  ## 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:

• The sign of the velocity will indicate the direction of travel. If traveling away from the reference point (i.e. where $s=0$), the velocity will be positive -- if traveling towards it, the velocity will be negative. So for example, in looking at falling bodies, we typically use the ground as our reference point (height of zero) thus requiring bodies traveling down to have negative velocities, and those traveling up to have positive velocities.

• The speed of an object simply indicates how fast it is traveling, not its direction of travel. Consequently, the speed of an object is the absolute value of its velocity.

• The sign of the acceleration indicates whether the moving body is speeding up (positive) or slowing down (negative).

• Often, one uses $h(x)$ and $h_0$ in place of $s(x)$ and $s_0$, when the position of an object refers to its height.

• On our planet, the acceleration due to gravity is roughly $32$ ft/s$^2$. Thus for free-falling bodies, we observe that the velocity (whose derivative must give the acceleration) must take the form $$v(t) = -32t + v_0$$ where $v_0$ is the initial velocity. Similarly, the height/position (whose derivative must give the velocity) must take the form $$h(t)=-16t^2+v_0 t + h_0$$ where $h_0$ is the initial height -- both of these presume one is measuring distance in feet and time in seconds.