Sometimes nobody timed anything. A braking test gives you a speed and a distance; a chute gives you a drop and an arrival speed. For those, eliminate $t$ between the two relations you already have and what survives is $v^2 = v_0^2 + 2ad$ — read aloud, v squared equals v-nought squared plus two a d. The cast is unchanged: $v$ is the final speed and $v_0$ the starting speed, both in m/s; $a$ is the acceleration in $\mathrm{m/s^2}$; $d$ is the distance in metres over which it acted. No $t$ anywhere — that absence is the entire reason it exists.

Everything here is a SQUARED speed, and that is the trap the relation sets. $v_0^2$ is not $v_0$; $(v - v_0)^2$ is not $v^2 - v_0^2$. Square each speed first, do the arithmetic in the currency of $\mathrm{m^2/s^2}$, and take the root only at the very end, when you are ready to be handed a speed back.

The engineering payoff is the one every braking chart is built on. Set $v = 0$ and the relation becomes $d = \dfrac{v_0^2}{2a}$: stopping distance grows with the SQUARE of the speed you started at. Twenty per cent faster is forty-four per cent more room. And notice the mass is nowhere in it — it cancelled between $F = ma$ and the friction that does the braking, which is why a laden truck and an empty one stop in the same distance at the same deceleration.