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 determine the time it takes for car A to increase its lead from \( l_1 = 10 \, \text{m} \) to \( l_2 = 20 \, \text{m} \) while maintaining the condition that the relative velocity between the two cars is directly proportional to the distance between them.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relative velocity \( V_R \) between the two cars is given by:
\[ V_R = V_A - V_B = k \cdot l \]
where \( l \) is the distance (lead) between the two cars and \( k \) is a proportionality constant.

2. **Finding the Proportionality Constant \( k \)**:
When the lead \( l_1 = 10 \, \text{m} \), car A is running \( 10 \, \text{m/s} \) faster than car B. Therefore:
\[ V_R = 10 \, \text{m/s} \quad \text{and} \quad l = l_1 = 10 \, \text{m} \]
Substituting these values into the equation gives:
\[ 10 = k \cdot 10 \]
From this, we can solve for \( k \):
\[ k = 1 \]

3. **Setting Up the Differential Equation**:
Now we have:
\[ V_R = l \]
The change in lead over time can be expressed as:
\[ \frac{dL}{dt} = l \]
This means that the rate of change of lead is equal to the current lead.

4. **Integrating the Equation**:
We can separate variables and integrate:
\[ \int \frac{dL}{L} = \int dt \]
The limits for \( L \) will be from \( l_1 \) to \( l_2 \) and for \( t \) from \( 0 \) to \( t \):
\[ \int_{l_1}^{l_2} \frac{1}{L} dL = \int_0^t dt \]
This results in:
\[ \ln(L) \bigg|_{l_1}^{l_2} = t \]

5. **Calculating the Time**:
Substituting the limits:
\[ \ln(l_2) - \ln(l_1) = t \]
Using the properties of logarithms:
\[ \ln\left(\frac{l_2}{l_1}\right) = t \]
Substituting \( l_2 = 20 \, \text{m} \) and \( l_1 = 10 \, \text{m} \):
\[ t = \ln\left(\frac{20}{10}\right) = \ln(2) \]

6. **Final Calculation**:
The value of \( \ln(2) \) is approximately \( 0.693 \, \text{s} \).

### Conclusion:
The time it takes for car A to increase its lead from \( 10 \, \text{m} \) to \( 20 \, \text{m} \) is approximately \( 0.693 \, \text{s} \).