Speeds of two identical cars are u and 4u at at specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is

Options

A. `1:1`
B. ` 1:4`
C. ` 1: 8`
D. ` 1: 16`

Explanation

To solve the problem of finding the ratio of the distances in which two identical cars stop, given their speeds \( u \) and \( 4u \), we can follow these steps:

### Step 1: Understand the Problem
We have two identical cars. Car A is moving with speed \( u \) and Car B is moving with speed \( 4u \). Both cars will apply brakes and come to a stop. We need to find the ratio of the distances \( s_1 \) (for Car A) and \( s_2 \) (for Car B) that they travel before stopping.

### Step 2: Use the Kinematic Equation
We will use the third kinematic equation which 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 as it is retardation)
- \( s \) = distance traveled before stopping

### Step 3: Set Up the Equations
For Car A (speed \( u \)):
\[ 0 = u^2 + 2(-a)s_1 \quad \Rightarrow \quad u^2 = 2as_1 \quad \Rightarrow \quad s_1 = \frac{u^2}{2a} \]

For Car B (speed \( 4u \)):
\[ 0 = (4u)^2 + 2(-a)s_2 \quad \Rightarrow \quad 16u^2 = 2as_2 \quad \Rightarrow \quad s_2 = \frac{16u^2}{2a} = \frac{8u^2}{a} \]

### Step 4: Find the Ratio of Distances
Now we can find the ratio \( \frac{s_1}{s_2} \):
\[ \frac{s_1}{s_2} = \frac{\frac{u^2}{2a}}{\frac{8u^2}{a}} = \frac{u^2}{2a} \cdot \frac{a}{8u^2} = \frac{1}{16} \]

### Conclusion
Thus, the ratio of the distances in which the two cars stop is:
\[ \frac{s_1}{s_2} = \frac{1}{16} \]