A displacement vector $D$ is given as $40.0 \mathrm{~m}$ at an angle of 60.0 degrees EAST of NORTH. The $D_{x}$ component and the $\mathrm{D}_{\mathrm{y}}$ component of the vector are
A) $D_{\mathrm{x}}=+34.6 \mathrm{~m}, \mathrm{D}_{\mathrm{y}}=+20.0 \mathrm{~m}$.
B) $\mathrm{D}_{\mathrm{x}}=+34.6 \mathrm{~m}, \mathrm{D}_{\mathrm{y}}=-20.0 \mathrm{~m}$.
C) $\mathrm{D}_{\mathrm{x}}=-34.6 \mathrm{~m}, \mathrm{D}_{\mathrm{y}}=+20.0 \mathrm{~m}$.
D) $\mathrm{D}_{\mathrm{x}}=+20.0 \mathrm{~m}, \mathrm{D}_{\mathrm{y}}=-34.6 \mathrm{~m}$.
E) $\mathrm{D}_{\mathrm{x}}=+20.0 \mathrm{~m}, \mathrm{D}_{\mathrm{y}}=+34.6 \mathrm{~m}$.

Question

Quizplus · Accepted Answer

The displacement vector's components can be found using trigonometry. $D_x = D \cos(\theta)$ and $D_y = D \sin(\theta)$, where $\theta$ is the angle from the north, making it 30 degrees from the east (90 - 60). Thus, $D_x = 40 \cos(30^\circ) = 34.6$ m, and $D_y = 40 \sin(30^\circ) = 20.0$ m, with $D_x$ being eastward and $D_y$ northward, both positive.

A Displacement Vector DDD Is Given As 40.0 m40.0 \mathrm{~m}40.0 m At an Angle of 60

A Displacement Vector $D$ Is Given As $40.0 \mathrm{~m}$ At an Angle of 60