A car is moving with speed of `2ms^(-1)` on a circular path of radius 1 m and its speed is increasing at the rate of `3ms(-1)` The net acceleration of the car at this moment in `m//s^2` is

Explanation

To find the net acceleration of the car moving on a circular path, we need to consider both the centripetal acceleration and the tangential acceleration. Here’s how we can solve the problem step by step:

### Step 1: Identify the given values
- Speed of the car, \( V = 2 \, \text{m/s} \)
- Radius of the circular path, \( R = 1 \, \text{m} \)
- Rate of increase of speed (tangential acceleration), \( a_t = 3 \, \text{m/s}^2 \)

### Step 2: Calculate the centripetal acceleration
Centripetal acceleration (\( a_c \)) is given by the formula:
\[ a_c = \frac{V^2}{R} \]
Substituting the values:
\[ a_c = \frac{(2 \, \text{m/s})^2}{1 \, \text{m}} = \frac{4 \, \text{m}^2/\text{s}^2}{1 \, \text{m}} = 4 \, \text{m/s}^2 \]

### Step 3: Identify the tangential acceleration
The tangential acceleration (\( a_t \)) is already given as:
\[ a_t = 3 \, \text{m/s}^2 \]

### Step 4: Calculate the net acceleration
The net acceleration (\( a \)) is the vector sum of the centripetal and tangential accelerations. Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem:
\[ a = \sqrt{a_c^2 + a_t^2} \]
Substituting the values:
\[ a = \sqrt{(4 \, \text{m/s}^2)^2 + (3 \, \text{m/s}^2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m/s}^2 \]

### Final Answer
The net acceleration of the car at this moment is:
\[ \boxed{5 \, \text{m/s}^2} \]

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