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Question

Two particles A and B are located in X-Y plane at point (0,0) and 0,8m. If they start moving simultaneously with velocities VA=10^j ms1 and VB=10^i ms1, then

A
the distance between them remains constant
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B
the distance between them will first decrease and then increase
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C
the shortest distance between them will be 42 m
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D
the time after which they are at a minimum separation distance is 0.4 s
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Solution

The correct options are
B the distance between them will first decrease and then increase
C the shortest distance between them will be 42 m
D the time after which they are at a minimum separation distance is 0.4 s
Consider the relative motion of A w.r.t observer on B
relative velocity,
VAB=VAGVBG=10^j10^i (in second quadrant)
VAB=102ms1
Here, minimum distance between them will be BP which is equal to 8sin45=8(12)=42 m
Time after which their separation distance becomes a minimum will be
t=APVAB|=8cos45102=810[12]
t=0.4

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