Pls solve.this question step by step Share with your friends Share 0 Vipra Mishra answered this Solution After time t the coordinate of A and B arerA→ = 300 j^ - 30t j^ = (300-30t)j^rB→ = 300 i^ - 40t i^ = (300-40t)i^rAB → = rA→-rB→ = (300-30t)j^- (300-40t)i^rAB → = (40t-300)i^ + (300-30t)j^rAB =(40t-300)2+ (300-30t)2For rAB to be minimumddtrAB =0ddtrAB =40×2(40t-300)+2×-30 (300-30t)2(40t-300)2+ (300-30t)2 =040×2(40t-300)+2×-30 (300-30t) =080(40t-300)-60 (300-30t) =080(40t-300)=60 (300-30t)=04(40t-300)=3 (300-30t)160t -1200=900-90tt =2100250 =425sMinimum Distance =rAB =(40×425-300)2+ (300-30×425)2=362+482 = 60 m 0 View Full Answer
