Question:
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 as shown in the figure. The speed of A is \(100\sqrt{3}\) m/s. At time \(t=0\) s, an observer in A finds B at a distance of \(500\) m. This 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 second is
\(\{\sin (30)^{\circ }=0.5,\,\cos (30)^{\circ }=3^{0.5}/2\}\)
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 as shown in the figure. The speed of A is \(100\sqrt{3}\) m/s. At time \(t=0\) s, an observer in A finds B at a distance of \(500\) m. This 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 second is
\(\{\sin (30)^{\circ }=0.5,\,\cos (30)^{\circ }=3^{0.5}/2\}\)
\(\{\sin (30)^{\circ }=0.5,\,\cos (30)^{\circ }=3^{0.5}/2\}\)
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Use the condition that the velocity of B relative to A is perpendicular to the velocity of A to find the speed of B, then divide the distance by the relative speed.
Updated On: Nov 12, 2026
- \(2\)
- \(3\)
- \(4\)
- \(5\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
A moves at \(30^{\circ}\) and B at \(60^{\circ}\), so the angle between their velocities is \(30^{\circ}\).
The relative velocity \(\vec{v}_{BA}=\vec{v}_B-\vec{v}_A\) must be perpendicular to \(\vec{v}_A\).
Step 2: Find the speed of B
The component of \(\vec{v}_{BA}\) along \(\vec{v}_A\) is zero:
\(v_B\cos 30^{\circ}-v_A=0\)
\(v_B=\dfrac{100\sqrt{3}}{\sqrt{3}/2}=200\) m/s
Step 3: Relative speed
The perpendicular component is
\(v_{BA}=v_B\sin 30^{\circ}=200\times 0.5=100\) m/s
Step 4: Time
The separation of \(500\) m is covered at the relative speed:
\(t_0=\dfrac{500}{100}=5\) s
Final Answer:
The time t0 is 5 seconds. \[ \boxed{\text{(D) }5} \]
A moves at \(30^{\circ}\) and B at \(60^{\circ}\), so the angle between their velocities is \(30^{\circ}\).
The relative velocity \(\vec{v}_{BA}=\vec{v}_B-\vec{v}_A\) must be perpendicular to \(\vec{v}_A\).
Step 2: Find the speed of B
The component of \(\vec{v}_{BA}\) along \(\vec{v}_A\) is zero:
\(v_B\cos 30^{\circ}-v_A=0\)
\(v_B=\dfrac{100\sqrt{3}}{\sqrt{3}/2}=200\) m/s
Step 3: Relative speed
The perpendicular component is
\(v_{BA}=v_B\sin 30^{\circ}=200\times 0.5=100\) m/s
Step 4: Time
The separation of \(500\) m is covered at the relative speed:
\(t_0=\dfrac{500}{100}=5\) s
Final Answer:
The time t0 is 5 seconds. \[ \boxed{\text{(D) }5} \]
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