A-(b); B-(c); C-(a).

$textbf{The graph begins with a straight line since the rate at which Jan walks is constant. The graph has a negative slope since she walks toward from the sensor}$, and her distance from the sensor decreases as time increases.Jan starts $5$ m from the sensor.Use $(0, 5)$ as the distance intercept.Jan walks $4$ m toward the sensor at a
constant rate for $5$ s, so use the point $(5,1)$.

$textbf{Jan then walks away from the sensor. The line has a postive slope, because her distance from the sensor inreases as time increases.}$

$textbf{The graph ends with a horizontal line that has a slope of 0 because Jan is not moving}$. The slope indicates that her distance from the sensor does not change.

Here we $textbf{have graph of data}$ given in the table in the task:

Rachel’s average speed over each part of her climb is:

On interval $0leq{t}leq40$ we have:

$textbf{average speed}$ = $dfrac{6-5}{40-0}=dfrac{1}{40}=0.025$

On interval $40leq{t}leq130$ we have:

$textbf{average speed}$ = $dfrac{7-6}{130-40}=dfrac{1}{90}=0.011$

On interval $130leq{t}leq250$ we have:

$textbf{average speed}$ = $dfrac{7-7}{250-130}=0$

On interval $250leq{t}leq290$ we have:

$textbf{average speed}$ = $dfrac{8-7}{290-250}=dfrac{1}{40}=0.025$

On interval $290leq{t}leq335$ we have:

$textbf{average speed}$ = $dfrac{9-8}{335-290}=dfrac{1}{45}=0.022$

On interval $335leq{t}leq395$ we have:

$textbf{average speed}$ = $dfrac{10-9}{395-335}=dfrac{1}{60}=0.0167$

Here we have $textbf{a speed versus time graph}$ of Rachel’s climb:

$textbf{The bottle have a constant diameter so the water level increases at a constant rate}$. The water level will rise the fastest in the container with the smallest diameter.The graph is on following picture:

$textbf{The diameter of the vase varies, so the water level will increase at different rates}$. As the water level rises, the diameter of each crosssection gets smaller, causing the water level to increase more rapidly.So the graph must be nonlinear. The graph is on following picture:

$textbf{The average speed for Kommy’s first length of the pool is}$:

Speed = $|dfrac{Delta{d}}{Delta{t}}|=|dfrac{50-0}{45-0}|=|dfrac{50}{45}|=1.11$ m/s

#### (b)

$textbf{The average speed for Kommy’s second length of the pool is}$:

Speed = $|dfrac{Delta{d}}{Delta{t}}|=|dfrac{0-50}{110-55}|=|-dfrac{50}{55}|=|-0.91|=0.91$ m/s

#### (c)

$$
textbf{For the first length, slope is positive and for the second length slope is negative.Also, slope of the second length is smaller than slope for the first one.}
$$

Here we have a $textbf{distance versus time graph for Kommy’s swim:}$

At $t=50$, he was resting, so, $textbf{his speed is}$ $0$ at that moment.

#### (f)

Here we have $textbf{a speed versus time graph for Kommy’s swim:}$

(a) – A

(b) – C

(c) – D

(d) – B

Explanation of those solutions are on page $103$ in this book.

(i)

Standing $5$ m i front of the sensor and then walking $3$ m toward it at a constant rate for $3$ s.Next, walking $3$ m away from the sensor for $3$ s.

(ii)

Standing $6$ m in front of the sensor and then walking $2$ m toward it at aconstant rate for $2$ s.Then, waiting there for $1$ s and then walking away from it $2$ m at a constant rate for $2$ s.
#### (b)

(i)

$textbf{speed}$ = $|dfrac{Delta{d}}{Delta{t}}|=|dfrac{2-5}{3-0}|=|-dfrac{3}{3}|=1$ m/s

$textbf{speed}$ = $|dfrac{Delta{d}}{Delta{t}}|=|dfrac{5-2}{6-3}|=|dfrac{3}{2}|=dfrac{3}{2}$ m/s

(ii)

$textbf{speed}$ = $|dfrac{Delta{d}}{Delta{t}}|=|dfrac{6-4}{2-0}|=|dfrac{2}{2}|=1$ m/s

$textbf{speed}$ = $|dfrac{Delta{d}}{Delta{t}}|=|dfrac{4-4}{3-2}|=0$ m/s

$textbf{speed}$ = $|dfrac{Delta{d}}{Delta{t}}|=|dfrac{4-2}{5-3}|=|dfrac{2}{2}|=1$ m/s

$textbf{speed}$ = $|dfrac{Delta{d}}{Delta{t}}|=|dfrac{2-4}{6-5}|=|-dfrac{2}{1}|=2$ m/s

Here we have $textbf{a speed versus time graph}$ to represent this situation given in this task:

Equation of line that passes through points $(10,5)$ and $(11,10)$ is:

$y-5=dfrac{10-5}{11-10}(x-10)$ $Rightarrow$ $y=5x-45$

So, in order to calculate instaneous rate of change for $x=10.5$, let’s take two values $x=10.4$ and $x=10.6$, and their values of $y$ are:

$x=10.4: y=5cdot{10.4}-45=7$

$x=10.6:y=5cdot{10.6}-45=8$

So, finally,$textbf{ the instantaneous rate of change}$ in the runner’s speed at
$10.5$ min is:

$$
dfrac{Delta{y}}{Delta{x}}=dfrac{8-7}{10.6-10.4}=5
$$

#### (c)

The slopes of lines passes throgh points $(11,10$ and $(49,3)$ are, respectively:

$0$ and $-2$. so, $textbf{the average rate of change is}$:

$$
dfrac{-2-0}{49-11}=-dfrac{2}{38}=-0.31
$$

#### (d)

Answer for part (c) does not accurately represent the runner’s training schedule from minute $11$ to minute $49$ because we didn’t include all lines between this two points in our calculating average rate of change.