Solution

Let's move to the frame of reference associated with the $1$th car at the moment when it enters the damaged area and starts moving with velocity $v_2$. The second car will move with a velocity equal to the velocity difference $$v_\text{from}=v_1-v_2$$

The time taken by the second car in the reference frame of the "stopped" first car is given by:

$$t=\frac{L}{v_1–v_2},\tag{1}$$

where $L$ is the distance between the cars.

During the time $t$, the first car, moving with the modified speed $v_2$, must cover at least a distance equal to its own length:

$$t=\frac{l}{v_2} \tag{2}$$

Equating $(1)$ and $(2)$:

$$\frac{L}{v_1–v_2}=\frac{l}{v_2}$$

From this, the minimum distance $L$ between the cars, such that they do not collide, is given by:

$$\fbox{$L \geq \frac{v_1–v_2}{v_2} \cdot l$}$$