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PhysicsInteger+3Question 11 of 60

JEE Advanced 2014 · Paper 1, Physics question 11

Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30° and 60° with respect to the horizontal respectively as shown in figure. The speed of A is 1003 m s−1100\sqrt3\ \mathrm{m\,s^{-1}}. At time t=0t=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=t0t=t_0, A just escapes being hit by B, t0t_0 in seconds is

Plane A flying up along a dashed line at 30° to the horizontal; plane B, lower and to the right, flying up along a dashed line at 60° to the horizontal, its path heading towards A's line.
Answer and worked solution

Official answer: 5

Relative velocity. B's velocity makes 30° with A's. For B to appear to move perpendicular to A's line, the components along A's direction must match:
vBcos⁡30°=vA=1003  ⟹  vB=200 m s−1.v_B\cos30°=v_A=100\sqrt3\implies v_B=200\ \mathrm{m\,s^{-1}}.

Relative speed. vBA=vBsin⁡30°=100 m s−1v_{BA}=v_B\sin30°=100\ \mathrm{m\,s^{-1}}, perpendicular to A's path.

Time. B, 500 m away along that perpendicular, closes the gap at 100 m s⁻¹:
t0=500100=5 s.t_0=\frac{500}{100}=5\text{ s}.

Answer: 5

Topics:Relative velocity
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