Hello, samtheman17!

At 11 a.m., ship $\displaystyle b$ leaves the harbor, sailing due south at 6 km/hr,

At noon, ship $\displaystyle a$ passes 9 km west of the harbor, heading due south at 8 km/h.

Give an expression for $\displaystyle d$, the distance between the two ships,

in terms of $\displaystyle t$, the number of hours after noon. First, *make a sketch!* Code:

9
Q * - - - - - - - - * H
| |
8t | |
| | 6
A * |
: * |
: * d * P
6-2t : * |
: * | 6t
: * |
C * - - - - - - - - * B
9

At 11 a.m. ship $\displaystyle b$ starts at $\displaystyle H$, sails south at 6 km/hr for one hour.

At noon, ship $\displaystyle b$ is at $\displaystyle P\!:\;HP = 6$

In the next $\displaystyle t$ hours, it sails $\displaystyle 6t$ km to point $\displaystyle B\!:\;\;PB = 6t$

At noon, ship $\displaystyle a$ starts at $\displaystyle Q\!:\;HQ = 9$ km,

. . and sails south at 8 km/hr,

In the next $\displaystyle t$ hours, it sails $\displaystyle 8t$ km to point $\displaystyle A\!:\;QA = 8t$

Note that: .$\displaystyle AC \:=\:(6 + 6t) - 8t \:=\:6 - 2t$

In right triangle $\displaystyle ACB\!:\;d^2 \;=\;(6-2t)^2 + 9^2$

Therefore: .$\displaystyle d \;=\;\sqrt{4t^2 - 24t + 117}$