Step-by-Step Solution

1. Define a Coordinate System and Express Velocities in Vector Form

Let's set up a 2D Cartesian coordinate system where the x-axis is horizontal and the y-axis is vertical. The velocities of the airplanes A and B can be expressed as vectors.

• Velocity of Airplane A ($v_A$): The speed of A is $|v_A| = 100\sqrt{3}$ m/s, and its direction is $30^\circ$ with the horizontal. $v_A = |v_A| (\cos(30^\circ) \hat{i} + \sin(30^\circ) \hat{j})$ $v_A = 100\sqrt{3} \left(\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}\right)$ $v_A = (150 \hat{i} + 50\sqrt{3} \hat{j}) \text{ m/s}$

• Velocity of Airplane B ($v_B$): Let the speed of B be $v_B$. Its direction is $60^\circ$ with the horizontal. $v_B = v_B (\cos(60^\circ) \hat{i} + \sin(60^\circ) \hat{j})$ $v_B = v_B \left(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}\right) \text{ m/s}$

2. Calculate the Relative Velocity of B with respect to A ($v_{BA}$)

The velocity of B as seen by an observer in A is the relative velocity $v_{BA}$. $v_{BA} = v_B - v_A$ $v_{BA} = \left(\frac{v_B}{2} \hat{i} + \frac{v_B\sqrt{3}}{2} \hat{j}\right) - (150 \hat{i} + 50\sqrt{3} \hat{j})$ $v_{BA} = \left(\frac{v_B}{2} - 150\right) \hat{i} + \left(\frac{v_B\sqrt{3}}{2} - 50\sqrt{3}\right) \hat{j}$

3. Apply the Perpendicularity Condition

The problem states that the observer in A sees B moving with a constant velocity perpendicular to the line of motion of A. This means the relative velocity vector $v_{BA}$ is perpendicular to the velocity vector of A, $v_A$.

For two vectors to be perpendicular, their dot product must be zero. $v_{BA} \cdot v_A = 0$ $\left[\left(\frac{v_B}{2} - 150\right) \hat{i} + \left(\frac{v_B\sqrt{3}}{2} - 50\sqrt{3}\right) \hat{j}\right] \cdot (150 \hat{i} + 50\sqrt{3} \hat{j}) = 0$ $\left(\frac{v_B}{2} - 150\right)(150) + \left(\frac{v_B\sqrt{3}}{2} - 50\sqrt{3}\right)(50\sqrt{3}) = 0$ $75v_B - 22500 + \left(\frac{v_B\sqrt{3}}{2}\right)(50\sqrt{3}) - (50\sqrt{3})(50\sqrt{3}) = 0$ $75v_B - 22500 + \frac{150v_B}{2} - (2500 \times 3) = 0$ $75v_B - 22500 + 75v_B - 7500 = 0$ $150v_B - 30000 = 0$ $150v_B = 30000$ $v_B = \frac{30000}{150} = 200 \text{ m/s}$

4. Calculate the Magnitude of the Relative Velocity

Now that we have the speed of B, we can find the relative velocity vector $v_{BA}$. $v_{BA} = \left(\frac{200}{2} - 150\right) \hat{i} + \left(\frac{200\sqrt{3}}{2} - 50\sqrt{3}\right) \hat{j}$ $v_{BA} = (100 - 150) \hat{i} + (100\sqrt{3} - 50\sqrt{3}) \hat{j}$ $v_{BA} = (-50 \hat{i} + 50\sqrt{3} \hat{j}) \text{ m/s}$

The magnitude of the relative velocity (the relative speed) is: $|v_{BA}| = \sqrt{(-50)^2 + (50\sqrt{3})^2}$ $|v_{BA}| = \sqrt{2500 + 2500 \times 3} = \sqrt{2500 + 7500}$ $|v_{BA}| = \sqrt{10000} = 100 \text{ m/s}$

5. Determine the Time to Near-Collision ($t_0$)

The phrase "A just escapes being hit by B" implies that the airplanes are on a direct collision course. In the reference frame of A, airplane B is moving directly towards A.

At $t=0$, the distance between A and B is given as 500 m. This is the initial relative distance, $|r_{BA}(0)| = 500$ m.

Since they are on a collision course, the time $t_0$ it takes for them to meet (or nearly meet) is the initial distance divided by their relative speed of approach. $t_0 = \frac{\text{Initial distance}}{|\text{Relative velocity}|} = \frac{|r_{BA}(0)|}{|v_{BA}|}$ $t_0 = \frac{500 \text{ m}}{100 \text{ m/s}}$ $t_0 = 5 \text{ s}$

Thus, at $t = 5$ s, A just escapes being hit by B.