To determine Tia's velocity in vector form, we first note that she is traveling at a speed of 16 miles per hour at an angle of [tex]\(30^\circ\)[/tex] west of north.
1. Convert the Angle:
- Since Tia's direction is [tex]\(30^\circ\)[/tex] west of north, we can consider the angle relative to the north direction.
- The north direction is typically represented by the positive [tex]\(j\)[/tex] direction (the y-axis), and west would be in the negative [tex]\(i\)[/tex] direction (the x-axis).
2. Calculate the Components of the Velocity Vector:
- We break down the velocity into its components using trigonometry. Specifically, we use sine and cosine for the angle relative to the north direction.
3. Velocity in the North (y) Direction:
- The northward component (y-component) is given by the cosine of the angle times the speed because it is adjacent to the angle.
- So, this component is [tex]\( vy = 16 \cdot \cos(30^\circ) \)[/tex].
4. Velocity in the West (x) Direction:
- The westward component (x-component) is given by the sine of the angle times the speed because it is opposite to the angle. Note that this direction is negative in the standard Cartesian plane.
- So, this component is [tex]\( vx = -16 \cdot \sin(30^\circ) \)[/tex].
5. Values of Sine and Cosine for [tex]\(30^\circ\)[/tex]:
- [tex]\(\sin(30^\circ) = \frac{1}{2}\)[/tex]
- [tex]\(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)[/tex]
6. Calculate the Components:
- [tex]\( vx = -16 \cdot \frac{1}{2} = -8 \)[/tex]
- [tex]\( vy = 16 \cdot \frac{\sqrt{3}}{2} = 8\sqrt{3} \)[/tex]
7. Combine into a Velocity Vector:
- The vector representing Tia's velocity is therefore [tex]\( \vec{v} = -8i + 8\sqrt{3}j \)[/tex].
Hence, the correct answer is:
[tex]\[ B. \, -8i + 8\sqrt{3}j \][/tex]
