A car is moving with a speed of 30 `ms^(-1)` on a circular path of radius 500 m. If its speed is increasing at the rate of 2 `ms^(-2)` , the net acceleration of the car is

Options

A. `3.6 ms^(-2)`
B. `2.7 ms^(-2)`
C. `1.8 ms^(-2)`
D. `2 ms^(-2)`

Answer

To find the net acceleration of the car moving in a circular path with increasing speed, we need to consider both the tangential acceleration and the centripetal acceleration.

### 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:**
The formula for centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(30)^2}{500} = \frac{900}{500} = 1.8 \, \text{m/s}^2 \]

3. **Calculate the net acceleration:**
The net acceleration \( a_{net} \) 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:
\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a_{net} = \sqrt{(2)^2 + (1.8)^2} = \sqrt{4 + 3.24} = \sqrt{7.24} \]
Calculating the square root:
\[ a_{net} \approx 2.7 \, \text{m/s}^2 \]

4. **Conclusion:**
The net acceleration of the car is approximately \( 2.7 \, \text{m/s}^2 \).