A car is moving at a speed of 40 m/s on a circular track of radius 400 m. Its 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 tangential acceleration and the centripetal acceleration.

### Step-by-Step Solution:

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

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

3. **Determine the Direction of Accelerations:**
- The tangential acceleration (\( a_t \)) acts in the direction of the velocity (tangential to the circular path).
- The centripetal acceleration (\( a_c \)) acts towards the center of the circular path (radially inward).

4. **Combine the Accelerations:**
Since the tangential and centripetal accelerations are perpendicular to each other, we can find the net acceleration (\( a_{net} \)) using the Pythagorean theorem:
\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a_{net} = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

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