It's Green

Relevant wiki: One Dimensional Kinematics: Motion Along a Straight Line

Let $t = 0$ when the first car begins to move, and $t^\star$ is some time after the second car reaches the maximum speed.

During this time interval, the cars move in similar ways. The only differences are that the second car begins with $\SI{1.5}{s}$ standing still (waiting for the first car to move), and spending $\SI{1.5}{s}$ less time moving at the speed limit. Therefore the difference in distance traveled is $\Delta x = \SI{1.5}{s}\times \SI{20}{m/s} = \SI{30}{m}.$ Add this to the original $\SI{2}{m}$ between the cars, and you find the final distance to be $\boxed{\SI{32}{m}}$ .

For those who want to make a more detailed analysis:

$\begin{array}{c|cc|cc|c|cc|cc} \hline t & v_1 & v_2 & x_1 & x_2 & \Delta t & \Delta v_1 & \Delta v_2 & \Delta x_1 & \Delta x_2 \\ \hline 0 & 0 & 0 & 0 & -2 & & & & & \\ & & & & & 1.5 & 6 & 0 & 4.5 & 0 \\ 1.5 & 6 & 0 & 4.5 & -2 & & & & & \\ & & & & & 3.5 & 14 & 14 & 45.5 & 24.5 \\ 5.0 & 20 & 14 & 50 & 22.5 & & & & & \\ & & & & & 1.5 & 0 & 6 & 30 & 25.5 \\ 6.5 & 20 & 20 & 80 & 48 & & & & & \\ \hline & & & & & 6.5 & 20 & 20 & 80 & 50 \\ \hline \end{array}$

Here is a graphical illustration of Arjen's solution. The green curve is identical to the red curve, but shifted by 1.5s and 2m.

Each car starts moving $1.5\ \text{s}$ after the car in front. The distance travelled & time taken during acceleration is the same for every car, but the car in front will have travelled at $20\ \text{m/s}$ for $1.5\ \text{s}$ before the car behind reaches top speed. This means the car in front will have travelled $30\ \text{m}$ more than the car behind. adding this to the starting $2\ \text{m}$ gap gives us the total difference of $\boxed{32\ \text{m}}$ .

Consider the graph of speed as a function of time. At this rate of acceleration, each car starts accelerating 1.5 s after the one before. The colors of the lines are the same as the cars in the problem. Therefore the blue area is the distance between the blue and the red car, if they had started together, so 30 m. As they started 2 m apart, they end up being 32 m apart.

Ignore the acceleration, the cars do the same thing just offset by 1.5 seconds. When travelling at 20 m/s, 1.5 seconds translates to a 30 meter difference. They were also 2 meter apart at the start giving $\SI{20}{m/s} \times \SI{1.5}{s} + \SI{2}{m} = \SI{32}{m}$

In a velocity-time graph, distance covered equals area under the curve. The distance-time graph for two adjacent cars looks like this: The second car sets off at 1.5s (when the first car has reached 6m/s). The difference in distance travelled is the green area. This has a base of 1.5s and a height of 20m/s, so an area of 30m. Add on the initial separation of 2m to get 32m.

You can do it in your head: Taking the position of the first car as the origin, the first car will reach $x$ meters from the origin in $t$ seconds. The second car will reach $x-2$ meters from the origin in $t+1.5$ seconds. So the distance between them will be $20\times 1.5+2=32$ .

$\text{Evaluate}\left[ \\ \ \ \text{Assuming}\left[t\geq 0\land t\in \mathbb{R},\int _0^t\int _0^t\left( \begin{array}{cc} \{ & \begin{array}{cc} 4 & 0\leq x<5 \\ \end{array} \\ \end{array} \right)dxdt\right]- \\ \ \ \left(\text{Assuming}\left[t\geq 0\land t\in \mathbb{R},\int _0^t\int _0^t\left( \begin{array}{cc} \{ & \begin{array}{cc} 4 & \frac{3}{2}\leq x<5+\frac{3}{2} \\ \end{array} \\ \end{array} \right)dxdt\right]-2\right)\right]\text{/.}\, t\to 10 \Rightarrow 32$

The car separations:

$\text{Assuming}\left[t\geq 0\land t\in \mathbb{R},\int _0^t\int _0^t\left( \begin{array}{cc} \{ & \begin{array}{cc} 4 & 0\leq x<5 \\ \end{array} \\ \end{array} \right)dxdt\right] \Rightarrow \begin{array}{cc} \{ & \begin{array}{cc} 2 t^2 & 0<t\leq 5 \\ 10 (2 t-5) & t>5 \\ \end{array} \\ \end{array}$

$\text{Assuming}\left[t\geq 0\land t\in \mathbb{R},\int _0^t\int _0^t\left( \begin{array}{cc} \{ & \begin{array}{cc} 4 & \frac{3}{2}\leq x<5+\frac{3}{2} \\ \end{array} \\ \end{array} \right)dxdt\right]-2 \Rightarrow \left( \begin{array}{cc} \{ & \begin{array}{cc} 20 (t-4) & t>\frac{13}{2} \\ \frac{1}{2} \left(4 t^2-12 t+9\right) & \frac{3}{2}<t\leq \frac{13}{2} \\ \end{array} \\ \end{array} \right)-2$

$\text{FullSimplify}[10 (2 t-5)-(20 (t-4)-2)]\Rightarrow 32$

I just guessed 2^5 = 32 Not sure that can be put to practice.

The first car accelerates to $6m/s^2$ at 1.5 sec. Then the second car drives. So it takes 1.5 sec for the second car to get to $20m/s^2$ after the first car gets there. That means the first car can drive another 1.5 seconds of $20m/s$ . Which is 30 meters, plus the initial 2 meters, is 32 meters.

At an acceleration of $4m/s^2$ each car reaches a speed of $6m/s$ after $1.5s$ and a speed of $20m/s$ after $5s$ .

So we know the last car will drive for $5s$ , the middle car will drive for $5+1.5=6.5s$ , and the front car will drive $5+2(1.5) = 8s$

To calculate distance $D(t) = \begin{cases} \int_{0}^{t} 4t = 2t^2 & t\leq 5 \\ 20x & t\gt x \end{cases}$

First car traveled $110+4 = 114m$ ; second car traveled $80+2=82m$ ; last car traveled $50m$

$\Delta$ between cars $= \boxed{32m}$

The question is easier if we calculate it backwards, start considering the last car.

Buy the time the rear car accelerates to 20m/s, the front car already accelerates to 20m/s 1.5 seconds ago (6/4=1.5s), and keep constant speed at 20m/s for 1.5s.

We don't need to care the positive acceleration part, because both cars moves for same distance during their acceleration. It doesn't affect their separation.

So final distance between them is just 20*1.5+2 = 32m, where the 2m is their initial separation.