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. `5ms^(-2)`
B. `2ms^(-2)`
C. `3.2ms^(-2)`
D. `4.3ms^(-2)`

Explanation

To find the net acceleration of a particle moving in a circular path with given parameters, we can follow these steps:

### Step 1: Identify the given values
- Radius of the circular path, \( R = 10 \, \text{m} \)
- Speed of the particle, \( v = 5 \, \text{m/s} \)
- Rate of increase of speed (tangential acceleration), \( a_t = 2 \, \text{m/s}^2 \)

### Step 2: Calculate the centripetal acceleration
Centripetal acceleration (\( a_c \)) can be calculated using the formula:
\[ a_c = \frac{v^2}{R} \]
Substituting the known values:
\[ a_c = \frac{(5 \, \text{m/s})^2}{10 \, \text{m}} = \frac{25}{10} = 2.5 \, \text{m/s}^2 \]

### Step 3: Calculate the net acceleration
The net acceleration (\( a_{net} \)) is the vector sum of the tangential acceleration and the centripetal acceleration. 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 have:
\[ a_{net} = \sqrt{(2 \, \text{m/s}^2)^2 + (2.5 \, \text{m/s}^2)^2} \]
Calculating the squares:
\[ = \sqrt{4 + 6.25} = \sqrt{10.25} \]

### Step 4: Calculate the final value
Now, we find 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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