When a velocity changes at a CONSTANT rate, its average over the interval is simply the midpoint of its two ends: $\bar{v} = \dfrac{v_0 + v}{2}$ — read v-bar equals v-nought plus v, over two. That gives the chapter's first kinematic relation, $d = \dfrac{v_0 + v}{2}\,t$, read d equals v-nought plus v, over two, times t. The cast: $d$ is the displacement in metres, $v_0$ (v-nought) is the velocity at the start in m/s, $v$ is the velocity at the end in m/s, and $t$ is the elapsed time in seconds. Whichever of the four the question leaves blank is the one you solve for.

Where it comes from is worth more than the formula. Plot velocity against time and the displacement is the AREA under the line. Uniform acceleration makes that line straight, so the area is a trapezoid — and a trapezoid's area is the average of its two parallel sides times its width. This relation is the trapezoid rule, wearing a lab coat.

The condition is the fine print, and examiners love it: the midpoint shortcut is legal only for uniform acceleration. On a curved velocity graph the mean of the endpoints is not the mean of the journey, and next year's integral exists precisely to handle that case.