A car is moving in a circular path of radius `500m` with a speed of `30 m//s`. If the speed is increased at the rate of `2 m//s^(2)`, the resultant acceleration will be .

Options

A. `2 m//s^(2)`
B. `2.5 m//s^(2)`
C. `2.7 m//s^(2)`
D. `4 m//s^(2)`

Explanation

To solve the problem, we need to find the resultant acceleration of a car moving in a circular path with given parameters. Here’s a step-by-step solution:

### Step 1: Identify the given parameters
- Radius of the circular path (r) = 500 m
- Speed of the car (v) = 30 m/s
- Rate of increase of speed (tangential acceleration, \( a_t \)) = 2 m/s²

### Step 2: Calculate the radial (centripetal) acceleration
The formula for radial (centripetal) acceleration (\( a_r \)) is given by:
\[ a_r = \frac{v^2}{r} \]
Substituting the values:
\[ a_r = \frac{(30 \, \text{m/s})^2}{500 \, \text{m}} = \frac{900}{500} = 1.8 \, \text{m/s}^2 \]

### Step 3: Identify the tangential acceleration
The tangential acceleration (\( a_t \)) is given directly in the problem:
\[ a_t = 2 \, \text{m/s}^2 \]

### Step 4: Calculate the resultant acceleration
The resultant acceleration (\( a_{net} \)) can be calculated using the Pythagorean theorem since the radial and tangential accelerations are perpendicular to each other:
\[ a_{net} = \sqrt{a_r^2 + a_t^2} \]
Substituting the values we calculated:
\[ a_{net} = \sqrt{(1.8 \, \text{m/s}^2)^2 + (2 \, \text{m/s}^2)^2} \]
\[ = \sqrt{3.24 + 4} = \sqrt{7.24} \approx 2.69 \, \text{m/s}^2 \]

### Final Answer
The resultant acceleration of the car is approximately \( 2.69 \, \text{m/s}^2 \).

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