Airplanes A and B are flying with constant ve | vedclass

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(A) Let the velocity of airplane $A$ be $\vec{V}_A$ and that of $B$ be $\vec{V}_B$. The angle of $A$ with the horizontal is $30^{\circ}$ and that of $

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$5$

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(A) Let the velocity of airplane $A$ be $\vec{V}_A$ and that of $B$ be $\vec{V}_B$. The angle of $A$ with the horizontal is $30^{\circ}$ and that of $B$ is $60^{\circ}$.Given,$V_A = 100\sqrt{3} \ m/s$.The observer in $A$ sees $B$ moving perpendicular to the line of motion of $A$. This means the component of the relative velocity of $B$ with respect to $A$ along the direction of $A$'s motion is zero.Let $\vec{V}_{BA} = \vec{V}_B - \vec{V}_A$. The component of $\vec{V}_{BA}$ along the direction of $\vec{V}_A$ is $V_{B} \cos(60^{\circ} - 30^{\circ}) - V_A = 0$.$V_B \cos(30^{\circ}) = V_A = 100\sqrt{3}$.$V_B (\frac{\sqrt{3}}{2}) = 100\sqrt{3} \implies V_B = 200 \ m/s$.The relative velocity of $B$ perpendicular to the line of motion of $A$ is $V_{BA, \perp} = V_B \sin(60^{\circ} - 30^{\circ}) = V_B \sin(30^{\circ}) = 200 	imes \frac{1}{2} = 100 \ m/s$.The initial distance perpendicular to the line of motion is $d = 500 \ m$.The time taken to cover this distance is $t_0 = \frac{d}{V_{BA, \perp}} = \frac{500}{100} = 5 \ s$.

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