Answer
$(24\hat{i} - 336\hat{j})\,m/s^2$
Work Step by Step
We're given $\vec{r}(t) = (2t^3-5t)\hat{i} + (6-7t^4)\hat{j}$, where $\vec{r}(t)$ is in meters.
We know that the velocity vector is the derivative of the position vector. We can thus take derivatives of $\vec{r}(t)$ component-wise to obtain $\vec{v}(t)$:
$\vec{v}(t) = (6t^2-5)\hat{i} - 28t^3\hat{j}$, where $\vec{v}(t)$ is in meters per second.
Similarly, the acceleration vector is the derivative of the velocity vector:
$\vec{a}(t) = 12t\,\hat{i}-84t^2\,\hat{j}$, where $\vec{a}(t)$ is in meters per second squared.
We can plug in $t=2\,s$ to obtain:
$\vec{a}(2) = (12*2\hat{i} -84*2^2\,\hat{j})\,m/s^2 = (24\hat{i} - 336\hat{j})\,m/s^2$.
