Calculating stopping distance of a car on an inclined plane with friction

If you have the acceleration $a$, you can calculate the distance traveled $s$ using the equation

$v^2 = u^2 + 2as$

You have your initial velocity which is $u = 30 ms^{-1}$ and you need your final velocity to be $v = 0$.

And if this distance is shorter than the distance to the traffic light you will have your answer.

The trick to this question is to realize there is no need for a complicated dynamic rules. You took your time to evaluate the friction force acting on a car

$$F = mg sin(alpha) - mu mg cos(alpha) $$

and you can see that this force is contants, therefore your acceleration $a = F/m = g( sin(alpha) - mu cos(alpha))$ is also constant. Now you can work with uniformly accelerated motion. You know (for $t_f$ being final time)

$$s = v_0t_f + frac 12 a t_f^2, qquad 0 = v_0 + at_f$$

which leads to the time of breaking $t_f = -v_0/a$ and therefore the distance of breaking

$$ s = -frac{v_0^2}a + frac 12 a left( frac{v_0}a right)^2 = -frac{v_0^2}{2a}. $$

You can see that the minus is fine, because acceleration is negative in this notation.