Airplanes $A$ and $B$ are flying with constant velocity in the same vertical plane at angles $30^\circ$ and $60^\circ$ with respect to the horizontal respectively as shown in figure. The speed of $A$ is $100\sqrt{3}\,\text{m/s}$. At time $t = 0\,\text{s}$, an observer in $A$ finds $B$ at a distance of $500\,\text{m}$. The observer sees $B$ moving with a constant velocity perpendicular to the line of motion of $A$. If at $t = t_0$, $A$ just escapes being hit by $B$, $t_0$ in seconds is

Step 1: Define the velocity vector of airplane $A$.

Using the given speed and angle of $A$ with respect to the horizontal, we write its velocity vector.

$$\vec{v}_A = 100\sqrt{3} (\cos 30^\circ \hat{i} + \sin 30^\circ \hat{j})$$ $$\vec{v}_A = 150\hat{i} + 50\sqrt{3}\hat{j}$$

Step 2: Express the velocity vector of airplane $B$.

Let $v_B$ be the speed of airplane $B$, which flies at $60^\circ$ to the horizontal.

$$\vec{v}_B = v_B (\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j})$$ $$\vec{v}_B = \frac{v_B}{2}\hat{i} + \frac{\sqrt{3}v_B}{2}\hat{j}$$

Step 3: Apply the condition for perpendicular relative velocity.

The relative velocity of $B$ with respect to $A$, denoted as $\vec{v}_{BA}$, is perpendicular to the velocity of $A$.

$$\vec{v}_{BA} \cdot \vec{v}_A = 0$$ $$(\vec{v}_B - \vec{v}_A) \cdot \vec{v}_A = 0$$ $$\vec{v}_B \cdot \vec{v}_A = |\vec{v}_A|^2$$

Step 4: Solve for the speed of airplane $B$.

Substitute the velocity vectors into the dot product relation.

$$\left(\frac{v_B}{2}\hat{i} + \frac{\sqrt{3}v_B}{2}\hat{j}\right) \cdot (150\hat{i} + 50\sqrt{3}\hat{j}) = (100\sqrt{3})^2$$ $$75v_B + 75v_B = 30000$$ $$150v_B = 30000$$ $$v_B = 200\,\text{m/s}$$

Step 5: Calculate the relative velocity vector and its magnitude.

Substitute $v_B = 200\,\text{m/s}$ to find the relative velocity $\vec{v}_{BA}$.

$$\vec{v}_B = 100\hat{i} + 100\sqrt{3}\hat{j}$$ $$\vec{v}_{BA} = (100 - 150)\hat{i} + (100\sqrt{3} - 50\sqrt{3})\hat{j}$$ $$\vec{v}_{BA} = -50\hat{i} + 50\sqrt{3}\hat{j}$$ $$|\vec{v}_{BA}| = \sqrt{(-50)^2 + (50\sqrt{3})^2} = 100\,\text{m/s}$$

Step 6: Find the time $t_0$ for the closest approach.

Since $A$ just escapes being hit by $B$, the minimum separation is zero, meaning $B$ travels the initial distance of $500\,\text{m}$ directly towards $A$ in $A$'s frame of reference.

$$t_0 = \frac{AB}{|\vec{v}_{BA}|}$$ $$t_0 = \frac{500}{100}$$ $$t_0 = 5\,\text{s}$$

Answer: 5