Step 1: Understanding the Concept:
"Retardation" is simply negative acceleration; it indicates that the velocity is decreasing.
"Stopping distance" is the distance an object travels after the brakes are applied until its velocity becomes zero.
Since time is not given in the problem statement, we should use the kinematic equation that relates velocity, acceleration, and distance.
Step 2: Key Formula or Approach:
The third equation of motion is:
\[ v^{2} = u^{2} + 2as \]
Where:
- \( u \) = initial velocity.
- \( v \) = final velocity.
- \( a \) = acceleration (negative for retardation).
- \( s \) = stopping distance.
Step 3: Detailed Explanation:
1. Identify the values:
- Initial velocity (\(u\)) = 20 m/s.
- Final velocity (\(v\)) = 0 m/s (it comes to rest).
- Retardation = 5 m/s\(^2\), so acceleration (\(a\)) = \(-5 \text{ m/s}^2\).
2. Rearrange the formula to solve for \( s \):
\[ s = \frac{v^2 - u^2}{2a} \]
3. Substitute the values:
\[ s = \frac{0^2 - (20)^2}{2 \times (-5)} \]
\[ s = \frac{0 - 400}{-10} \]
\[ s = \frac{-400}{-10} = 40 \text{ m} \]
Step 4: Final Answer:
The stopping distance of the car is 40 m.
This corresponds to Option (C).