An object moves in two dimensions according to
\[\vec{r}(t) = (4.0t^2 - 9.0)\vec{i} + (2.0t - 5.0)\vec{j}\]
where $r$ is in meters and $t$ in seconds. When does the object cross the $x$-axis?

$ \textbf{(A)}\ 0.0 \text{ s}\qquad\textbf{(B)}\ 0.4 \text{ s}\qquad\textbf{(C)}\ 0.6 \text{ s}\qquad\textbf{(D)}\ 1.5 \text{ s}\qquad\textbf{(E)}\ 2.5 \text{ s}$
1. To determine when the object crosses the $x$-axis, we need to find the time $t$ when the $y$-coordinate of the position vector $\vec{r}(t)$ is zero. The position vector is given by:
   \[
   \vec{r}(t) = (4.0t^2 - 9.0)\vec{i} + (2.0t - 5.0)\vec{j}
   \]

2. The object crosses the $x$-axis when the $y$-component of $\vec{r}(t)$ is zero. Therefore, we set the $y$-component equal to zero and solve for $t$:
   \[
   2.0t - 5.0 = 0
   \]

3. Solving the equation for $t$:
   \[
   2.0t - 5.0 = 0 \implies 2.0t = 5.0 \implies t = \frac{5.0}{2.0} = 2.5
   \]

4. Thus, the object crosses the $x$-axis at $t = 2.5$ seconds.

The final answer is $\boxed{2.5 \text{ s}}$