A car has linear velocity v on a circular road of radius r. IF r it incrasing its speed at the rate of a `ms^(-2)`, then the resultant acceleration will be

Options

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

Explanation

To find the resultant acceleration of a car moving on a circular road while increasing its speed, we can break down the problem into steps.

### Step-by-Step Solution:

1. **Identify the Types of Acceleration**:
- The car experiences two types of acceleration:
- **Tangential Acceleration (a_t)**: This is due to the change in speed and is given as \( a = 1 \, \text{m/s}^2 \).
- **Radial (Centripetal) Acceleration (a_r)**: This is due to the circular motion and is given by the formula:
\[ a_r = \frac{v^2}{r} \]

2. **Calculate Radial Acceleration**:
- Substitute the linear velocity \( v \) and radius \( r \) into the centripetal acceleration formula:
\[ a_r = \frac{v^2}{r} \]

3. **Calculate Resultant Acceleration**:
- The resultant acceleration \( a_{\text{net}} \) is the vector sum of the tangential and radial accelerations. Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem:
\[ a_{\text{net}} = \sqrt{a_t^2 + a_r^2} \]
- Substitute \( a_t = a \) and \( a_r = \frac{v^2}{r} \):
\[ a_{\text{net}} = \sqrt{(1)^2 + \left(\frac{v^2}{r}\right)^2} \]

4. **Simplify the Expression**:
- This simplifies to:
\[ a_{\text{net}} = \sqrt{1 + \frac{v^4}{r^2}} \]

5. **Final Result**:
- Thus, the resultant acceleration of the car is:
\[ a_{\text{net}} = \sqrt{1 + \frac{v^4}{r^2}} \]