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.
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
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.
Solution.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
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
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., $v(a) = 2a$.