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 find the magnitude of the acceleration of the car moving on a circular path, we need to consider both the tangential acceleration and the centripetal (or radial) acceleration.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Speed of the car, \( v = 20 \, \text{m/s} \)
- Radius of the circular path, \( R = 100 \, \text{m} \)
- Tangential acceleration, \( a_t = 3 \, \text{m/s}^2 \)

2. **Calculate Centripetal Acceleration**:
- The formula for centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{R} \]
- Substitute 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 Tangential and Centripetal Accelerations**:
- 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:
\[ a = \sqrt{a_t^2 + a_c^2} \]
- Substitute the values:
\[ a = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

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