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. `4 m//s^(2)`
C. `5 m//s^(2)`
D. `7 m//s^(2)`

Answer

To find the total acceleration of the motor car traveling on a circular road while increasing its speed, we need to consider both the tangential acceleration and the centripetal acceleration. Here’s how we can solve the problem step by step:

### Step 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, \( a_t = 4 \, \text{m/s}^2 \)

### Step 2: Calculate the centripetal acceleration
Centripetal acceleration (\( a_c \)) can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(60 \, \text{m/s})^2}{1200 \, \text{m}} = \frac{3600 \, \text{m}^2/\text{s}^2}{1200 \, \text{m}} = 3 \, \text{m/s}^2 \]

### Step 3: Determine the resultant acceleration
The total acceleration (\( a \)) of the car 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:
\[ a = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a = \sqrt{(4 \, \text{m/s}^2)^2 + (3 \, \text{m/s}^2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m/s}^2 \]

### Conclusion
The total acceleration of the car is \( 5 \, \text{m/s}^2 \).

---