Two insects, A and B, start at the origin at the same time. Insect A moves with a constant velocity of (i + 4j + 3k) m/s and insect B moves with a constant velocity of (4i + 2j - 4k) m/s. What is the

Assuming both insects start at the origin $(0,0,0)$ at $t=0$, the position vector $R_A$ at time $t$ is given by $R_A = V_A t$.
Given $V_A = \hat{i} + 4\hat{j} + 3\hat{k}$ m/s and $t = 5$ s.

$$R_A = (\hat{i} + 4\hat{j} + 3\hat{k}) \times 5$$ $$R_A = 5\hat{i} + 20\hat{j} + 15\hat{k} \, \text{m}$$

Similarly, the position vector $R_B$ at time $t$ is given by $R_B = V_B t$.
Given $V_B = 4\hat{i} + 2\hat{j} - 4\hat{k}$ m/s and $t = 5$ s.

$$R_B = (4\hat{i} + 2\hat{j} - 4\hat{k}) \times 5$$ $$R_B = 20\hat{i} + 10\hat{j} - 20\hat{k} \, \text{m}$$

The relative position vector $R_{AB}$ (position of B relative to A) is $R_{AB} = R_B - R_A$.

$$R_{AB} = (20\hat{i} + 10\hat{j} - 20\hat{k}) - (5\hat{i} + 20\hat{j} + 15\hat{k})$$ $$R_{AB} = (20 - 5)\hat{i} + (10 - 20)\hat{j} + (-20 - 15)\hat{k}$$ $$R_{AB} = 15\hat{i} - 10\hat{j} - 35\hat{k} \, \text{m}$$

The distance apart is the magnitude of the relative position vector $R_{AB}$.

$$|R_{AB}| = \sqrt{(15)^2 + (-10)^2 + (-35)^2}$$ $$|R_{AB}| = \sqrt{225 + 100 + 1225}$$ $$|R_{AB}| = \sqrt{1550}$$ $$|R_{AB}| \approx 39.3700 \, \text{m}$$

Comparing this value with the given options, the closest match is $39.37$m.

The final answer is $\text{39.37m}$.