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Question

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 (Figure). 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?
984317_db11bce0a56c49759fd0ac23154b9d64.png


Solution

Net movement along $$x$$-direction,
$$S_x = (6-4) \cos 45^o \, \hat{i} = 2\times \dfrac{1}{\sqrt{2}} = \sqrt{2}km$$
Net movement along $$y$$-direction,
$$S_y = (6+4) \sin\, 45^o \, \hat{j} = 10\times \dfrac{1}{\sqrt{2}} = 5\sqrt{2}km$$
Net movement from startinf 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,
$$\tan \theta = \dfrac{\text{Y-component}}{\text{X-component}} = \dfrac{5\sqrt{2}}{\sqrt{2}}$$
$$\therefore \theta = \tan^{-1}(5)$$

Physics

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