A man moves $4 m$ along east, then $3 m$ along north direction, after the climbs up a pole to a height $12 m$ . Find the distance covered by him and his displacement.

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Solution

Distance is just $a l g e b r a i c$ sum of the individual distance so distance $d = 4 m + 3 m + 12 m = 19 m$
The horizontal displacement will be $x = \sqrt{x_{e}^{2} + x_{N}^{2}} = \sqrt{16 + 9} = 5 m$
The vertical displacement will be $y = 12 m$
so the $n e t$ displacement from starting point will be $\sqrt{x^{2} + y^{2}} = \sqrt{25 + 144} = 13 m$

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A man moves 4 malong east, then 3 m along north direction, after the climbs up a pole to a height 12 m. Find the distance covered by him and his displacement.

Distance is just algebraic sum of the individual distance so distance d=4m+3m+12m=19mThe horizontal displacement will be x=√x2e+x2N=√16+9=5mThe vertical displacement will be y=12mso the net displacement from starting point will be √x2+y2=√25+144=13m

A man moves $4 m$ along east, then $3 m$ along north direction, after the climbs up a pole to a height $12 m$ . Find the distance covered by him and his displacement.