The minimum stopping distance of a car moving with velocity `u` is `x`. If the car is moving with velocity 2v, then the minimum stopping distance will be.

Options

A. `2x`
B. `4x`
C. 3x`
D. `8x`

Explanation

To solve the problem, we need to determine the minimum stopping distance of a car when its velocity changes from \( u \) to \( 2v \). We will use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance.

### Step-by-Step Solution:

1. **Understanding the Stopping Distance Formula**:
The stopping distance \( x \) for a car moving with an initial velocity \( u \) can be derived from the equation:
\[ v^2 = u^2 + 2as \]
where \( v \) is the final velocity (0 when the car stops), \( u \) is the initial velocity, \( a \) is the acceleration (negative for deceleration), and \( s \) is the stopping distance.

2. **Setting Up the Equation**:
Since the car comes to a stop, we set \( v = 0 \):
\[ 0 = u^2 + 2(-a)x \]
Rearranging gives:
\[ u^2 = 2ax \]
From this, we can express the stopping distance \( x \):
\[ x = \frac{u^2}{2a} \]

3. **Finding the Stopping Distance for Velocity \( 2v \)**:
Now, we need to find the stopping distance when the car is moving with a velocity of \( 2v \). Let’s denote this stopping distance as \( x' \):
\[ x' = \frac{(2v)^2}{2a} \]

4. **Calculating \( x' \)**:
Simplifying \( x' \):
\[ x' = \frac{4v^2}{2a} = \frac{4}{2} \cdot \frac{v^2}{a} = 2 \cdot \frac{2v^2}{2a} = 2 \cdot x \]
This shows that the stopping distance when the car is moving at \( 2v \) is \( 4x \).

5. **Conclusion**:
Therefore, the minimum stopping distance when the car is moving with velocity \( 2v \) is:
\[ x' = 4x \]

### Final Answer:
The minimum stopping distance of the car moving with velocity \( 2v \) will be \( 4x \).

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