On this screen we introduce the concept of “average velocity for a trip.”

This is a brief step away from the various functions we considered in the preceding section, where we saw how we can use Desmos to estimate the rate at which a function changes at a particular point—provided we have a function that we’re able to graph using Desmos. But relying on Desmos is a major limitation since a function can be represented in other ways, including verbally, or by providing a data set. More generally, we don’t want our development of mathematical knowledge to be dependent upon any particular tool such as Desmos (invaluable as Desmos and other graphing programs may be).

So to develop a deeper understanding and a better set of tools, let’s take one of the most frequently used problem-solving steps (that you should absolutely have in your repertoire): we start more simply with an everyday situation we can reason about. In particular, instead of trying immediately to find the rate at which a function changes at a particular point, let’s first determine the rate at which a function changes over an *interval* of whatever size we choose. We can then think about what happens as we shrink that interval, just as you did as you zoomed in more and more and moved the second point closer and closer to the first in the preceding topic.

To begin, we again return to the everyday situation of a car trip.

Scenario 1: Sophia's Roadtrip

Sofia takes a road trip, driving from her house to a destination 120 miles away.

She hits some stop-and-go traffic along the way, and stops to get gas at some point.

Later still she stops again to grab a snack.

Upon her arrival, having traveled the full **120 miles**, Sofia notes that exactly **3 hours** have passed since she departed. What was her average velocity for the trip, in miles per hour?

*Solution.*

Probably without thinking about it much, you did a calculation in your head: Sofia traveled 120 miles in 3 hours, so her average velocity for the trip was

$$\text{average velocity } = \frac{120 \text{ miles}}{3 \text{ hours}} = 40 \text{ miles/hour} \, \cmark$$

That everyday calculation is perfectly correct: on average, Sofia traveled 40 miles for each hour of her trip. That doesn’t mean that at every instant her speedometer showed 40 mph (miles per hour), or that every hour she actually traveled 40 miles. Indeed, we know that she stopped several times, and at those times her speedometer showed 0 mph. And then at other times she must have been going faster than 40 mph. The crucial point is that none of those details matter: her *average rate of travel *was simply her change in position (120 miles) divided by her trip-time (3 hours), for an average rate of 40 mph.

Said differently, if Sofia had traveled at exactly 40 mph for the entire trip, she would have arrived at exactly the same moment, 3 hours after leaving, as she did in her actual trip.

Let’s consider a second, related scenario.

Scenario 2: Samuel's Road Trip

Samuel departs for a trip at 2:00 pm, starting near the 10-mile marker on a highway. His position along the highway, as indicated by mile markers, is shown in the table.

$$ \begin{array}{c|c}

\text{time} & \text{position [miles]} \\

\hline

\text{2:00 pm} & 10 \\

\text{2:20 pm} & 28 \\

\text{2:40 pm} & 42 \\

\text{3:00 pm} & 68 \\

\text{3:20 pm} & 68 \\

\text{3:40 pm} & 75 \\

\text{4:00 pm} & 89 \\

\text{4:20 pm} & 107 \\

\text{5:00 pm} & 130 \\

\end{array} $$

He arrives at his destination at 5:00 pm.

What was Samuel’s average velocity for the trip? What are the units of your answer?

*Solution.*

We’re making the same calculation for Samuel’s trip as we did for Sofia’s in Scenario 1, and once again the details of the trip don’t matter. All that matters is his change in position over the entire trip, 120 miles (= 130 miles – 10 miles), and how long the trip took, 3 hours (= 5:00 pm – 2:00 pm). Samuel’s average velocity was thus

$$\text{average velocity } = \frac{120 \text{ miles}}{3 \text{ hours}} = 40 \text{ miles/hour} \, \cmark$$

The exact details of Sofia’s and Samuel’s trip were different: they drove at different speeds at different times, got stuck in different traffic, stopped for different reasons. But since they both changed their positions by 120 miles in exactly 3 hours, they both had the same average velocity of 40 mph. None of the other details matter.

Let’s consider a third and final related scenario.

Scenario 3: Yasmin's Road Trip

At 1:00 pm, Yasmin departs on a trip. She starts at her house, near a “mile 25” marker on the highway. She hits some stop-and-go traffic on her journey, and about halfway through her trip she stops for a snack. Her position, as determined by mile-markers along the highway, is shown as a function of time in the graph.

She arrives at her destination at 4:00 pm as shown.

What was Yasmin’s average velocity for the trip? What are the units of your answer?

*Solution.*

We’re making the same calculation for Sofia’s and Samuel’s trip above, and once again the details of the trip don’t matter. All we care about is his change in Yasmin’s position over the entire trip, 120 miles (= 145 miles – 25 miles), and how long the trip took, 3 hours (= 4:00 pm – 1:00 pm). Yasmin’s average velocity was thus

$$\text{average velocity } = \frac{120 \text{ miles}}{3 \text{ hours}} = 40 \text{ miles/hour} \, \cmark$$

The exact details of Yasmin’s trip don’t matter. Her position, like Sofia’s and Samuel’s, changed by 120 miles in exactly 3 hours, and so Yasmin’s average velocity was also 40 miles/hour (or 40 mph).

We’re going to explore the preceding example, about Yasmin’s trip, a bit further to draw some broader conclusions about average velocity, and then generalize to all average rates of change.

First, play the GIF below to view some details of Yasmin’s trip:

Crucially, note that — just as for Sofia’s and Samuel’s trips — we don’t care about any of those details when we calculate Yasmin’s average velocity for her trip. Instead, we just look at her starting and ending points, and complete our calculation using *only* those values and ignoring everything else.

The definition of average velocity for a trip captures these ideas:

\begin{align*}

\text{average velocity for a trip} &= \frac{\text{total change in position}}{\text{trip duration}} \\[8px] &= \frac{\text{final position}\,- \text{initial position}}{\text{final time}\,- \text{initial time}} \\[8px] &= \frac{s(t_f)\,-\,s(t_i)}{t_f\, -\, t_i}

\end{align*}

where the trip begins at time $t_i$ at position $s(t_i),$ and ends at time $t_f$ at position $s(t_f).$

Let’s recast our everyday reasoning for Yasmin’s trip as a calculation using the notation in the blue-box immediately above:

Example 1: Calculate Average Velocity for Yasmin's Trip

Yasmin’s starts a trip at $t_i =$ 1:00 pm with initial position $s(t_i=\text{1:00 pm})$ = 25 miles.

Her final position at $t_f =$ 4:00 pm is $s(t_f=\text{4:00 pm})$ = 145 miles.

Compute Yasmin’s average velocity for her trip.

*Solution*.

\begin{align*}

\text{average velocity for Yasmin’s trip} &= \frac{\text{total change in position}}{\text{trip duration}} \\[8px]
&= \frac{s(t_f)\,-\,s(t_i)}{t_f\, -\, t_i} \\[8px]
&= \frac{\text{145 miles – 25 miles}}{\text{4:00 pm – 1:00 pm}} \\[8px]
&= \frac{120 \text{ miles}}{3 \text{ hours} } \\[8px]
&= 40 \text{ miles/hour} \, \cmark

\end{align*}

In the following set of slides, we step through a way of visualizing the calculations for Yasmin’s trip by focusing on the graph — which will lead to a key conclusion:

The preceding slides illustrate this key point:

that connects the initial and final points on the position-versus-time graph

We call the line that passes through Yasmin’s initial and final position-points a **secant line** because it intersects Yasmin’s motion curve in two places. (By contrast, a few Topics from now we’ll develop the “tangent line” to a curve at a given point. This tangent line just grazes the curve at the single point of interest.) The word *secant* comes from the Latin word *secare*, which means “to cut”: the line “cuts” the curve in at least two places.

The GIF below illustrates one way to interpret the average velocity for a trip: if you were to leave at exactly the same moment as Yasmin and (somehow) maintained the constant rate of 40 miles/hour throughout your journey, following exactly the same path, you would arrive at the destination at the same moment as Yasmin does.

We can extend this interpretation of average velocity to apply to *any* motion:

Average velocity is the *constant* rate at which an object would travel

in order to arrive at the final position in the same amount of time as that of the original motion,

regardless of the details of that original motion.

in order to arrive at the final position in the same amount of time as that of the original motion,

regardless of the details of that original motion.

Let’s consider a quick Check Question regarding average velocity for an entire “trip”:

CHECK QUESTION 1: Matt's Swim

If this is all seeming straightforward: great! On the next screen, we’ll extend the ideas we’ve introduced here to develop a general expression for average velocity.

- Average velocity for a trip depends
*only*on the initial and final positions, and trip duration; it ignores any other details.

\begin{align*}

\text{average velocity for a trip} &= \frac{\text{total change in position}}{\text{trip duration}} \\[8px] &= \frac{\text{final position}\,- \text{initial position}}{\text{final time}\,- \text{initial time}} \\[8px] &= \frac{s(t_f)\,-\,s(t_i)}{t_f\, -\, t_i}

\end{align*}

where the trip begins at time $t_i$ at position $s(t_i),$ and ends at time $t_f$ at position $s(t_f).$ - Graphically, average velocity for a trip equals the
*slope*of the secant line that passes through the initial and final points on a position-versus-time graph. - The average velocity for a trip is the
*constant*rate at which an object would travel to leave and arrive at the exact same moments as the actual motion.

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