The problem requires finding vector $\vec{D} = 4.10 \vec{A} + \vec{B}$, where vectors $\vec{A}$ and $\vec{B}$ are given in the image. Let's break this down:

Step 1: Decompose the vectors $\vec{A}$ and $\vec{B}$ into their components.

• Vector $\vec{A}$ is given as 2.0 m in magnitude and directed at 15° above the positive $y$-axis.

  • The components of $\vec{A}$: A_x = 2.0 \sin(15^\circ) \quad \text{(since it's tilted towards the x-axis)} $A_y = 2.0 \cos(15^\circ)$

• Vector $\vec{B}$ is given as 4.0 m in magnitude and directed at 15° below the positive $x$-axis.

  • The components of $\vec{B}$: $B_x = 4.0 \cos(15^\circ)$ B_y = -4.0 \sin(15^\circ) \quad \text{(since it's below the x-axis)}

Step 2: Calculate the components of $4.10 \vec{A}$.

Since we are asked for $4.10 \vec{A}$, we multiply the components of $\vec{A}$ by 4.10: $(4.10 A_x, 4.10 A_y)$

Step 3: Add the vectors $\vec{D} = 4.10 \vec{A} + \vec{B}$.

To find the components of $\vec{D}$, sum the $x$- and $y$-components of $4.10 \vec{A}$ and $\vec{B}$ separately: $D_x = 4.10 A_x + B_x$ $D_y = 4.10 A_y + B_y$

Now, I will calculate the components using these expressions.The components of vector $\vec{D} = 4.10 \vec{A} + \vec{B}$ are approximately:

$D_x = 5.99 \, \text{m}, \quad D_y = 6.89 \, \text{m}$

Would you like further details or explanations on any step? Here are some related questions that might expand on this:

1. How are vector components calculated from magnitudes and angles?
2. Why are trigonometric functions used to find vector components?
3. How does the direction of a vector affect the signs of its components?
4. What is the significance of multiplying a vector by a scalar?
5. How can we find the magnitude and direction of the resultant vector $\vec{D}$?

Tip: When adding vectors, always ensure you resolve them into components to make the process easier and more accurate.