Motion is frequently modeled using calculus. A building block for this application is the concept of average velocity. Average velocity is defined to be net displacement divided by elapsed time. More precisely,
Definition1.1.1Average Velocity
If \(p\) is a position function for something moving along a numbered line, then we define the average velocity over the time interval \(\cinterval{t_0}{t_1}\) to be: \begin{equation*}\frac{\fe{p}{t_1}\fe{p}{t_0}}{t_1t_0}\text{.}\end{equation*}
According to simplified Newtonian physics, if an object is dropped from an elevation of \(200\) meters (m) and allowed to free fall to the ground, then the elevation of the object is given by the position function \begin{equation*}\fe{p}{t}=200\,\text{m}\left(4.9\,\text{m}/\text{s}^2\right)t^2\end{equation*} where \(t\) is the amount of time that has passed since the object was dropped. (The “\(200\) m” in the formula represents the elevation from which the object was dropped, and the “\(4.9\) ^{m}⁄_{s2}” represents one half of the acceleration due to gravity near the surface of Earth.)
1
Calculate \(\fe{p}{3\,\text{s}}\) and \(\fe{p}{5\,\text{s}}\). Make sure that you include the units specified in the formula for \(\fe{p}{t}\), and that you replace \(t\) with, respectively, \(3\) s and \(5\) s. Make sure that you simplify the units, as well as the numerical part of the function value.
2
Calculate \(\frac{\fe{p}{5\,\text{s}}\fe{p}{3\,\text{s}}}{{5\,\text{s}}{3\,\text{s}}}\); include units while making the calculation. Note that you already calculated \(\fe{p}{3\,\text{s}}\) and \(\fe{p}{5\,\text{s}}\) in Exercise 1.1.1.1. What does the result tell you in the context of this problem?
3
Use Definition 1.1.1 to find a formula for the average velocity of this object over the general time interval \(\cinterval{t_0}{t_1}\). The first couple of lines of this process are shown below. Copy these lines onto your paper and continue the simplification process.\begin{align*}
\frac{\fe{p}{t_1}\fe{p}{t_0}}{t_1t_0}&=\frac{\left[2004.9t_1^2\right]\left[2004.9t_0^2\right]}{t_1t_0}\\
&=\frac{2004.9t_1^2200+4.9t_0^2}{t_1t_0}\\
&=\frac{4.9t_1^2+4.9t_0^2}{t_1t_0}\\
&=\cdots
\end{align*}
HintFactor \(4.9\) from the numerator; the remaining factor will factor further.
4
Check the formula you derived in Exercise 1.1.1.3 using \(t_0=3\) and \(t_1=5\); that is, compare the value generated by the formula to that you found in Exercise 1.1.1.2.
5
Now explore some average velocities in tabular form.
Using the formula found in Exercise 1.1.1.3, replace \(t_0\) with \(3\) but leave \(t_1\) as a variable; simplify the result. Then copy Table 1.1.3 onto your paper and fill in the missing entries.
\(t_1\) (s) 
\(\frac{\fe{p}{t_1}\fe{p}{3}}{t_13}\) (^{m}⁄_{s}) 
2.9 

2.99 

2.999 

3.001 

3.01 

3.1 

Table1.1.3Values of \(\frac{\fe{p}{t_1}\fe{p}{3}}{t_13}\)
6
As the value of \(t_1\) gets closer to \(3\), the values in the \(y\)column of Table 1.1.3 appear to be converging on a single quantity; what is this quantity? What does it mean in the context of this problem?