An acceleration has units of length per time squared (or inverse length in natural units), and the vector field you describe has units of length, so there needs to be some factor of proportionality connecting the two. Let's say the acceleration is
$\ddot{\vec{r}}=\lambda^2 \vec{r}\;,$
where $\vec{r}$ is the position vector with components $x$ and $y$, a dot indicates differentiation with respect to time, and $\lambda^2$ is some positive proportionality constant which I've written as a square for later convenience. This is a second-order linear ordinary differential equation for $\vec{r}$, whose solution is
$\vec{r}(t)=\vec{r}_1\mathrm e^{\lambda t}+\vec{r}_2\mathrm e^{-\lambda t}$
with arbitrary vectors $\vec{r}_1$ and $\vec{r}_2$ which need to be determined using the initial conditions. Substituting your initial values $\vec{r}(0)=\vec{p}$ and $\dot{\vec{r}}(0)=\vec{v}$ yields
$ \begin{eqnarray} \vec{r}_1+\vec{r}_2&=&\vec{p}\;,\\ \vec{r}_1-\vec{r}_2&=&\frac{\vec{v}}{\lambda}\;, \end{eqnarray} $
with solution
$ \begin{eqnarray} \vec{r}_1&=&\frac{1}{2}\left(\vec{p}+\frac{\vec{v}}{\lambda}\right)\;,\\ \vec{r}_2&=&\frac{1}{2}\left(\vec{p}-\frac{\vec{v}}{\lambda}\right)\;, \end{eqnarray} $
so the overall solution is
$\vec{r}(t)=\frac{1}{2}\left[ \left(\vec{p}+\frac{\vec{v}}{\lambda}\right)\mathrm e^{\lambda t} + \left(\vec{p}-\frac{\vec{v}}{\lambda}\right)\mathrm e^{-\lambda t} \right]\;. $
This gives you the position at time $t$. To obtain the velocity at time $t$, you can differentiate with respect to time:
$\dot{\vec{r}}(t)=\frac{1}{2}\left[ \left(\lambda\vec{p}+\vec{v}\right)\mathrm e^{\lambda t} - \left(\lambda\vec{p}-\vec{v}\right)\mathrm e^{-\lambda t} \right]\;. $
To determine $\lambda$, take the square root of $\lambda^2$. Your units appear to be such that $\lambda^2=1$; in that case, you'd have $\lambda=1$.
[Edit:] I just realized that the vector field in your image doesn't look like the vector field you defined -- the vectors seem to all have the same length. If this is what you want the acceleration would be
$\ddot{\vec{r}}=\lambda^2 \frac{\vec{r}}{|\vec{r}|}\;.$
This can also be solved in closed form, but it's quite a bit more complicated, so I'll only work it out if that's what you need.