Velocity of Rectilinear motion

The same secant-to-tangent limiting idea solves the velocity problem. Suppose an object moves along a straight line with position (signed distance from the origin) $s=f(t)$ at time $t$; $f$ is called the position function. Over the time interval from $t_0$ to $t_0+\Delta t$, the change in position is $f(t_0+\Delta t)-f(t_0)$, so the average velocity over that interval is

which is exactly the slope of the secant line $PQ$ on the position–time graph, where $P=(t_0,f(t_0))$ and $Q=(t_0+\Delta t,\ f(t_0+\Delta t))$.

A subtlety worth flagging: over one fixed time interval $\Delta t$ with one fixed net displacement, the object could in principle have followed any number of genuinely different motions $C_1, C_2, C_3,\ldots$ between $P$ and $Q$ (speeding up then slowing down, pausing, reversing briefly, etc.) — all of them sharing the same average velocity $\Delta s/\Delta t$, because average velocity only sees the two endpoints, not the path between them.

To recover the velocity at the single instant $t=t_0$, shrink the interval: compute average velocities over shorter and shorter windows $[t_0, t_0+\Delta t]$, i.e. let $\Delta t \to 0$. The instantaneous velocity is defined as the limit of these average velocities:

Geometrically, $v(t_0)$ is precisely the slope of the tangent line to the position–time graph at $P=(t_0,f(t_0))$ — the velocity problem and the tangent-line problem are literally the same limit, read two different ways.

Worked illustration — free fall (the book's own example). A body falling freely from rest obeys the law of free fall $s=\tfrac12 gt^2$ ($g$ = the gravitational constant, no initial velocity). Then $f(t_0+\Delta t)=\tfrac12 g\big(t_0^2+2t_0\Delta t+(\Delta t)^2\big)$, so

$\Delta s = f(t_0+\Delta t)-f(t_0) = \tfrac12 g\big(2t_0\Delta t+(\Delta t)^2\big) = g\Delta t\left(t_0+\tfrac12\Delta t\right),$ …