Answer
$120\rm\;m/s$
Work Step by Step
The car here has two stages of motion.
The first stage is speeding up from rest at a constant rate for 9 seconds and then sowing down for 3 seconds (since the whole time to the judges' point is 12 s).
In these two stages, the car travels 990 m.
Finding $x_1$;
$$x_1=\overbrace{ x_0}^{0} +\overbrace{ v_{0x}t_1}^{0} +\frac{1}{2}a_{x1}t_1^2$$
since the car accelerates from rest.
$$x_1= \frac{1}{2}a_{x1}t_1^2= \frac{1}{2}a_{x1}\cdot 9^2$$
$$x_1=40.5a_{x1}\tag 1$$
Finding $x_2$;
$$x_2=x_1+v_{x1}t_2+ \frac{1}{2}a_{x2}t_2^2$$
Plugging the known and $x_1$ from (1);
$$990=40.5a_{x1}+3v_{x1} -\frac{1}{2}\cdot 5\cdot 3^2$$
$$1012.5=40.5a_{x1}+3v_{x1} \tag 3$$
Now we need to find the final velocity of the car at the end of the first stage.
$$v_{x1}=v_{0x}+a_{x1}t_1=0+9a_{x1}\tag{Pluggin into (3)}$$
$$1012.5=40.5a_{x1}+27a_{x1} $$
Solving for $a_{x1}$;
$$a_{x1}=\color{blue}{\bf 15}\;\rm m/s^2$$
And hence,
$$v_{x1}=9\cdot 15=\color{blue}{\bf 135}\;\rm m/s$$
Thus, the final speed is given by
$$v_{x2}=v_{x1}+a_{x2}t_2=135-(5\cdot 3)$$
$$v_{x2}=\color{magenta}{\bf120}\;\rm m/s$$
