A motor car is travelling at `30 m//s` on a circular track of radius 450 m . If its speed is increasing at the rate of `2m//s^(2)`, then its acceleration will be

Options

A. `1//sqrt2 m//s^(2)`
B. `2 m//s^(2)`
C. `sqrt2 m//s^(2)`
D. `2sqrt2 m//s^(2)`

Explanation

To solve the problem, we need to find the total acceleration of a motor car moving on a circular track. The total acceleration consists of two components: tangential acceleration and centripetal (or radial) acceleration.

### Step-by-step Solution:

1. **Identify Given Values**:
- Speed of the car, \( v = 30 \, \text{m/s} \)
- Radius of the circular track, \( r = 450 \, \text{m} \)
- Rate of increase of speed (tangential acceleration), \( a_t = 2 \, \text{m/s}^2 \)

2. **Calculate Centripetal Acceleration**:
The formula for centripetal acceleration (\( a_c \)) is given by:
\[ a_c = \frac{v^2}{r} \]
Substituting the known values:
\[ a_c = \frac{(30 \, \text{m/s})^2}{450 \, \text{m}} = \frac{900}{450} = 2 \, \text{m/s}^2 \]

3. **Calculate Total Acceleration**:
The total acceleration (\( a \)) is the vector sum of tangential acceleration and 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:
\[ a = \sqrt{(2 \, \text{m/s}^2)^2 + (2 \, \text{m/s}^2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \, \text{m/s}^2 \]

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