A particle moving along the circular path with a speed v and its speed increases by g in one second. If the radius of the circular path be r, then the net acceleration of the particle is:

Options

A. `v^2/r+g`
B. `v^2/r^2+g^2`
C. `[v^4/r^2+g^2]^(1//2)`
D. `[v^2/r+g]^(1//2)`

Explanation

To find the net acceleration of a particle moving along a circular path with increasing speed, we can follow these steps:

### Step 1: Identify the types of acceleration
In circular motion, there are two components of acceleration:
1. **Tangential acceleration (A_t)**: This is due to the change in speed along the circular path. Given that the speed increases by \( g \) in one second, we have:
\[ A_t = g \, \text{(m/s}^2\text{)} \]
2. **Radial (centripetal) acceleration (A_r)**: This is due to the change in direction of the velocity vector as the particle moves along the circular path. It is given by the formula:
\[ A_r = \frac{v^2}{r} \]

### Step 2: Determine the angle between the accelerations
The tangential acceleration and radial acceleration are perpendicular to each other. Therefore, the angle \( \theta \) between them is \( 90^\circ \).

### Step 3: Calculate the net acceleration
The net acceleration \( A_{net} \) can be calculated using the Pythagorean theorem since the two components are perpendicular:
\[ A_{net} = \sqrt{A_t^2 + A_r^2} \]
Substituting the values we have:
\[ A_{net} = \sqrt{g^2 + \left(\frac{v^2}{r}\right)^2} \]

### Step 4: Simplify the expression
This can be rewritten as:
\[ A_{net} = \sqrt{g^2 + \frac{v^4}{r^2}} \]

### Final Answer
Thus, the net acceleration of the particle is:
\[ A_{net} = \sqrt{g^2 + \frac{v^4}{r^2}} \]

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