Two particles A and B are located at points (0,−10√3) and (0,0) in x-y plane. They start moving simultaneously at time t=0 with velocities vₐ=5î and vᵦ=−5√3ĵ respectively. If the time when they are closest to each other is found to be k/2, find k. (All values are in S.I. units)
Physics · Mixed Chapter
JEE Mains

Question

Two particles A and B are located at points (0,−10√3) and (0,0) in x-y plane. They start moving simultaneously at time t=0 with velocities vₐ=5î and vᵦ=−5√3ĵ respectively. If the time when they are closest to each other is found to be k/2, find k. (All values are in S.I. units)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Relative initial position A−B is r₀=(0,−10√3) and relative velocity v=vₐ−vᵦ=(5,5√3). Step 2: Squared separation is d²(t)=|r₀+vt|². Minimise it by setting derivative zero: tₘᵢₙ=−(r₀·v)/|v|². Step 3: r₀·v=−150 and |v|²=25+75=100. Thus tₘᵢₙ=150/100=3/2 s. Step 4: tₘᵢₙ=k/2, hence k=3.