A car travels a distance of 6 , km at an angl | vedclass

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Question

(C) Let the origin be $(0, 0)$.The first displacement vector $\vec{d_1}$ is $6 \, km$ at $45^\circ$ with the East (x-axis).$\vec{d_1} = 6 \cos 45^\cir

Vedclass · Accepted Answer

$\sqrt{52} \, km$

Vedclass · Answer

(C) Let the origin be $(0, 0)$.The first displacement vector $\vec{d_1}$ is $6 \, km$ at $45^\circ$ with the East (x-axis).$\vec{d_1} = 6 \cos 45^\circ \hat{i} + 6 \sin 45^\circ \hat{j} = 6 \left( \frac{1}{\sqrt{2}} \right) \hat{i} + 6 \left( \frac{1}{\sqrt{2}} \right) \hat{j} = 3\sqrt{2} \hat{i} + 3\sqrt{2} \hat{j}$.The second displacement vector $\vec{d_2}$ is $4 \, km$ at $135^\circ$ with the East.$\vec{d_2} = 4 \cos 135^\circ \hat{i} + 4 \sin 135^\circ \hat{j} = 4 \left( -\frac{1}{\sqrt{2}} \right) \hat{i} + 4 \left( \frac{1}{\sqrt{2}} \right) \hat{j} = -2\sqrt{2} \hat{i} + 2\sqrt{2} \hat{j}$.The resultant displacement vector $\vec{s} = \vec{d_1} + \vec{d_2} = (3\sqrt{2} - 2\sqrt{2}) \hat{i} + (3\sqrt{2} + 2\sqrt{2}) \hat{j} = \sqrt{2} \hat{i} + 5\sqrt{2} \hat{j}$.The magnitude of the displacement is $|\vec{s}| = \sqrt{(\sqrt{2})^2 + (5\sqrt{2})^2} = \sqrt{2 + 50} = \sqrt{52} \, km$.

$A$ car travels a distance of $6 \, km$ at an angle of $45^\circ$ with the East,and then travels a distance of $4 \, km$ at an angle of $135^\circ$ with the East. What is the distance from the origin?

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