The particle on the circular path has a speed VA = 5 m/s at point A, which is increasing at a constant rate of 1.5 m/s2. What is its speed at point B? B AVA = 5 m/s A 2.5 m OVB OVB 6.06 m/s 3.64 m/s VB 8.49 m/s UB = 6.97 m/s

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### Understanding Circular Motion: Speed Calculation

In this problem, we analyze the speed of a particle moving along a circular path. The particle has a speed \( v_A = 5 \, \text{m/s} \) at point \( A \), and the speed increases at a constant rate of \( 1.5 \, \text{m/s}^2 \). We aim to determine its speed at another point \( B \) on the circular path.

Consider the given data:

1. **Speed at point A (\( v_A \))**: \( 5 \, \text{m/s} \)
2. **Radial acceleration (\( \frac{dv_A}{dt} \))**: \( 1.5 \, \text{m/s}^2 \)
3. **Radius of the circular path (\( r \))**: \( 2.5 \, \text{m} \)

**Diagram Explanation:**

- The red circle represents the path of motion.
- Point \( A \) is where the initial speed is given.
- Point \( B \) is where we want to calculate the speed.
- The radius \( r \) is marked as \( 2.5 \, \text{m} \).

Given options for the speed at point \( B \) are:
- \( v_B = 6.06 \, \text{m/s} \)
- \( v_B = 3.64 \, \text{m/s} \)
- \( v_B = 8.49 \, \text{m/s} \)
- \( v_B = 6.97 \, \text{m/s} \)

To solve this, we can use the kinematic equations of motion, considering the constant tangential acceleration and the initial speed.

```markdown
v_B = \sqrt{v_A^2 + 2 \cdot a \cdot s}
```

Where \( s = \pi \cdot r \).

Using these, we'll determine the correct answer.

### Multiple Choice Options:
- \( v_B = 6.06 \, \text{m/s} \)
- \( v_B = 3.64 \, \text{m/s} \)
- \( v_B = 8.49 \, \text{m/s} \)
- \( v_B = 6.97 \, \text
Transcribed Image Text:### Understanding Circular Motion: Speed Calculation In this problem, we analyze the speed of a particle moving along a circular path. The particle has a speed \( v_A = 5 \, \text{m/s} \) at point \( A \), and the speed increases at a constant rate of \( 1.5 \, \text{m/s}^2 \). We aim to determine its speed at another point \( B \) on the circular path. Consider the given data: 1. **Speed at point A (\( v_A \))**: \( 5 \, \text{m/s} \) 2. **Radial acceleration (\( \frac{dv_A}{dt} \))**: \( 1.5 \, \text{m/s}^2 \) 3. **Radius of the circular path (\( r \))**: \( 2.5 \, \text{m} \) **Diagram Explanation:** - The red circle represents the path of motion. - Point \( A \) is where the initial speed is given. - Point \( B \) is where we want to calculate the speed. - The radius \( r \) is marked as \( 2.5 \, \text{m} \). Given options for the speed at point \( B \) are: - \( v_B = 6.06 \, \text{m/s} \) - \( v_B = 3.64 \, \text{m/s} \) - \( v_B = 8.49 \, \text{m/s} \) - \( v_B = 6.97 \, \text{m/s} \) To solve this, we can use the kinematic equations of motion, considering the constant tangential acceleration and the initial speed. ```markdown v_B = \sqrt{v_A^2 + 2 \cdot a \cdot s} ``` Where \( s = \pi \cdot r \). Using these, we'll determine the correct answer. ### Multiple Choice Options: - \( v_B = 6.06 \, \text{m/s} \) - \( v_B = 3.64 \, \text{m/s} \) - \( v_B = 8.49 \, \text{m/s} \) - \( v_B = 6.97 \, \text
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