A car is travelling with linear velocity v on a circular road of radius r. If it is increasing its speed at the rate of `a` metre/`sec^2`, then the resultant acceleration will be

Options

A. `sqrt({(v^2)/(r^2)-a^2})`
B. `sqrt({(v^4)/(r^2)+a^2})`
C. `sqrt({(v^4)/(r^2)-a^2})`
D. `sqrt({(v^2)/(r^2)+a^2})`

Explanation

To solve the problem, we need to find the resultant acceleration of a car moving in a circular path while increasing its speed. The car has two components of acceleration: tangential acceleration and centripetal acceleration.

### Step-by-Step Solution:

1. **Identify the Components of Acceleration**:
- The car is moving with a linear velocity \( v \) on a circular path of radius \( r \).
- It is increasing its speed at a rate of \( a \) m/s², which is the tangential acceleration \( a_t \).
- The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{v^2}{r} \]

2. **Determine the Type of Accelerations**:
- The tangential acceleration \( a_t \) acts along the direction of the velocity (tangential to the circular path).
- The centripetal acceleration \( a_c \) acts towards the center of the circular path.

3. **Calculate the Resultant Acceleration**:
- Since \( a_t \) and \( a_c \) are perpendicular to each other, we can find the resultant acceleration \( a_{net} \) using the Pythagorean theorem:
\[ a_{net} = \sqrt{a_c^2 + a_t^2} \]

4. **Substituting the Values**:
- Substitute \( a_c \) and \( a_t \) into the equation:
\[ a_{net} = \sqrt{\left(\frac{v^2}{r}\right)^2 + a^2} \]

5. **Final Expression**:
- Simplifying the expression gives:
\[ a_{net} = \sqrt{\frac{v^4}{r^2} + a^2} \]

### Final Result:
The magnitude of the resultant acceleration \( a_{net} \) is:
\[ a_{net} = \sqrt{\frac{v^4}{r^2} + a^2} \]