Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30° and \(60^{\circ}\) with respect to the horizontal respectively as shown in figure. The speed of A is \(100 \sqrt{3} m s^{-1}\). 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 \(t=t_{0}, \mathrm{~A}\) just escapes being hit by \(\mathrm{B}, t_{0}\) in seconds is

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Kinematics

Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30° and

6 0^{\circ}

with respect to the horizontal respectively as shown in figure. The speed of A is

1003 m s^{- 1}

. 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 seconds is

Answer: 5

Solution

In A's frame, B moves perpendicular to A's velocity. The relative velocity magnitude is

∣ v_{B} - v_{A} ∣ = v_{A} tan 3 0^{\circ} = 1003 \times \frac{1}{3} = 100

m/s.

Since B moves perpendicular to A's motion and they just escape collision, the initial separation (500 m) is traversed by the relative velocity. Time

t_{0} = \frac{500}{100} = 5

seconds.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Kinematics

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