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. (a) 1:1
B. (b) 1:4
C. (c) 1:8
D. (d) 1:16

Explanation

To solve the problem of finding the ratio of the distances traveled by two identical cars before they come to a stop, we can use the principles of physics, specifically the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have two identical cars. The first car is moving with a speed \( u \) and the second car is moving with a speed \( 4u \). We need to find the ratio of the distances \( S_1 \) and \( S_2 \) that each car travels before coming to a stop.

### Step 2: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy.

For the first car (speed \( u \)):
- Initial kinetic energy \( K_1 = \frac{1}{2} m u^2 \)
- Final kinetic energy \( K_{1f} = 0 \) (since it comes to rest)

The work done \( W_1 \) when the first car stops is:
\[ W_1 = K_{1f} - K_1 = 0 - \frac{1}{2} m u^2 = -\frac{1}{2} m u^2 \]
The work done is also equal to the force applied times the distance \( S_1 \) (in the opposite direction):
\[ W_1 = -F S_1 \]
Setting these equal gives:
\[ -F S_1 = -\frac{1}{2} m u^2 \quad \Rightarrow \quad F S_1 = \frac{1}{2} m u^2 \quad \text{(Equation 1)} \]

### Step 3: Repeat for the Second Car
For the second car (speed \( 4u \)):
- Initial kinetic energy \( K_2 = \frac{1}{2} m (4u)^2 = \frac{1}{2} m \cdot 16u^2 \)
- Final kinetic energy \( K_{2f} = 0 \)

The work done \( W_2 \) when the second car stops is:
\[ W_2 = K_{2f} - K_2 = 0 - \frac{1}{2} m \cdot 16u^2 = -\frac{1}{2} m \cdot 16u^2 \]
Setting this equal to the work done gives:
\[ -F S_2 = -\frac{1}{2} m \cdot 16u^2 \quad \Rightarrow \quad F S_2 = \frac{1}{2} m \cdot 16u^2 \quad \text{(Equation 2)} \]

### Step 4: Find the Ratio of Distances
Now we can find the ratio \( \frac{S_1}{S_2} \) by dividing Equation 1 by Equation 2:
\[ \frac{F S_1}{F S_2} = \frac{\frac{1}{2} m u^2}{\frac{1}{2} m \cdot 16u^2} \]
The \( F \) and \( \frac{1}{2} m \) terms cancel out:
\[ \frac{S_1}{S_2} = \frac{u^2}{16u^2} = \frac{1}{16} \]

### Step 5: Conclusion
Thus, the ratio of the distances \( S_1 \) and \( S_2 \) is:
\[ S_1 : S_2 = 1 : 16 \]