A moving car brakes and comes to a stop. How far does it travel in the meantime? No time appears anywhere in that question. The only thing wanted is a distance.
The formulas in hand, though, are v = v_0 + at and x = v_0t + \dfrac{1}{2}at^2, and t sits inside both of them. So the work turns into two stages: first find how long the stop takes, then use that to get the distance.
It can be solved that way. But routing the answer through a quantity nobody asked about is the long way around. We would rather never touch t at all.
The velocity equation v = v_0 + at can be solved for t. Move the terms across to get at = v - v_0, then divide both sides by a, which leaves t = \dfrac{v - v_0}{a}.
On the graph this is doing something obvious. The slope of the line is the acceleration, so dividing the rise by the slope returns the run, and the run is the time it took.
Now t is written with nothing but velocities and the acceleration. All that is left is to put it into the equation for distance.
The distance came out as the area of a trapezoid, x = \dfrac{v_0 + v}{2}t. Putting t = \dfrac{v - v_0}{a} into it gives x = \dfrac{v_0 + v}{2} \cdot \dfrac{v - v_0}{a}.
The numerator is a sum times a difference. Expanded, it becomes v^2 - v_0^2, so x = \dfrac{v^2 - v_0^2}{2a}, and clearing the denominator leaves v^2 - v_0^2 = 2ax.
There is no t in that equation. Velocity, acceleration and distance are tied together directly, with no need to know the time. It goes by the name of the third formula, but it is not a new law. It is the first two with t removed.
At the stop v = 0, so 0 - v_0^2 = 2ax. Writing the size of the deceleration as a, the distance travelled before stopping is x = \dfrac{v_0^2}{2a}.
What matters here is that the initial speed enters squared. With the same brakes, twice the initial speed means four times the stopping distance. A small gain in speed stretches the room needed to stop by a great deal.
The two lanes in the figure differ in initial speed alone, one of them twice the other. The brakes are equally strong, yet the lower lane runs four times as far before it comes to rest.