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_1-t_0}\text{.} \end{equation*}

# Subsection1.1.1Exercises

According to simplified Newtonian physics, if an object is dropped from an elevation 1  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\,\mathrm{m}-\left(4.9\,\sfrac{\mathrm{m}}{\mathrm{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 “$4.9$ ms2” represents one half of the acceleration due to gravity near the surface of Earth.)

##### 1

Calculate $\fe{p}{3\,\mathrm{s}}$ and $\fe{p}{5\,\mathrm{s}}\text{.}$ Make sure that you include the units specified in the formula for $\fe{p}{t}\text{,}$ 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 $\frac{\fe{p}{5\,\text{s}}-\fe{p}{3\,\text{s}}}{{5\,\text{s}}-{3\,\text{s}}}\text{;}$ 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 $\cinterval{t_0}{t_1}\text{.}$ 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_1-t_0}&=\frac{\left[200-4.9t_1^2\right]-\left[200-4.9t_0^2\right]}{t_1-t_0}\\ &=\frac{200-4.9t_1^2-200+4.9t_0^2}{t_1-t_0}\\ &=\frac{-4.9t_1^2+4.9t_0^2}{t_1-t_0}\\ &=\cdots \end{align*} Hint
##### 4

Check the formula you derived in Exercise 1.1.1.3 using $t_0=3$ and $t_1=5\text{;}$ 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) $\frac{\fe{p}{t_1}-\fe{p}{3}}{t_1-3}$ (m⁄s) 2.9 2.99 2.999 3.001 3.01 3.1
##### 6

As the value of $t_1$ gets closer to $3\text{,}$ 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?