In Instantaneous Velocity you found the velocity at an instant as the slope of the tangent line to an $x$–$t$ graph, and in Position, Velocity and Acceleration you met the derivative as that slope. Now we put the two together. When the position is given as a formula $x(t)$, the velocity at every instant comes from one derivative.
🔑 Key idea
The velocity is the derivative of position with respect to time: $$v(t) = \dfrac{dx}{dt} = \lim_{\Delta t \to 0}\dfrac{x(t + \Delta t) - x(t)}{\Delta t}.$$ Its sign gives the direction of motion along the axis, and its size is the speed. Units: meters per second.
💡 Why it works
An average velocity $\Delta x/\Delta t$ over a long interval blurs together fast and slow stretches. Shrink the interval around one instant and the average stops changing: it settles on one number, the limit. That limit is exactly how the derivative is defined, so "velocity at an instant" and "derivative of position" are the same thing. On the graph, the secant line through two nearby points turns into the tangent line.
$x(t) = t^3 - 6t^2 + 9t$. The slope of the tangent, in m/s, is the velocity $v(t) = 3t^2 - 12t + 9$.
Working with a position formula
Every rule from calculus applies: the power rule, the chain rule for $\sin(\omega t)$ or $e^{-t/\tau}$, and so on. Keep the units in view. If $x$ is in meters and $t$ in seconds, then $dx/dt$ is in m/s, because a derivative is a ratio of a tiny change in $x$ to a tiny change in $t$.
✏️ Worked example: Velocity at an instant
A particle moves along the $x$-axis with $x(t) = 2t^3 - 9t^2 + 12t$ ($x$ in m, $t$ in s). Find its velocity at $t = 3\ \text{s}$, and say which way it moves.
Differentiate term by term: $v(t) = 6t^2 - 18t + 12$.
The sign is positive, so at that instant the particle moves in the $+x$ direction at 12 m/s.
Answer: $v(3) = 12\ \text{m/s}$, in the $+x$ direction.
✏️ Worked example: When is it at rest?
For the same particle, when is it momentarily at rest?
At rest means $v = 0$: $6t^2 - 18t + 12 = 0$.
Divide by 6 and factor: $t^2 - 3t + 2 = (t - 1)(t - 2) = 0$.
So $v = 0$ at $t = 1\ \text{s}$ and $t = 2\ \text{s}$. Between those times $v < 0$: the particle backs up before moving forward again.
Answer: At $t = 1\ \text{s}$ and $t = 2\ \text{s}$.
⚠️ Common mistake
Dividing the position by the time. For the particle above, $x(3)/3 = 9/3 = 3\ \text{m/s}$, but the true velocity at 3 s is 12 m/s. The ratio $x(t)/t$ is the average velocity from $t = 0$ (and only if the particle started at $x = 0$); it is not the velocity at the instant $t$. Check: the velocity at an instant must come from positions just before and just after that instant, which is what the derivative uses.
A derivative of a function you can't differentiate by hand can still be estimated: compute $\dfrac{x(t + h) - x(t - h)}{2h}$ with a small $h$, such as 0.001 s. That is how a motion sensor turns position readings into velocity.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 1 XP.