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. `1 m//s^2`
B. `3 m//s^2`
C. `4 m//s^2`
D. `5 m//s^2`

Answer

To find the magnitude of the acceleration of a car moving on a circular path, we need to consider both the tangential acceleration and the centripetal acceleration. Here’s how to solve the problem step by step:

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

### Step 2: Calculate the Centripetal Acceleration
Centripetal acceleration (\( a_c \)) can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the known values:
\[ a_c = \frac{(20 \, \text{m/s})^2}{100 \, \text{m}} = \frac{400 \, \text{m}^2/\text{s}^2}{100 \, \text{m}} = 4 \, \text{m/s}^2 \]

### Step 3: Combine the Accelerations
The total acceleration (\( a \)) 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 we found:
\[ a = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

### Step 4: Conclusion
The magnitude of the acceleration of the car at that moment is:
\[ \boxed{5 \, \text{m/s}^2} \]

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