Velocity is the rate of change of position. Take the rate of change of THAT and you have acceleration: $a = \dfrac{\Delta v}{\Delta t}$, and in the notation waiting for you in MCV4U, $a = \dfrac{dv}{dt}$. It is the slope of the velocity–time graph — the very quotient you computed in the last lesson, with velocity sitting on the y-axis. Its unit, $\mathrm{m/s^2}$, is best read as metres per second, per second: how much speed is gained every second.

Rearranged forward, that gives $v = v_0 + at$ — read v equals v-nought plus a t. Naming the letters: $v$ is the final velocity in m/s, $v_0$ the initial velocity in m/s, $a$ the acceleration in $\mathrm{m/s^2}$, and $t$ the elapsed time in seconds; solve for whichever one the question withholds.

Now look at its shape, because this is the nugget: $v = v_0 + at$ is $y = b + mx$ in disguise. The starting velocity is the intercept, the acceleration is the slope, and time is the horizontal axis. Uniform acceleration means a straight velocity line, and everything you know about straight lines transfers intact. Push the units through any rearrangement you try: if they refuse to cancel into the answer's unit the rearrangement is wrong, no appeal — though units that do work out never prove you right. The check runs one way only.