Question: User: y
B\⃗ (15.0 m)
30.0°
D\⃗ (10.0 m)
53.0°
25.0°
C\⃗ (12.0 m)
O
x
A\⃗ (8.00 m)
Graph shapes and math:
A\⃗ (8.00 m) is a vector drawn from the origin straight downward along the negative y-axis, position_hint = bottom-center.
B\⃗ (15.0 m) is a vector drawn from the origin into the top-right quadrant, making 30.0° with the positive y-axis, position_hint = top-right.
D\⃗ (10.0 m) is a vector drawn from the origin into the top-left quadrant, making 53.0° with the positive y-axis, position_hint = top-left.
C\⃗ (12.0 m) is a vector drawn from the origin into the bottom-left quadrant, making 25.0° below the negative x-axis, position_hint = bottom-left.
Use your answers to find the magnitude of negative ModifyingAbove Upper A with right arrow minus ModifyingAbove Upper B with right arrow.
Express your answer in meters.
1. **State the problem:** We need to find the magnitude of the vector $-\vec{A} - \vec{B}$ given vectors $\vec{A}$ and $\vec{B}$.
2. **Given vectors:**
- $\vec{A}$ has magnitude $8.00$ m and points straight downward along the negative y-axis.
- $\vec{B}$ has magnitude $15.0$ m and makes an angle of $30.0^\circ$ with the positive y-axis in the top-right quadrant.
3. **Express vectors in components:**
- For $\vec{A}$ pointing downward along negative y-axis:
$$\vec{A} = (0, -8.00)$$
- For $\vec{B}$, angle $30^\circ$ with positive y-axis means its components are:
$$B_x = 15.0 \sin 30^\circ = 15.0 \times 0.5 = 7.5$$
$$B_y = 15.0 \cos 30^\circ = 15.0 \times \frac{\sqrt{3}}{2} \approx 15.0 \times 0.866 = 12.99$$
So,
$$\vec{B} = (7.5, 12.99)$$
4. **Calculate $-\vec{A}$:**
$$-\vec{A} = (0, 8.00)$$
5. **Calculate $-\vec{A} - \vec{B}$:**
$$-\vec{A} - \vec{B} = (0 - 7.5, 8.00 - 12.99) = (-7.5, -4.99)$$
6. **Find magnitude:**
$$| -\vec{A} - \vec{B} | = \sqrt{(-7.5)^2 + (-4.99)^2} = \sqrt{56.25 + 24.90} = \sqrt{81.15} \approx 9.01$$
**Final answer:** The magnitude of $-\vec{A} - \vec{B}$ is approximately $9.01$ meters.