When the speed of the car is v, the minimum distance over which it can be stopped is x. If the speed becomes nv, what will be the minimum distance over which it can be stopped during same time:

Options

A. `x//n`
B. `nx`
C. `x//n^(2)`
D. `n^(2)x`

Answer

To solve the problem step by step, we need to analyze the relationship between the speed of the car, the distance required to stop, and the acceleration (or retardation) involved.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- When the speed of the car is \( v \), the minimum distance over which it can be stopped is \( x \).
- We can use the kinematic equation for uniformly accelerated motion:
\[ v^2 = u^2 + 2as \]
- Here, \( v \) is the final velocity (0 when the car stops), \( u \) is the initial velocity (which is \( v \)), \( a \) is the acceleration (negative in this case since it is retardation), and \( s \) is the distance (which is \( x \)).

2. **Applying the Kinematic Equation**:
- Rearranging the equation for our case (where the final velocity \( v = 0 \)):
\[ 0 = v^2 + 2(-a)x \]
- This simplifies to:
\[ v^2 = 2ax \]
- From this, we can express the acceleration \( a \):
\[ a = \frac{v^2}{2x} \]

3. **Considering the New Condition**:
- Now, if the speed of the car becomes \( nv \), we need to find the new minimum distance \( s \) over which it can be stopped.
- Using the same kinematic equation:
\[ (nv)^2 = (nv)^2 + 2(-a)s \]
- This becomes:
\[ n^2v^2 = 2as \]

4. **Substituting the Expression for Acceleration**:
- Substitute \( a \) from the earlier step:
\[ n^2v^2 = 2\left(\frac{v^2}{2x}\right)s \]
- Simplifying this gives:
\[ n^2v^2 = \frac{v^2}{x}s \]

5. **Solving for the New Distance \( s \)**:
- We can cancel \( v^2 \) from both sides (assuming \( v \neq 0 \)):
\[ n^2 = \frac{s}{x} \]
- Rearranging gives:
\[ s = n^2x \]

6. **Conclusion**:
- Therefore, the minimum distance over which the car can be stopped when its speed is \( nv \) is:
\[ s = n^2x \]

### Final Answer:
The minimum distance over which the car can be stopped when its speed is \( nv \) is \( n^2x \).