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Standard XII

Physics

Vector Addition

A particle ha...

Question

A particle has displacement of 12 m towards east and 5 m towards north then 6 m vertically upward. The sum of these displacements is

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10.04 m

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14.31 m

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None of these

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Solution

The correct option is C
14.31 m

$R = \sqrt{12^{2} + 5^{2} + 6^{2}} = \sqrt{144 + 25 + 36} = \sqrt{205} = 14.31 m$

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A particle has displacement of 12 m towards east and 5 m towards north then 6 m vertically upward. The sum of these displacements is

The correct option is C 14.31 m R=√122+52+62=√144+25+36=√205=14.31m

The correct option is C
14.31 m

$R = \sqrt{12^{2} + 5^{2} + 6^{2}} = \sqrt{144 + 25 + 36} = \sqrt{205} = 14.31 m$