Section I. B. [Motivation] Estimating Instantaneous Velocity (revised 2/1/2014)

The fascination with motion, at least since the Greeks, has led to many important mathematical concepts attempting to analyze and explain its nature. The key element of motion is the measurable change in the position of an object over an interval of time. We usually measure the change in position in units of length such as centimeters, inches, feet, meters, yards, kilometers, and miles, while time is measured in seconds, minutes, hours, days, years, and even centuries. It is interesting to note that sometimes a distance  is measured with time units, as  we talk about a location being about 10 minutes from school, New York City  being a three hour train ride from Albany, or the distance to the nearest star to our solar being measured in light year.

Average Velocity and Speed:
We compare the change in position, denoted $\Delta s$, to the duration of the time interval, denoted Δt , using the ratio, $\frac {\Delta s}{\Delta t}$. This ratio is called the average velocity of the object for the given time interval. The magnitude (or absolute value) of this ratio, $|\frac {\Delta s}{\Delta t}|$, is called the average speed of the object, referring only to the size of the change in position, not the direction in which the change has occurred. Common measures of velocity and speed are centimeters per second, feet per second, meters per second, miles per hour, and kilometers per hour.

Most moving objects (cars, trucks, buses, ships and airplanes, as well as humans) do not travel at a constant speed. Video of fall from over 24 miles above surface of the earth!
Instead,the speed of these objects varies almost every instant,even though the changes in speed may not be noticeable.This last statement may seem a little vague, but consider your experience with automobile speedometers; this should support your understanding of the concept of instantaneous speed for a moving object. For example, suppose a car travels 5 miles through city traffic in 20 minutes with a speed that varies during the trip. The instantaneous speed of the car during the trip will be reported on the cars speedometer, while the average speed will be $\frac {\Delta s}{\Delta t} = \frac {5 miles}{\frac 1 3 hours} = 15 \frac m {hr}$.



Example I.B.1.
Table I.B.1
Time Runner's Position
0
0
1
4
6
39
11.5
100
In studying where a particular runner could improve her performance, her coach has examined some information about her performance in running a trial for a 100 meters race. Though the coach had more detailed information, let's suppose for our discussion that at time one second the runner was 4 meters from the starting line, while five seconds later, at time six seconds, she was 39 meters from the starting line. The runner finished the race in 11.5 seconds.


See Table I.B.1 and Figure I.B.2.

Figure I.B.2
Mapping Diagram
Let's compare the runner's speed during the three time intervals, $ [0,1], [1,6], $and $[6,11.5]$, to see in which time interval she was traveling with the fastest average speed and when she was slowest.
For the first second the runner's speed is $ 4 \frac m s$.
During the second time interval, between 1 and 6 seconds the runner has moved $(\Delta s=) 35$ meters in $(\Delta t=) 5$ seconds giving an average speed of  $ \frac {\Delta s}{\Delta t} = \frac {35 meters}{5  seconds} = 7 \frac m s$.
Finally, between 6  and 11.5 seconds the runner has moved $(\Delta s=) 61$ meters in $(\Delta t=) 5.5$ seconds giving an average speed of $\frac {\Delta s}{\Delta t} = \frac {61  meters}{5.5 seconds} = 11 \frac 1 {11} \frac m s$. 

So the runner's average speed was greatest at the end of the race while her slowest average speed was during the first second of the race.
[Since the time intervals for these averages were different it may be that the coach will still want to examine information about the second interval to see whether the runner can improve her performance at that stage of the race.]

Figure I.B.3
GeoGebra
Spreadsheet
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
Example I.B.4. Average Velocity Estimates of Instantaneous Velocity.
Assume the position of an object moving on a coordinate line at time $t$ seconds is $s(t)$ feet where $s(t) = t^2$. Estimate the instantaneous velocity of the object when $t = 3$ seconds  by finding the average velocity for the intervals $[2,3], [2.9,3], [2.99,3], [3,4], [3,3.1],$ and $[3,3.01]$. Find a formula for the average velocity of the object for any interval between $t = 3$ and $t = x$. Using this formula, discuss wat the instantaneous velocity of this object is at $3$ seconds.

S
olution.Let $\overline{v}(x)$ denote the average velocity for the time interval determined by times $t = 3$ and $t=x$ seconds. We initially want to determine $ \overline{v}(2), \overline{ v}(2.9),  \overline{v}(2.99), \overline{v}(4),\overline{ v}(3.1),\overline{v}(3.01),$ and finally a formula for $\overline{v}(x)$.

Figure I.B.3 created with GeoGebra shows a table with the relevant information for the computation of these average velocities along with the results. Based on this information it would seem reasonable to estimate the instantaneous velocity, which we denote by $v$, with a number very close to $6$, i.e., $v \approx 6$.

Next we find $\overline{v}(x)$ in general  (with $x \ne 3)$ using some algebra to simplify the expression: $\overline v (x) = \frac {\Delta S} {\Delta t} = \frac {s(x) - s(3)}{x-3} = \frac {x^2 - 9} {x-3} =\frac {(x + 3)(x -3)} {(x~-~3)} = x + 3$.
Notice that this formula for the average velocity at time $x$ is consistent with results we had from the direct computations.

Figure I.B.4
GeoGebra
Mapping Diagram
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
Using the arrow notation for estimation introduced in section I.A.2 to continue our discussion, consider what happens to $\overline v (x) = x + 3$  as  $x \rightarrow 3$ with $x \ne 3$.
See Figure I.B.4.

You can think of this either statically with $x$ chosen very close to $3$ ordynamically, with $x$ approaching the number $3$. With either point of view it should not be too hard to see that $\overline v (x) = x + 3 \rightarrow  6$. 

Since the average velocities will be better estimates of the instantaneous velocity when $x$ is close to $3$, we conclude that the instantaneous velocity at $3$ seconds must be $6 \frac {meters} { second}$.

Comment: Notice the similarity between this example and the tangent problem estimates in Example I.A.3.

We continue to find the instantaneous velocity for an object with position $s(t)=t^2$  meters at time $t=a$ seconds. We denote this instantaneous velocity by $v(a)$, and the average velocity determined on the time interval between $t = a$ and $t = x$  by $\overline{v_a}(x)$. We simplify the expression for $\overline{v_a}(x)$, namely

$\overline {v_a }(x) = \frac {\Delta s} {\Delta t} = \frac {s(x) - s(a)}{x-a} = \frac {x^2 - a^2} {x-a} =\frac {(x + a)(x -a)} {(x~-~a)} = x + a$.

Then as $x \rightarrow a$ with $x \ne a$, $\overline {v_a }(x) = x + a \rightarrow  2a$.

As in the previous discussion, when $x$ approaches ( or, is close to) $a$, $\overline{v_a}(x)$ should be a good estimate for $v(a)$, i.e., as $x \rightarrow a$ with $x \ne a$, $\overline {v_a }(x) \rightarrow  v(a)$ .
Therefore $v(a)$ must equal $2a$, i.e., $v(a) = 2a$.



Alternative Algebra.


Graphs and Motion: Secant and Tangent Lines.


Tool for Estimating Tangent Line Slopes and Instantaneous Velocity


Exercises I.B.