The text in the image is asking about a velocity vector that forms a 45° angle with the positive x-axis, and its magnitude is 60 m/s. The task is to find the components of the vector.

To solve this, we can use trigonometry to find the components of the velocity vector. The vector can be represented in terms of its horizontal (x) and vertical (y) components using the following formulas:

• $v_x = v \cdot \cos(\theta)$
• $v_y = v \cdot \sin(\theta)$

Where:

• $v$ is the magnitude of the vector (60 m/s),
• $\theta$ is the angle (45°),
• $v_x$ and $v_y$ are the x and y components of the vector.

Now, plugging in the values:

1. For $v_x$: $v_x = 60 \cdot \cos(45^\circ) = 60 \cdot \frac{\sqrt{2}}{2} \approx 60 \cdot 0.707 \approx 42.43 \, \text{m/s}$

2. For $v_y$: $v_y = 60 \cdot \sin(45^\circ) = 60 \cdot \frac{\sqrt{2}}{2} \approx 60 \cdot 0.707 \approx 42.43 \, \text{m/s}$

Thus, the components of the velocity vector are approximately:

• $v_x \approx 42.43 \, \text{m/s}$
• $v_y \approx 42.43 \, \text{m/s}$

If you need further explanations or have more questions, feel free to ask!

Here are 5 related questions you might find helpful:

1. How do I interpret vector components in physics?
2. What is the significance of the angle between vectors?
3. How can I find the magnitude of a vector given its components?
4. What other methods can be used to decompose vectors into components?
5. What would happen if the angle were 90° or 0°?

Tip: When working with vectors, always double-check the angle's orientation relative to the axes, as that will affect which trigonometric functions to use (sine for y-components, cosine for x-components).