Two particle A and B are located at points ( 0 , − 10 3 ) and ( 0 , 0 ) in xy plane. They start moving simultaneously at time t = 0 with constant velocities V A = 5 i ^ m / s and V B = − 5 3 j

Question

Two particle A and B are located at points ( 0 , − 10 3 ​ ) and ( 0 , 0 ) in xy plane. They start moving simultaneously at time t = 0 with constant velocities V A ​ = 5 i ^ m / s and V B ​ = − 5 3 ​ j ^ ​ m / s , respectively. Time when they are closest to each other is found to be K /2 second. Find K . All distances are given in meter.

TestHub · Accepted Answer

As observed by B motion of A is along AM and BM is the shortest distance between them. Relative displacement of A w.r.t to B is $AM = AB cos 3 0^{\circ} V_{A B} = 5 \hat{i} + 53 \hat{j}$

$Time taken (t) = \frac{S _{A B}}{V _{A B}} = \frac{A M}{V _{A B}} = \frac{10 3 COS 3 0 ^{\circ}}{10} = 1.5 sec$

Physics - Kinematics Question with Solution | TestHub

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