A position function $s(t)$ gives where an object is, in metres, at each time $t$ in seconds. Its average velocity over an interval is a secant slope, $\dfrac{\Delta s}{\Delta t} = \dfrac{s(t_2) - s(t_1)}{t_2 - t_1}$, read aloud delta s over delta t: the change in position over the time it took. Its instantaneous velocity at one moment is the tangent slope there, the derivative: $v(t) = s'(t) = \lim_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t}$, read aloud v of t equals s prime of t. That is the speedometer reading, and with a polynomial position it is one power rule away. For $s(t) = 5t^2 + 2t$: $v(t) = 10t + 2$, and $v(3) = 32$ m/s.

The trap is to divide position by time: $s(3) = 51$ m, so $\dfrac{s(3)}{3} = 17$ m/s, which is the average velocity over the first three seconds, not the velocity at $t = 3$. It comes out in m/s, and units guide without confessing: the right unit never proves a quantity right. The wrong unit does prove one wrong, so $s(3) = 51$ m is not a velocity of any kind. Velocity carries a sign, the direction of motion along the line; speed is its size, $|v|$, and is never negative.