Question

Commute to Work Jade and her roommate Jari commute to work each morning, traveling west on $\text{I-10}$. One morning Jade left for work at $\text{6:50A.M.}$, but Jari left $\text{10}$minutes later. Both drove at a constant speed. The following graphs show the distance (in miles) each of them has traveled on $\text{I-10}$ at time t (in minutes). where $t=0$is $\text{7:00A.M.}$.

(a) Use the graph to decide which of them is travelling faster.

(b) Find the speed $\left(\text{in\hspace{0.33em}mi}/\text{h}\right)$at which each of them is driving.

(c) Find linear functions f and g that model the distances that Jade and Jari travel as functions of t (in minutes).

Short answer

(a) Jari travels faster than Jade.

(b) The speed of Jade is $\text{60mi}/\text{hr}$and the speed of Jari is $70\text{mi}/\text{hr}$.

(c) The function that models the distance travelled by Jari is $f\left(t\right)=\frac{7}{6}t$and the function that models the distance travel by Jade is $f\left(t\right)=t+10$.

Step-by-step solution

Part a. Step 1. List out the given information.

The given graph is,

Part a. Step 2. Interpret the graph.

The graph shows the distance travelled by Jade and Jari,

From the graph it can be inferred that the graph of Jari is steeper as compared to the graph of Jarry.

Part a. Step 3. Find the result.

The velocity of steeper graph is more as compared to a flat graph.

Since, the graph of Jari is more steep as compare to Jade. Hence, the velocity of Jade is more as compared to Jade.

Therefore, Jari travels faster than Jade.

Part b. Step 1. List the given data.

The given graph is shown below.

The graph represent the distance travelled by Jade and Jari.

Part b. Step 2. Apply the concept of slope.

The slope of a line is given by,

$$\begin{gathered} \text{Slope}=\frac{\text{Changeinverticaldistance}}{\text{Changeinhorizontaldistance}} \\ =\frac{y_2-y_1}{x_2-x_1} \end{gathered}$$

Here,$x_1,x_2,y_1,y_2$ are points on the lines.

Part b. Step 3. Calculate the speed of Jade and Jari.

Any two points on the graph of Jari are $\left(0,0\right)\text{and}\left(6,7\right)$.

Substitute the values in the expression for slope.

role="math" localid="1646897643901" $$\begin{gathered} {\text{Slope}}_{Jari}=\frac{7-0}{6-0} \\ =\frac{7}{6}\text{mi}/\text{min} \\ =\frac{7}{6}\times 60\text{mi}/\text{hr} \\ =70\text{mi}/\text{hr} \end{gathered}$$

Any two points on the graph of Jade are $\left(0,10\right)\text{and}\left(6,16\right)$.

Substitute the values in the expression for slope.

role="math" localid="1646897717948" $$\begin{gathered} {\text{Slope}}_{Jari}=\frac{16-10}{6-0} \\ =\frac{6}{6}\text{mi}/\text{min} \\ =\frac{6}{6}\times 60\text{mi}/\text{hr} \\ =60\text{mi}/\text{hr} \end{gathered}$$

Therefore, the speed of Jade is $60\text{mi}/\text{hr}$and speed of Jari is $\text{70mi}/\text{hr}$.

Part c. Step 1. Apply the concept of linear function.

Let a function f that models the distance that Jade travels as a function of$t$ be,

$f\left(t\right)=at+b\mathrm{...}\left(1\right)$

Here,$a$ is the slope of the equation and b is the constant or value at initial condition.

Part c. Step 2. Find the equation for Jade.

The slope of Jade is,

$$\begin{gathered} a=60\text{mi}/\text{hr} \\ =60\times \frac{1}{60}\text{\hspace{0.33em}mi}/\min \\ =1\text{\hspace{0.33em}mi}/\text{min} \end{gathered}$$

And,

$b=10$

Substitute the values of a and b in equation (1).

role="math" localid="1646897946012" $\begin{array}{c}f\left(t\right)=\left(1\right)t+10\\ =t+10\end{array}$

Part c. Step 3. Find the equation of Jari.

The slope of Jari is,

$$\begin{gathered} a=70\text{mi}/\text{hr} \\ =70\times \frac{1}{60}\text{\hspace{0.33em}mi}/\min \\ =\frac{7}{6}\text{\hspace{0.33em}mi}/\text{min} \end{gathered}$$

And,

$b=0$

Substitute the values of a and b in equation (1).

role="math" localid="1646898073710" $$\begin{gathered} f\left(t\right)=\left(\frac{7}{6}\right)t+0 \\ =\frac{7}{6}t \end{gathered}$$

Thus, the function that models the distance travelled by Jari is$f\left(t\right)=\frac{7}{6}t$ and the function that models the distance travel by Jade is $f\left(t\right)=t+10$.