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Question

A particle A starts from rest and moves with an acceleration of 5 m/s2 due East while another particle B moves with a constant velocity of 5 m/s due North from the same position. Then the magnitude of relative position of A w.r.t B at 2 second is
  1. 10 m 
  2. 102 m
  3. 200 m 
  4. 102 m


Solution

The correct option is B 102 m
 For A: xA=ut+12at2xA=(0×2)+12×5×22=10 m
For B: xB=5×2=10 m
Relative position of A w.r.t B, xAB=xAxB=10^i10^j m
xAB=102+102=102 m

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