There are two particles A and B, separated by 10m. B has constant velocity of 4m/s, perpendicular to line joining AB at t=0. The angle making with line AB at t=0 A should move with constant velocity at 2m/s, such that separation between A & B becomes minimum after some time is π/n, find n.
Physics · Mixed Chapter
JEE Mains

Question

There are two particles A and B, separated by 10m. B has constant velocity of 4m/s, perpendicular to line joining AB at t=0. The angle making with line AB at t=0 A should move with constant velocity at 2m/s, such that separation between A & B becomes minimum after some time is π/n, find n.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Put initial separation AB=d along the x-axis. Let A move with speed 2 at angle θ to AB; B moves perpendicular to AB with speed 4. The relative velocity of B with respect to A has components (−2cosθ, 4−2sinθ). Step 2: The shortest distance between the relative-motion line and the origin is dₘᵢₙ=d(4−2sinθ)/√[(2cosθ)²+(4−2sinθ)²]. Step 3: Minimise dₘᵢₙ² with respect to s=sinθ. The stationary condition for (4−2s)²/(20−16s) yields s=1/2. Step 4: Therefore θ=30°=π/6, and π/n=π/6 gives n=6. An…