A car moving with a velocity v is stopped within a distance x by applying a retarding force F. If the speed of the car is doubled, then the force required to stop the car with the same distance is

Options

A. 2F
B. 3F
C. 4F
D. `F//4`

Explanation

To solve the problem step by step, we will analyze the situation using the equations of motion and the relationship between force, mass, and acceleration.

### Step 1: Understand the Initial Situation
The car is moving with an initial velocity \( v \) and is stopped by applying a retarding force \( F \) over a distance \( x \).

### Step 2: Apply the Equation of Motion
Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity (0, since the car stops),
- \( u \) is the initial velocity (\( v \)),
- \( a \) is the acceleration (which will be negative since it is a retarding force),
- \( s \) is the distance traveled (\( x \)).

Substituting the values:
\[ 0 = v^2 + 2(-a)x \]
This simplifies to:
\[ v^2 = 2ax \]
From this, we can express acceleration \( a \):
\[ a = \frac{v^2}{2x} \]

### Step 3: Consider the New Situation with Doubled Speed
Now, if the speed of the car is doubled, the new initial velocity \( u' = 2v \). We need to find the new force \( F' \) required to stop the car within the same distance \( x \).

### Step 4: Apply the Equation of Motion Again
Using the same equation of motion for the new situation:
\[ 0 = (2v)^2 + 2(-a')x \]
This simplifies to:
\[ 0 = 4v^2 + 2(-a')x \]
Rearranging gives:
\[ 4v^2 = 2a'x \]
From this, we can express the new acceleration \( a' \):
\[ a' = \frac{4v^2}{2x} = \frac{2v^2}{x} \]

### Step 5: Relate the New Acceleration to the Old Acceleration
We previously found that:
\[ a = \frac{v^2}{2x} \]
Now, we can relate \( a' \) to \( a \):
\[ a' = 4a \]

### Step 6: Find the New Force
Using Newton's second law, the force is given by:
\[ F' = ma' \]
Substituting \( a' \):
\[ F' = m(4a) = 4(ma) \]
Since \( F = ma \), we can substitute:
\[ F' = 4F \]

### Conclusion
Thus, the force required to stop the car when its speed is doubled, while stopping within the same distance \( x \), is:
\[ \boxed{4F} \]

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