A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is `5ms^(-1)` and the speed is increasing at a rate of `2ms^(-2)`. At this instant, the magnitude of the net acceleration will be

Options

A. `3.2ms^(-2)`
B. `2ms^(-2)`
C. `1.2ms^(-2)`
D. `4.3ms^(-2)`

Explanation

To find the magnitude of the net acceleration of a particle moving on a circular path, we need to consider both the tangential acceleration and the centripetal acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of the circular path, \( R = 10 \, \text{m} \)
- Speed of the particle, \( V = 5 \, \text{m/s} \)
- Tangential acceleration, \( a_t = 2 \, \text{m/s}^2 \)

2. **Calculate Centripetal Acceleration:**
- The formula for centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{V^2}{R} \]
- Substituting the known values:
\[ a_c = \frac{(5 \, \text{m/s})^2}{10 \, \text{m}} = \frac{25}{10} = 2.5 \, \text{m/s}^2 \]

3. **Determine the Net Acceleration:**
- The net acceleration \( a_{net} \) is the vector sum of the tangential acceleration \( a_t \) and the centripetal acceleration \( a_c \). Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem:
\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]
- Substituting the values we calculated:
\[ a_{net} = \sqrt{(2 \, \text{m/s}^2)^2 + (2.5 \, \text{m/s}^2)^2} \]
- Calculating the squares:
\[ a_{net} = \sqrt{4 + 6.25} = \sqrt{10.25} \]

4. **Calculate the Final Value:**
- Taking the square root:
\[ a_{net} \approx 3.2 \, \text{m/s}^2 \]

### Final Answer:
The magnitude of the net acceleration is approximately \( 3.2 \, \text{m/s}^2 \).

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