Particle A starts from (0,0) with velocity 6î m/s and B starts from (40,30) m with velocity (-2î-4ĵ) m/s. The minimum distance between them for t≥0 is:
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Particle A starts from (0,0) with velocity 6î m/s and B starts from (40,30) m with velocity (−2î−4ĵ) m/s. The minimum distance between them for t≥0 is:

G1 Physics · Accepted Answer

Relative position r⃗=(40−8t)î+(30−4t)ĵ and relative velocity v⃗=−8î−4ĵ. Closest approach occurs at t=−r⃗₀·v⃗/v²=440/80=5.5 s. Then r⃗=(−4î+8ĵ) m, so d_min=√80=4√5 m.