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 in a circular path with an increasing speed, we need to consider both the tangential acceleration and the centripetal (or radial) acceleration.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of the car (v) = 2 m/s
- Radius of the circular path (r) = 1 m
- Tangential acceleration (a_t) = 3 m/s²

2. **Calculate the Centripetal Acceleration (a_c):**
The formula for centripetal acceleration is given by:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(2 \, \text{m/s})^2}{1 \, \text{m}} = \frac{4 \, \text{m}^2/\text{s}^2}{1 \, \text{m}} = 4 \, \text{m/s}^2 \]

3. **Combine the Accelerations:**
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:
\[ a_{net} = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} \]
\[ a_{net} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

4. **Conclusion:**
The net acceleration of the car at this moment is \(5 \, \text{m/s}^2\).