Two cars `A` and` B` are racing along straight line. Car `A` is leading, such that their relative velocity is directly proportional to the distance between the two cars. When the lead of car `A` is `l_(1)=10m` , its running `10m//s` faster than car `B`. Determine the time car `A` will take to increase its lead to `l_(2)=20m` from car `B`.

Explanation

To solve the problem, we need to analyze the relationship between the velocities of the two cars and the distance between them. Let's break it down step by step.

### Step 1: Understand the relationship between relative velocity and distance
We are given that the relative velocity of car A with respect to car B is directly proportional to the distance between them. Mathematically, this can be expressed as:
\[ v_A - v_B = k \cdot (x_A - x_B) \]
where \( v_A \) is the velocity of car A, \( v_B \) is the velocity of car B, \( x_A \) is the position of car A, \( x_B \) is the position of car B, and \( k \) is a proportionality constant.

### Step 2: Set up the initial conditions
From the problem, we know:
- When the lead \( l_1 = 10 \, m \), the relative velocity \( v_A - v_B = 10 \, m/s \).
- Thus, we can write:
\[ 10 = k \cdot 10 \]
From this, we can solve for \( k \):
\[ k = 1 \]

### Step 3: Write the equation for relative velocity
Now we can rewrite the equation for relative velocity:
\[ v_A - v_B = (x_A - x_B) \]
This tells us that the difference in velocities is equal to the distance between the two cars.

### Step 4: Define the distance variable
Let \( y = x_A - x_B \). Therefore, we can express the equation as:
\[ \frac{dy}{dt} = y \]

### Step 5: Separate variables and integrate
We can separate the variables:
\[ \frac{dy}{y} = dt \]
Now we integrate both sides. The limits for \( y \) will change from \( 10 \, m \) to \( 20 \, m \) and for \( t \) from \( 0 \) to \( T \):
\[ \int_{10}^{20} \frac{1}{y} \, dy = \int_{0}^{T} 1 \, dt \]

### Step 6: Perform the integration
The left side integrates to:
\[ \ln(y) \bigg|_{10}^{20} = \ln(20) - \ln(10) = \ln\left(\frac{20}{10}\right) = \ln(2) \]
The right side integrates to:
\[ T - 0 = T \]

### Step 7: Set the equations equal
Now we can set the two sides equal:
\[ \ln(2) = T \]

### Step 8: Calculate the time \( T \)
Using the value of \( \ln(2) \):
\[ T \approx 0.693 \, s \]

### Final Answer
The time car A will take to increase its lead from 10 m to 20 m is approximately \( 0.693 \, s \).

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