A car travels 6 , km at an angle of 45^circ n | vedclass

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(C) Let the starting point be the origin $(0,0)$.First displacement vector $\vec{d_1} = 6 \cos(45^\circ) \hat{i} + 6 \sin(45^\circ) \hat{j} = 6(\frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the starting point be the origin $(0,0)$.First displacement vector $\vec{d_1} = 6 \cos(45^\circ) \hat{i} + 6 \sin(45^\circ) \hat{j} = 6(\frac{1}{\sqrt{2}}) \hat{i} + 6(\frac{1}{\sqrt{2}}) \hat{j} = 3\sqrt{2} \hat{i} + 3\sqrt{2} \hat{j} \, km$.Second displacement vector $\vec{d_2} = 4 \cos(135^\circ) \hat{i} + 4 \sin(135^\circ) \hat{j} = 4(-\frac{1}{\sqrt{2}}) \hat{i} + 4(\frac{1}{\sqrt{2}}) \hat{j} = -2\sqrt{2} \hat{i} + 2\sqrt{2} \hat{j} \, km$.Resultant displacement $\vec{R} = \vec{d_1} + \vec{d_2} = (3\sqrt{2} - 2\sqrt{2}) \hat{i} + (3\sqrt{2} + 2\sqrt{2}) \hat{j} = \sqrt{2} \hat{i} + 5\sqrt{2} \hat{j} \, km$.Magnitude $R = \sqrt{(\sqrt{2})^2 + (5\sqrt{2})^2} = \sqrt{2 + 50} = \sqrt{52} \, km$.Angle $	heta$ with east: $	an 	heta = \frac{R_y}{R_x} = \frac{5\sqrt{2}}{\sqrt{2}} = 5 \implies 	heta = 	an^{-1}(5)$.

$A$ car travels $6 \, km$ at an angle of $45^\circ$ north of east and then travels $4 \, km$ at an angle of $135^\circ$ north of east. How far is the final point from the starting point? What angle does the straight line joining its initial and final position make with the east?

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