im} \gamma \subset M$. In my case this is enough to conclude, so this is just a curiosity :) Maybe something in the spirit of calculus of variations?



Almost geodesic on non complete manifolds
Start with the plane $\mathbb R^2$ and remove a slab, but keep a line going through the slab:


$$ Slab = \{(x, y) \in \mathbb R^2 : 0 < y < 1, x \neq 0\} $$
$$ M = \mathbb R^2 - Slab$$



```
  y

--------o-------------
--------o-------------
--------o-------------

x

```

Note that $M\_1$ is connected but curves going from one side of the slab to the other have a fixed direction for some time.


Now cut away a line-with-a-hole:


$$ Line\_\delta =\{(x, y) \in \mathbb R^2 : y = 1 + \delta, x \neq 50\}$$
$$ N\_\delta = \mathbb R^2 - Slab - Line\_\delta $$



```
  y
-----------------------------o--     -
                                     | δ
--------o-----------------------     -
--------o-----------------------
--------o-----------------------

x

```

At the top of the slab (the point $(1, 0)$) we will always have $\gamma'/|\gamma| = (0, 1)$. If $\delta$ is small enough compared to your $\epsilon$, you shouldn't be able to turn fast enough to avoid crashing into $Line\_\delta$.


Edit: Another answer is to take the manifold


$$ ThickenedCircle\_{r, \delta} = \{ p \in \mathbb R^2 : r-\delta < |p| < r+\delta \}.$$
First chose a sufficiently small $r$ so that the circle of radius $r$ does not obey your condition on the curve for $\epsilon/2$. Then if you chose $\delta$ small enough you get a flat incomplete 2-manifold where geodesics still must accelerate too much.


