A particle moves from the point $$(2.0\hat{i} + 4.0\hat{j})$$ m, at $$t = 0$$, with an initial velocity $$(5.0\hat{i} + 4.0\hat{j})$$ ms$$^{-1}$$. It is acted upon by a constant force which produces a constant acceleration $$(4.0\hat{i} + 4.0\hat{j})$$ ms$$^{-2}$$. What is the distance of the particle from the origin at time 2 s?

We have to find the position of the particle after $$t = 2\ \text{s}$$ and then its distance from the origin.

The standard kinematic relation for constant acceleration in vector form is stated first:

$$\vec r(t) \;=\; \vec r_0 \;+\; \vec v_0\,t \;+\; \tfrac12\,\vec a\,t^{\,2}.$$

Here

$$\vec r_0 = 2.0\,\hat i + 4.0\,\hat j\ \text{m},\qquad \vec v_0 = 5.0\,\hat i + 4.0\,\hat j\ \text{m s}^{-1},\qquad \vec a = 4.0\,\hat i + 4.0\,\hat j\ \text{m s}^{-2}.$$

Now we substitute $$t = 2\ \text{s}$$ step by step.

First, the displacement due to initial velocity:

$$\vec v_0\,t \;=\; (5.0\,\hat i + 4.0\,\hat j)(2) \;=\; 10\,\hat i + 8\,\hat j\ \text{m}.$$

Next, the displacement due to acceleration:

$$\tfrac12\,\vec a\,t^{\,2} \;=\; \tfrac12\,(4.0\,\hat i + 4.0\,\hat j)\,(2)^{2} \;=\; \tfrac12\,(4.0\,\hat i + 4.0\,\hat j)\,(4) \;=\; 8\,\hat i + 8\,\hat j\ \text{m}.$$

Now we add all three contributions:

$$\vec r(2) \;=\; \vec r_0 + \vec v_0\,t + \tfrac12\,\vec a\,t^{\,2}$$

$$\phantom{\vec r(2)} \;=\; (2\,\hat i + 4\,\hat j) + (10\,\hat i + 8\,\hat j) + (8\,\hat i + 8\,\hat j).$$

Combining the $$\hat i$$ components: $$2 + 10 + 8 = 20,$$ so we get $$20\,\hat i.$$

Combining the $$\hat j$$ components: $$4 + 8 + 8 = 20,$$ so we get $$20\,\hat j.$$

Thus the position vector at $$t = 2\ \text{s}$$ is

$$\vec r(2) = 20\,\hat i + 20\,\hat j\ \text{m}.$$

The distance from the origin is the magnitude of this vector:

$$|\vec r(2)| \;=\; \sqrt{(20)^{2} + (20)^{2}} \;=\; \sqrt{400 + 400} \;=\; \sqrt{800} \;=\; 20\sqrt{2}\ \text{m}.$$

Hence, the correct answer is Option B.