What is the maximum distance between the particle and the origin?

(A) $\text{2.00 m}$
(B) $\text{2.50 m}$
(C) $\text{3.50 m}$
(D) $\text{5.00 m}$
(E) $\text{7.00 m}$
1. **Determine the distance function**: The distance \(d(t)\) from the origin at any time \(t\) is given by the magnitude of the position vector \(\mathbf{r}(t)\):
   \[
   d(t) = \sqrt{x(t)^2 + y(t)^2}
   \]

2. **Substitute the given functions**: Substitute \(x(t) = 3\cos(t)\) and \(y(t) = 4\sin(t)\) into the distance function:
   \[
   d(t) = \sqrt{(3\cos(t))^2 + (4\sin(t))^2}
   \]

3. **Simplify the expression**: Simplify the expression inside the square root:
   \[
   d(t) = \sqrt{9\cos^2(t) + 16\sin^2(t)}
   \]

4. **Use trigonometric identities**: Recognize that \(\cos^2(t) + \sin^2(t) = 1\):
   \[
   d(t) = \sqrt{9\cos^2(t) + 16(1 - \cos^2(t))}
   \]
   \[
   d(t) = \sqrt{9\cos^2(t) + 16 - 16\cos^2(t)}
   \]
   \[
   d(t) = \sqrt{16 - 7\cos^2(t)}
   \]

5. **Find the maximum value**: The maximum distance occurs when the expression inside the square root is maximized. Since \(\cos^2(t)\) ranges from 0 to 1, the minimum value of \(7\cos^2(t)\) is 0 (when \(\cos^2(t) = 0\)), and the maximum value is 7 (when \(\cos^2(t) = 1\)):
   \[
   d_{\text{max}} = \sqrt{16 - 0} = \sqrt{16} = 4
   \]

Therefore, the maximum distance between the particle and the origin is \(4\) meters.

The final answer is \(\boxed{4.00 \text{ m}}\).