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:

Options

A. `3 m//s^(2)`
B. `5 m//s^(2)`
C. `5 m//s^(2)`
D. `7 m//s^(2)`

Explanation

To solve the problem, we need to find the resultant acceleration of the motor car that is traveling on a circular road while also increasing its speed. The total acceleration in this case consists of two components: tangential acceleration and centripetal acceleration.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of the car, \( v = 60 \, \text{m/s} \)
- Radius of the circular road, \( r = 1200 \, \text{m} \)
- Tangential acceleration, \( a_t = 4 \, \text{m/s}^2 \)

2. **Calculate Centripetal Acceleration:**
The formula for centripetal acceleration (\( a_c \)) is given by:
\[ a_c = \frac{v^2}{r} \]
Substituting the known values:
\[ a_c = \frac{(60)^2}{1200} = \frac{3600}{1200} = 3 \, \text{m/s}^2 \]

3. **Calculate the Resultant Acceleration:**
The resultant acceleration (\( a_r \)) is found by combining the tangential and centripetal accelerations using the Pythagorean theorem:
\[ a_r = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a_r = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m/s}^2 \]

4. **Final Result:**
The resultant acceleration of the car is:
\[ a_r = 5 \, \text{m/s}^2 \]