A motor car travelling at `30 m//s` on a circular road of radius `500m`. It is increasing its speed at the rate of `2 ms^(-2)`. What its accleration at that instant ?

Options

A. `2(m)/(sec^2)`
B. `2.7(m)/(sec^2)`
C. `1.8(m)/(sec^2)`
D. `9.8(m)/(sec^2)`

Explanation

To find the acceleration of the motor car at that instant, we need to consider both the tangential acceleration and the centripetal acceleration since the car is moving in a circular path and its speed is increasing.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Speed of the car, \( v = 30 \, \text{m/s} \)
- Radius of the circular path, \( R = 500 \, \text{m} \)
- Tangential acceleration, \( a_t = 2 \, \text{m/s}^2 \)

2. **Calculate Centripetal Acceleration:**
Centripetal acceleration (\( a_c \)) is given by the formula:
\[ a_c = \frac{v^2}{R} \]
Substituting the values:
\[ a_c = \frac{(30 \, \text{m/s})^2}{500 \, \text{m}} = \frac{900}{500} = 1.8 \, \text{m/s}^2 \]

3. **Determine the Net Acceleration:**
The net acceleration (\( a_{\text{net}} \)) 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:
\[ a_{\text{net}} = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a_{\text{net}} = \sqrt{(2 \, \text{m/s}^2)^2 + (1.8 \, \text{m/s}^2)^2} \]
\[ a_{\text{net}} = \sqrt{4 + 3.24} = \sqrt{7.24} \approx 2.69 \, \text{m/s}^2 \]

4. **Final Result:**
Therefore, the acceleration of the motor car at that instant is approximately:
\[ a_{\text{net}} \approx 2.7 \, \text{m/s}^2 \]