A particle A starts from rest and moves with an acceleration of 5m/s2 due East while another particle B moves with a constant velocity of 5m/s due North from the same position. Then the magnitude of relative position of A w.r.t B at 2 second is
A
10m
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B
10√2m
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C
200m
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D
10√2m
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Solution
The correct option is B10√2m For A:xA=ut+12at2xA=(0×2)+12×5×22=10m For B:xB=5×2=10m Relative position of A w.r.t B, →xAB=xA−xB=10^i−10^jm ∣∣→xAB∣∣=√102+102=10√2m