Acceleration, Velocity, and Speed

Suppose you have a particle traveling in a straight line, and the position of the particle relative to some reference point is given by $s(t)$ at any time $t$. This might describe the motion of a falling body, or a car on a road, or a ball rolling down a hill, etc...

Recall that one's average velocity over a particular interval is given by the distance traveled (i.e., the change in position) divided by the time it took to travel that distance (i.e., the change in time). $$v_{avg} = \frac{\Delta s}{\Delta t}$$ Similarly, we define instantaneous velocity to be $$v = \lim_{\Delta t \rightarrow 0} \frac{\Delta s}{\Delta t}$$ There is clearly a strong parallel to slopes of secants and tangents here. Recall, to find the slope of a secant to a function we find the ratio of "rise" (i.e., the change in $y$) to the "run" (i.e., the change in $x$), or equivalently: $$m_{secant} = \frac{\Delta y}{\Delta x}$$ and to find the slope of a tangent to a function at a given $x$-value, we consider the limiting slope seen by secants passing through $(x,f(x))$ and some other near-by point $(x+h, f(x+h))$. $$m_{tangent} = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$$ In this way, one can see that the instantaneous velocity of a particle turns out to be the derivative of the related position function: $$v(t)=s'(t)$$ Similarly, since acceleration describes the ratio of changes in velocity per unit time, we can see that acceleration in this context should be the derivative of velocity: $$a(t)=v'(t)$$

A couple of notes are in order: