Two persons A and B are located in X - Y plane at the points ( 0 , 0 ) and ( 0 , 10 ) respectively. (The distances are measured in MKS unit). At a time t = 0 , they start moving simultaneously with ve

Question

Two persons A and B are located in X - Y plane at the points ( 0 , 0 ) and ( 0 , 10 ) respectively. (The distances are measured in MKS unit). At a time t = 0 , they start moving simultaneously with velocities v a ​ = 2 j ^ ​ m s − 1 and v b ​ = 2 i ^ m s − 1 respectively. The time after which A and B are at their closest distance is

TestHub · Accepted Answer

Let after the time ( t ) the position of A is ( 0 , v A ​ t ) and position of B = ( v B ​ t , 10 ) . Distance between them y = ( 0 − v B ​ t ) 2 + ( v A ​ t − 10 ) 2 ​ or y 2 = ( 2 t ) 2 + ( 2 t − 10 ) 2 or y 2 = l = 4 t 2 + 4 t 2 + 100 − 40 t ⇒ l = 8 t 2 + 100 − 40 t Now, d t d l ​ = ( 16 t − 40 ) = 0 or t = 16 40 ​ = 2.5 s ∵ d t 2 d 2 l ​ = 16 = ( + ve ) So, l will be minimum.

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