A car is moving on a circular road of radius `100m`. At some instant its speed is `20m//s` and is increasing at the rate of `3m//s^(2)`. The magnitude of its acceleration is

Options

A. `2m//s^(2)`
B. `3m//s^(2)`
C. `5m//s^(2)`
D. `4m//s^(2)`

Explanation

To find the magnitude of the net acceleration of the car moving on a circular road, we need to consider both the tangential acceleration and the centripetal acceleration. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Radius of the circular road, \( R = 100 \, \text{m} \)
- Speed of the car, \( v = 20 \, \text{m/s} \)
- 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: Calculate the net acceleration
The net 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:
\[ a = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

### Step 4: Conclusion
The magnitude of the net acceleration of the car is \( 5 \, \text{m/s}^2 \).

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