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 100√3 ms⁻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 = t₀, A just escapes being hit by B, t₀ in seconds is:

Question

G1 Physics · Accepted Answer

The angle between the velocity directions of A and B is 30°. Let B's speed be vᴮ. Since B is seen from A to move perpendicular to A's direction, the component of B's velocity along A must equal A's speed: vᴮcos30° = 100√3, so vᴮ = 200 m/s. The relative velocity is then perpendicular to A with magnitude vᴮsin30° = 100 m/s. With initial separation 500 m along this relative-motion line, t₀ = 500/100 = 5 s.