Displacement from Average Velocity

Under constant acceleration, velocity changes along a straight line, so the average velocity over the interval is the plain arithmetic mean of the value at the start and the value at the end. Multiply that by the elapsed time and you have the displacement: $d = \tfrac{v_0 + v}{2}\,t$. The reason this works is best seen on a velocity-versus-time graph, where displacement is the area underneath the curve. With constant acceleration the curve is a straight line, so the area is a trapezoid, and the area of a trapezoid is the mean of the two parallel sides times the width. The half in this formula is the half in the trapezoid rule.

A train easing from 30 m/s down to 10 m/s over 20 s covers $d = \tfrac{30 + 10}{2} \times 20 = 400$ m. Notice what you never needed: the acceleration. That is what makes this equation worth having as a separate page rather than treating it as a corollary: when the two speeds and the duration are what you have measured, it answers directly.

It is the bridge between the other kinematic equations rather than an independent fact. Substitute $v = v_0 + at$ into it and the algebra collapses to $d = v_0 t + \tfrac{1}{2}at^2$; substitute $v_0 = v - at$ instead and you recover $d = vt - \tfrac{1}{2}at^2$. All five SUVAT relations are the same two facts (velocity changes at a steady rate, displacement is the area under the velocity curve) rearranged to suit whichever variable is missing.

The failure mode is applying it when the acceleration was not constant, and it fails silently. Consider a car that sits at 10 m/s for 55 s and then accelerates hard to 30 m/s in the last 5 s. The starting velocity is 10, the final is 30, the time is 60 s, and this formula confidently reports 1200 m. The true distance is about 650 m. Nothing in the arithmetic warns you, because the shortcut assumes a straight line between the endpoints and the real motion was nothing of the kind. Whenever the acceleration varies, go back to the definition, total displacement over total time, or split the trip into segments where it genuinely is constant. One further distinction: this is average velocity, not average speed. If the motion reverses within the interval, the two are different numbers, and it is the velocity version that this formula computes.