A motor car is travelling 20m/s on a circular road of radius 400 m. If it increases its speed at the rate of `1 m//s^(2)`, then its acceleration will be

Options

A. `2sqrt2m//s^(2)`
B. `sqrt3m//s^(2)`
C. `sqrt 2 m//s^(2)`
D. `3 sqrt3 m//s^(2)`

Explanation

To solve the problem, we need to find the total acceleration of the motor car that is moving in a circular path while also accelerating tangentially. The total acceleration is the vector sum of the centripetal acceleration and the tangential acceleration.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial speed of the car, \( v = 20 \, \text{m/s} \)
- Radius of the circular road, \( r = 400 \, \text{m} \)
- Tangential acceleration, \( a_t = 1 \, \text{m/s}^2 \)

2. **Calculate Centripetal Acceleration:**
The formula for centripetal acceleration (\( a_c \)) is given by:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(20 \, \text{m/s})^2}{400 \, \text{m}} = \frac{400}{400} = 1 \, \text{m/s}^2 \]

3. **Calculate Resultant Acceleration:**
The resultant acceleration (\( a_r \)) 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:
\[ a_r = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a_r = \sqrt{(1 \, \text{m/s}^2)^2 + (1 \, \text{m/s}^2)^2} = \sqrt{1 + 1} = \sqrt{2} \, \text{m/s}^2 \]

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

### Summary:
The total acceleration of the motor car as it travels around the circular road while increasing its speed is \( \sqrt{2} \, \text{m/s}^2 \).