Calculate Time for Relative Motion Near Collision of Two Airplanes

To find the time $t_0$ at which airplane A narrowly escapes colliding with airplane B, we can analyze the motion using a relative reference frame attached to airplane A.

Step 1: Velocity Analysis in Vector Form

Let the horizontal direction be the x-axis ($\hat{i}$) and the vertical direction be the y-axis ($\hat{j}$).

The velocity of airplane A, flying at an angle of $30^\circ$ relative to the horizontal with speed $v_A = 100\sqrt{3}\text{ m s}^{-1}$, is given by: $\vec{v}_A = v_A \cos 30^\circ \hat{i} + v_A \sin 30^\circ \hat{j}$ $\vec{v}_A = (100\sqrt{3}) \left(\frac{\sqrt{3}}{2}\right)\hat{i} + (100\sqrt{3})\left(\frac{1}{2}\right)\hat{j} = 150\hat{i} + 50\sqrt{3}\hat{j} \text{ m s}^{-1}$

The velocity of airplane B, flying at an angle of $60^\circ$ relative to the horizontal with speed $v_B$, is given by: $\vec{v}_B = v_B \cos 60^\circ \hat{i} + v_B \sin 60^\circ \hat{j} = \frac{v_B}{2}\hat{i} + \frac{\sqrt{3}v_B}{2}\hat{j}$

Step 2: Relative Velocity

The velocity of airplane B relative to airplane A is: $\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A = \left(\frac{v_B}{2} - 150\right)\hat{i} + \left(\frac{\sqrt{3}v_B}{2} - 50\sqrt{3}\right)\hat{j}$

Step 3: Utilizing the Perpendicularity Condition

An observer on airplane A sees airplane B moving perpendicular to A's direction of motion. Therefore, the dot product of $\vec{v}_{B/A}$ and $\vec{v}_A$ must be zero: $\vec{v}_{B/A} \cdot \vec{v}_A = 0$

Alternatively, resolving the component of $\vec{v}_{B/A}$ along the direction of motion of A: $v_{B/A, \parallel} = v_B \cos(60^\circ - 30^\circ) - v_A = 0$ $v_B \cos 30^\circ = v_A$ $v_B \left(\frac{\sqrt{3}}{2}\right) = 100\sqrt{3} \implies v_B = 200\text{ m s}^{-1}$

Step 4: Magnitude of Relative Velocity

The magnitude of the component of relative velocity perpendicular to A's line of motion is: $v_{B/A} = v_B \sin(60^\circ - 30^\circ) = v_B \sin 30^\circ = 200 \times \frac{1}{2} = 100\text{ m s}^{-1}$

Step 5: Calculation of Time $t_0$

At $t = 0\text{ s}$, the initial distance between A and B in the relative frame is $d = 500\text{ m}$. For a potential collision (or a near miss) at $t = t_0$, B travels this relative distance along the line of relative motion at constant relative speed $v_{B/A}$: $d = v_{B/A} \times t_0$ $500 = 100 \times t_0$ $t_0 = 5\text{ s}$