A car is moving with speed `30m//sec` on a circular path of radius 500 m . Its speed is increasing at the rate of , `2m//sec^(2)`, What is the acceleration of the car

Options

A. `2m//sec^(2)`
B. `2.7 m//sec^(2)`
C. `1.8 m//sec^(2)`
D. `9.8 m//sec^(2)`

Answer

To solve the problem, we need to find the total acceleration of the car moving in a circular path with both tangential and centripetal components.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of the car, \( v = 30 \, \text{m/s} \)
- Radius of the circular path, \( r = 500 \, \text{m} \)
- Tangential acceleration, \( a_t = 2 \, \text{m/s}^2 \)

2. **Calculate the Centripetal Acceleration:**
Centripetal acceleration (\( a_c \)) can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(30 \, \text{m/s})^2}{500 \, \text{m}} = \frac{900}{500} = 1.8 \, \text{m/s}^2 \]

3. **Determine the Total Acceleration:**
The total acceleration (\( a \)) of the car 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:
\[ a = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a = \sqrt{(2 \, \text{m/s}^2)^2 + (1.8 \, \text{m/s}^2)^2} \]
\[ a = \sqrt{4 + 3.24} = \sqrt{7.24} \approx 2.69 \, \text{m/s}^2 \]

4. **Final Result:**
The total acceleration of the car is approximately:
\[ a \approx 2.69 \, \text{m/s}^2 \]