Velocity-Displacement Relation (v² = v₀² + 2ad)

This is the SUVAT equation with the clock taken out of it. Every other member of the family needs a time; this one relates the starting speed, the finishing speed, the acceleration and the distance directly, which makes it the right tool whenever you know where something ended up but not how long it took getting there. You obtain it by solving $v = v_0 + at$ for $t$ and substituting into $d = v_0 t + \tfrac{1}{2}at^2$; the time cancels and $v^2 = v_0^2 + 2ad$ is what survives.

A car braking from 25 m/s — 90 km/h — at a firm $-8$ m/s² comes to rest in $d = (0 - 25^2)/(2 \times -8) \approx 39$ m. That is the distance the tyres need after the brakes are on, and adding the driver's reaction time at 25 m/s puts another 25 m or so in front of it. Braking-distance tables in road-safety pamphlets are this equation, run once per speed.

Multiply the whole thing by $\tfrac{1}{2}m$ and it turns into something familiar: $\tfrac{1}{2}mv^2 = \tfrac{1}{2}mv_0^2 + (ma)d$, which is the work–energy theorem, final kinetic energy equals initial kinetic energy plus the work done by the net force. That is not a coincidence — it is the same statement written twice, once in the language of kinematics and once in the language of energy. It also explains the squares: energy has always gone as $v^2$, so a relation involving distance and force had to.

Signs are where this equation punishes carelessness. Pick a positive direction, then stay in it: a car braking while travelling in the positive direction has a negative $a$, and entering $+8$ instead of $-8$ returns a stopping distance of $-39$ m, which is the calculator telling you the car would have had to be reversing. The physical consequence of the squares is the one every driving instructor tries to convey and this formula proves: stopping distance goes as the square of speed. The 39 m from 90 km/h becomes about 156 m from 180 km/h. A 20% increase in speed is a 44% increase in the distance you need. Two smaller points: solving for $v$ or $v_0$ takes a square root, and this page returns the principal non-negative branch — if the object actually reversed direction, the negative root is the physical one and you should supply the sign yourself. And $d$ is displacement along the direction of motion, not path length, so it is not the right tool for a curved route.