Answer:
Given:
- Magnitude of [tex]\( \mathbf{A} \)[/tex]:[tex]\( 28 \, \text{m/s} \)[/tex]
- Direction: [tex]\( 30^\circ \)[/tex] northeast
To find the components, we use trigonometry:
### East (Horizontal) Component [tex](\( A_x \))[/tex]:
[tex] A_x = A \cos(\theta) [/tex]
[tex] A_x = 28 \, \text{m/s} \cos(30^\circ) [/tex]
[tex] A_x = 28 \, \text{m/s} \cdot \frac{\sqrt{3}}{2} [/tex]
[tex] A_x = 14\sqrt{3} \, \text{m/s} \approx 24.25 \, \text{m/s} [/tex]
### North (Vertical) Component [tex](\( A_y \))[/tex]:
[tex] A_y = A \sin(\theta) [/tex]
[tex] A_y = 28 \, \text{m/s} \sin(30^\circ) [/tex]
[tex] A_y = 28 \, \text{m/s} \cdot \frac{1}{2} [/tex]
[tex] A_y = 14 \, \text{m/s} [/tex]
Thus, the components of vector [tex]\( \mathbf{A} \) are:
- East (Horizontal) Component: [tex] \approx 24.25 \, \text{m/s} [/tex]
- North (Vertical) Component: [tex] 14 \, \text{m/s} [/tex]
In vector form, we can express [tex]\( \mathbf{A} \)[/tex] as:
[tex] \mathbf{A} = (24.25 \, \text{m/s} \hat{i}) + (14 \, \text{m/s} \hat{j}) [/tex]
Where [tex] \hat{i} [/tex] and [tex] \hat{j} [/tex] are the unit vectors in the east (horizontal) and north (vertical) directions, respectively.
