A car of mass `m` moves in a horizontal circular path of radius `r` metre. At an instant its speed is `V m//s` and is increasing at a rate of `a m//sec^(2)`.then the acceleration of the car is

Options

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

Answer

To find the total acceleration of the car moving in a horizontal circular path, we need to consider both the tangential acceleration and the radial (centripetal) acceleration.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of the car: \( m \) (not needed for acceleration calculation)
- Radius of the circular path: \( r \) meters
- Speed of the car: \( V \) m/s
- Rate of increase of speed (tangential acceleration): \( a \) m/s²

2. **Determine the Radial Acceleration**:
- The radial acceleration (centripetal acceleration) for an object moving in a circle is given by the formula:
\[ a_r = \frac{V^2}{r} \]

3. **Tangential Acceleration**:
- The tangential acceleration is given directly as:
\[ a_t = a \, \text{(m/s²)} \]

4. **Combine the Accelerations**:
- The total acceleration \( A \) of the car is the vector sum of the tangential acceleration and the 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:
\[ A = \sqrt{a^2 + \left(\frac{V^2}{r}\right)^2} \]

5. **Final Expression**:
- Thus, the total acceleration of the car is:
\[ A = \sqrt{a^2 + \left(\frac{V^2}{r}\right)^2} \]