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.