A man walks 30m towards north ,then 20m towards east and in last 30√2m towards south-west the displacement from origin is....
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Use vector form to solve the solution.
First the man walks towards north. So it is 30 j^
Then he walks in east direction, So 20i^
Then finally 30✓2 towards South-west. resolve in x and y direction, which is = -30✓2cos thita + (-30✓2 sin thita)= -30i^ - 30j^.
For net displacement add, +30j^ -30j^ + 20i^ - 30i^ = - 10i^.So it is 10m towards west..
First the man walks towards north. So it is 30 j^
Then he walks in east direction, So 20i^
Then finally 30✓2 towards South-west. resolve in x and y direction, which is = -30✓2cos thita + (-30✓2 sin thita)= -30i^ - 30j^.
For net displacement add, +30j^ -30j^ + 20i^ - 30i^ = - 10i^.So it is 10m towards west..
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