A car can be stopped within a distance of s when its speed is v. What is the minimum distance within which the car can be stopped during the same time when its speed is nv ?

Options

A. ns
B. s/n
C. `n^(2)s`
D. `s//n^(2)`

Explanation

To solve the problem, we need to find the minimum distance within which a car can be stopped when its speed is increased to \( nv \). Here's a step-by-step breakdown of the solution:

### Step 1: Understand the given information
- The initial speed of the car is \( v \).
- The car can stop within a distance \( s \) when traveling at speed \( v \).
- We need to find the stopping distance when the speed is \( nv \).

### Step 2: Use the kinematic equation
We will use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity (0 when the car stops)
- \( u \) = initial velocity
- \( a \) = acceleration (negative in this case since it's deceleration)
- \( s \) = distance

### Step 3: Find the acceleration when the speed is \( v \)
For the initial case where the speed is \( v \):
- Final velocity \( v = 0 \)
- Initial velocity \( u = v \)
- Distance \( s \)

Using the equation:
\[ 0 = v^2 + 2(-a)s \]
This simplifies to:
\[ v^2 = 2as \]
From this, we can express the acceleration \( a \):
\[ a = \frac{v^2}{2s} \]

### Step 4: Find the stopping distance when the speed is \( nv \)
Now, we consider the case where the speed is \( nv \):
- Final velocity \( v = 0 \)
- Initial velocity \( u = nv \)
- Let the stopping distance be \( s_1 \).

Using the same kinematic equation:
\[ 0 = (nv)^2 + 2(-a)s_1 \]
This simplifies to:
\[ (nv)^2 = 2as_1 \]
Substituting for \( a \) from the previous step:
\[ n^2v^2 = 2\left(\frac{v^2}{2s}\right)s_1 \]
This simplifies to:
\[ n^2v^2 = \frac{v^2}{s}s_1 \]

### Step 5: Solve for \( s_1 \)
Now, we can solve for \( s_1 \):
\[ s_1 = n^2s \]

### Conclusion
The minimum distance within which the car can be stopped when its speed is \( nv \) is:
\[ \boxed{n^2s} \]