Chapter 2, Problem 13P

Distance Two ships leave port at the same time. One sails south at 15 mi/h, and the other sails east at 20 mi/h. Find a function that models the distance D between the ships in terms of the time t (in hours) elapsed since their departure.

To determine

To find: The function that models the distance $\left(D\right)$ between the ships in terms of time $t\left(\text{in}\text{hours}\right)$ elapsed since their departure.

Answer to Problem 13P

The function that models distance between ships in terms of time (in hours) is $D\left(t\right)=25t,t>0$.

Explanation of Solution

Given:

Speed of ship sailing south is $15\text{mi/h}$

Speed of ship sailing east is $20\text{mi/h}$

Calculation:

Draw the Figure (1) as shown below after time t since departure of ships.

Figure (1)

In Figure (1) $O$ is starting point of ships, $A,B$ are the points where ship sailing towards south and east reach after time $t$ respectively.

Calculate distance $OA$ and $OB$ shown in Figure (1) as,

$OA=15\cdot t\left[\because \text{Distance}=\text{Speed}\cdot \text{time}\right]$

And,

$OB=20t\left[\because \text{Distance}=\text{Speed}\cdot \text{time}\right]$

Apply Pythagorean theorem in triangle $OAB$ shown in Figure (1)

${\left(AB\right)}^2={\left(OA\right)}^2+{\left(OB\right)}^2$

Substitute $D$ for $AB$, $15t$ for $OA$ and $20t$ for $OB$ in above equation.

$\begin{array}{l}{\left(D\right)}^2={\left(15t\right)}^2+{\left(20t\right)}^2\\ D^2=225t^2+400t^2\end{array}$

Factor $\left(t^2\right)$ from right hand side of the above equation.

$\begin{array}{c}{D}^2=t^2\left(225+400\right)\\ =t^2\left(625\right)\end{array}$

Take square root of above equation.

$D=25t$

Thus, function that models distance between ships in terms of time (in hours) is $D\left(t\right)=25t,t>0$.