Two relations from school physics come back, and they will not leave again. First, $v = v_0 + at$ — read aloud, v equals v-nought plus a t. Here $v$ is the final speed in m/s, $v_0$ (say “v-nought”) is the starting speed in m/s, $a$ is the acceleration in $\mathrm{m/s^2}$, and $t$ is the elapsed time in seconds. Subscript zero always means “at the beginning of the interval”; a bare symbol means “at the end”. That convention holds for the whole chapter.

Second, $d = v_0 t + \tfrac{1}{2} a t^2$ — d equals v-nought t plus one-half a t-squared — where $d$ is the distance covered in metres over that same interval. Read its two terms as two jobs: $v_0 t$ is the ground you would have covered by simply coasting, and $\tfrac{1}{2} a t^2$ is the extra the acceleration piles on top. The ½ is not decoration — the speed climbs steadily, so that second term is the TRIANGLE under the speed line, exactly half its rectangle.

What is new at this level is the numbers. Plants quote speeds in km/h and specify in metres per second, and the bridge is $\div 3.6$ going to m/s (3600 seconds per hour over 1000 metres per kilometre). Cross it BEFORE the socket, never inside the formula — a conversion done halfway through an expression is a conversion done wrong.