A car is moving at a speed of 40 m/s on a circular track of radius 400 m. this speed is increasing at the rate of 3 m/`s^(2)`. The acceleration of car is

Options

A. 4 m/`s^(2)`
B. 7 m/`s^(2)`
C. 5 m/`s^(2)`
D. 3 m/`s^(2)`

Answer

To find the acceleration of the car moving on a circular track, we need to consider both the centripetal acceleration and the tangential acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Speed of the car, \( v = 40 \, \text{m/s} \)
- Radius of the circular track, \( r = 400 \, \text{m} \)
- Rate of increase of speed (tangential acceleration), \( a_t = 3 \, \text{m/s}^2 \)

2. **Calculate Centripetal Acceleration:**
Centripetal acceleration (\( a_c \)) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(40)^2}{400} = \frac{1600}{400} = 4 \, \text{m/s}^2 \]

3. **Determine the Resultant Acceleration:**
The resultant acceleration (\( a \)) can be found using the Pythagorean theorem since the centripetal and tangential accelerations are perpendicular to each other:
\[ a = \sqrt{a_c^2 + a_t^2} \]
Substituting the values:
\[ a = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m/s}^2 \]

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