A car moving with a speed v is stopped by applying brakes with in a distance s and time t. If it would have been travelling with Kv then by applying same breaking force it can be stopped with in

Options

A. ks, kt
B. `k^(2)s, k^(2)t`
C. `k^(2)s, kt`
D. `ks,k^(2)t`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have a car moving with an initial speed \( v \) that comes to a stop after traveling a distance \( s \) in time \( t \) by applying brakes. We need to determine how far the car will travel if it is initially moving at a speed of \( Kv \) while applying the same braking force.

### Step 2: Use the equations of motion
We can use 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,
- \( a \) is the acceleration (deceleration in this case, so it will be negative),
- \( s \) is the distance traveled.

For the first case (initial speed \( v \)):
\[ 0 = v^2 + 2(-a)s \]
This simplifies to:
\[ v^2 = 2as \quad \text{(1)} \]

### Step 3: Express deceleration in terms of \( v \) and \( s \)
From equation (1), we can express \( a \):
\[ a = \frac{v^2}{2s} \quad \text{(2)} \]

### Step 4: Analyze the second case with speed \( Kv \)
Now, if the car is moving at speed \( Kv \), we apply the same braking force, which means the deceleration \( a \) remains the same. We can use the same equation of motion:
\[ 0 = (Kv)^2 + 2(-a)s' \]
This simplifies to:
\[ (Kv)^2 = 2as' \quad \text{(3)} \]

### Step 5: Substitute \( a \) from equation (2) into equation (3)
Substituting \( a \) from equation (2) into equation (3):
\[ K^2v^2 = 2\left(\frac{v^2}{2s}\right)s' \]
This simplifies to:
\[ K^2v^2 = \frac{v^2}{s}s' \]

### Step 6: Solve for \( s' \)
Rearranging gives:
\[ s' = K^2s \]

### Conclusion
Thus, the distance \( s' \) that the car will travel when moving at speed \( Kv \) and applying the same braking force is:
\[ s' = K^2s \]