A car travels 6 , km at an angle of 45^o nort | vedclass

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(C) Let the starting point be the origin $(0,0)$.First displacement vector $\vec{d_1} = 6 \cos(45^o) \hat{i} + 6 \sin(45^o) \hat{j} = 6(\frac{1}{\sqrt

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

Vedclass · Answer

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

Column $-I$	Column $-II$
$(a) \vec a + \vec b = \vec c$	$(i)$ Vector $\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ connects the origin to the tip of $\vec c$
$(b) \vec a - \vec c = \vec b$	$(ii)$ Vector $\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ connects the origin to the tip of $\vec b$
$(c) \vec b - \vec a = \vec c$	$(iii)$ Vector $\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$
$(d) \vec a + \vec b + \vec c = 0$	$(iv)$ Vector $\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ connects the origin to the tip of $\vec b$

$A$ car travels $6 \, km$ at an angle of $45^o$ north of east and then travels a distance of $4 \, km$ at an angle of $135^o$ 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 direction?

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