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Question

A particle undergoes three successive displacements given by S1=2 m north-east, S2=2 m due south and S3=4 m, 30 north of west, then magnitude of net displacement is

A
14+43 m
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B
1443 m
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C
4 m
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D
123 m
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Solution

The correct option is B 1443 m
Let us first write down all the three displacements in vector form
S1=(2 cos 45)^i+(2 sin 45)^j
=1^i+1^j
S2=2^j
and S3=(4 cos 30)^i+(4 sin 30)^j
S3=23^i+2^j
Total displacement =S1+S2+S3=(123)^i+1^j
Hence, magnitude of total displacement = (123)2+12=1443 m

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