A car is moving with speed `20 ms ^(-1)` on a circular path of radius 100 m. Its speed is increasing at a rate of `3 ms^(-2)` . The magnitude of the acceleration of the car at that moment is

Options

A. `1ms^(-2)`
B. `3ms^(-2)`
C. `4ms^(-2)`
D. `5ms^(-2)`

Answer

To solve the problem, we need to find the magnitude of the acceleration of a car moving in a circular path with given parameters. The car is moving with a speed of \(20 \, \text{m/s}\) on a circular path of radius \(100 \, \text{m}\), and its speed is increasing at a rate of \(3 \, \text{m/s}^2\).

### Step-by-step Solution:

1. **Identify the Types of Acceleration**:
- In non-uniform circular motion, there are two components of acceleration:
- **Tangential Acceleration (\(a_t\))**: This is due to the change in speed along the circular path. Given as \(3 \, \text{m/s}^2\).
- **Centripetal Acceleration (\(a_c\))**: This is due to the change in direction of the velocity vector as the car moves along the circular path.

2. **Calculate Centripetal Acceleration**:
- The formula for centripetal acceleration is given by:
\[ a_c = \frac{v^2}{R} \]
- Where \(v\) is the speed of the car and \(R\) is the radius of the circular path.
- Substituting the values:
\[ a_c = \frac{(20 \, \text{m/s})^2}{100 \, \text{m}} = \frac{400 \, \text{m}^2/\text{s}^2}{100 \, \text{m}} = 4 \, \text{m/s}^2 \]

3. **Combine the Accelerations**:
- Since the tangential acceleration and centripetal acceleration are perpendicular to each other, we can find the resultant acceleration (\(a_{net}\)) using the Pythagorean theorem:
\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]
- Substituting the values:
\[ a_{net} = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

4. **Conclusion**:
- The magnitude of the acceleration of the car at that moment is \(5 \, \text{m/s}^2\).