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A car makes a displacement 100m towards east and then 200m towards north. Find the magnitude and direction of the resultant

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$223.7 m \tan^{- 1} (2) N o f E$

$223.7 m \tan^{- 1} (2) E o f N$

$300 m \tan^{- 1} (2) N o f E$

$100 m \tan^{- 1} (2) N o f E$

answer is A.

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Detailed Solution

$| \vec{O B} | = \sqrt{O A^{2} + A B^{2}}$
$\vec{O B} = \sqrt{{(100)}^{2} + {(200)}^{2}}$
$\vec{O B} = 100 \sqrt{5} = 223.7 m$
$\tan α = \frac{200}{100} \Rightarrow α = \tan^{- 1} (2) N o f E$

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