The vector \( \vec{v} \) is given as \( 110 \, \text{m/s} \) at a direction of \( 40^\circ \) West of South (S40°W). To break this down:

### Understanding the Direction:
- The notation \( \text{S40}^\circ\text{W} \) means that the vector points 40° West of the South direction.

### Decomposition into Components:
We can decompose the vector into its **South** and **West** components using trigonometry:

1. The **Southward component** (\( v_y \)) is along the negative \( y \)-axis.
2. The **Westward component** (\( v_x \)) is along the negative \( x \)-axis.

For a vector of magnitude \( 110 \, \text{m/s} \):

\[
v_y = 110 \cdot \cos(40^\circ)
\]
\[
v_x = 110 \cdot \sin(40^\circ)
\]

Let's calculate these components.The components of the velocity vector \( \vec{v} \) are:

- Southward component (\( v_y \)): \( 84.26 \, \text{m/s} \)
- Westward component (\( v_x \)): \( 70.71 \, \text{m/s} \)

This means the object is moving at a speed of \( 84.26 \, \text{m/s} \) southward and \( 70.71 \, \text{m/s} \) westward.

Would you like more details on this, or do you have any further questions?

Here are some related questions for practice:
1. How would the components change if the direction was \( \text{N40}^\circ\text{E} \)?
2. What is the magnitude of the vector if only the Southward component is doubled?
3. What is the total displacement in 5 seconds along this velocity?
4. How would the components be affected if the magnitude of the velocity increased to \( 150 \, \text{m/s} \)?
5. Can you express the vector in terms of unit vectors \( \hat{i} \) and \( \hat{j} \)?

**Tip:** For vectors, always check the reference direction when decomposing to avoid sign errors in your components.

Learn how to decompose a velocity vector of 110 m/s at S40°W into its southward and westward components using trigonometry. Detailed breakdown of the components as 84.26 m/s south and 70.71 m/s west. Suitable for Grades 10-12.

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