The third machine: $d = v_0 t + \tfrac{1}{2}at^2$ (read aloud: d equals v-nought t, plus one-half a t-squared — and yes, $v_0$ is said “v-nought”), for anything that enters at one speed and accelerates as it goes. The picture explains the half: distance is the area under the speed line. Steady $v_0$ draws a rectangle, area $v_0 t$. The gained speed draws a triangle on top — and a triangle holds HALF its box, $\tfrac{1}{2}at^2$.

Forget the half and every answer lands too far. And run the L3 check before computing: $\mathrm{(m/s)\,s = m}$ and $\mathrm{(m/s^2)\,s^2 = m}$ — both terms cancel to metres, so the rearrangement can face the numbers.