A car is travelling at 20 m/s on a circular road of radius 100 m. It is increasing its speed at the rate of `3" m/s"^(2)`. Its acceleration is

Options

A. (a)`3" m/s"^(2)`
B. (b)`4" m/s"^(2)`
C. (c)`5" m/s"^(2)`
D. (d)`7" m/s"^(2)`

Answer

To solve the problem, we need to find the total acceleration of a car moving in a circular path while also increasing its speed. The total acceleration consists of two components: tangential acceleration and radial (centripetal) acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity of the car, \( v = 20 \, \text{m/s} \)
- Radius of the circular road, \( r = 100 \, \text{m} \)
- Tangential acceleration, \( a_t = 3 \, \text{m/s}^2 \) (the rate of increase of speed)

2. **Calculate Radial (Centripetal) Acceleration:**
- The formula for radial acceleration \( a_r \) is given by:
\[ a_r = \frac{v^2}{r} \]
- Substituting the values:
\[ a_r = \frac{(20 \, \text{m/s})^2}{100 \, \text{m}} = \frac{400 \, \text{m}^2/\text{s}^2}{100 \, \text{m}} = 4 \, \text{m/s}^2 \]

3. **Calculate Total Acceleration:**
- The total acceleration \( a \) is the vector sum of tangential acceleration \( a_t \) and radial acceleration \( a_r \). Since these two components are perpendicular to each other, we can use the Pythagorean theorem:
\[ a = \sqrt{a_t^2 + a_r^2} \]
- Substituting the values:
\[ a = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

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

### Final Answer:
The acceleration of the car is \( 5 \, \text{m/s}^2 \).

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