A motor car is travelling at `60m//s` on a circular road of radius 1200m. It is increasing its speed at the rate of `4m//s^(2)`. The acceleration of the car is:

Explanation

To solve the problem, we need to find the total acceleration of the motor car, which is moving in a circular path while also increasing its speed. The total acceleration consists of two components: tangential acceleration and radial (centripetal) acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity (V) = 60 m/s
- Radius of the circular path (r) = 1200 m
- Tangential acceleration (a_t) = 4 m/s²

2. **Calculate Radial (Centripetal) Acceleration:**
- Radial acceleration (a_r) can be calculated using the formula:
\[ a_r = \frac{V^2}{r} \]
- Substituting the values:
\[ a_r = \frac{(60)^2}{1200} = \frac{3600}{1200} = 3 \text{ m/s}^2 \]

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

4. **Final Answer:**
- The total acceleration of the car is:
\[ a = 5 \text{ m/s}^2 \]