A car travels 6 km towards north at an angle of 45^o to the east and then travels distance of 4 km towards north at an angle of 135^o to the east. How far is the point from the
starting point? What angle does the straight line joining its initial and final position makes with the east?

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$\sqrt{50} km and \tan^{- 1} (5)$

$10 km and \tan^{- 1} (\sqrt{5})$

$\sqrt{52} km and \tan^{- 1} (5)$

$\sqrt{52} km and \tan^{- 1} (\sqrt{5})$

answer is C.

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Detailed Solution

Net movement along x-direction

$S_{x} = (6 - 4) \cos 45^{\circ} \hat{i} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} km$

Net movement along y-direction

$S_{y} = (6 + 4) \sin 45^{\circ} \hat{j} = 10 \times \frac{1}{\sqrt{2}} = 5 \sqrt{2} km$

Net movement from starting point

$| \vec{S} | = \sqrt{S_{x}^{2} + S_{y}^{2}} = \sqrt{(\sqrt{2})^{2} + (5 \sqrt{2})^{2}} = \sqrt{52} km$

Angle which makes with the east direction

$\tan θ = \frac{y -component}{x -component} = \frac{5 \sqrt{2}}{\sqrt{2}}$

$∴ θ = \tan^{- 1} (5)$

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