1. ## rate problem

I'm not very good with word problems. I find putting words into equations and diagrams difficult. Can anyone help out with this question?

Ship A is sailing south at 24 km/h while ship B, which is 48 km due south of A, is sailing east at 18 km/h.

i)At what rate are they approaching or separating after one hour?

ii) when do they cease to approach one another, how far apart are they at that time, and what is the significance of that distance?

2. Hello, dcfi6052!

Ship A is sailing south at 24 km/h while ship B,
which is 48 km due south of A, is sailing east at 18 km/h.

i) At what rate are they approaching or separating after one hour?

ii) When do they cease to approach one another?
How far apart are they at that time?
What is the significance of that distance?
Code:
-        P*
:         |
:     24t |
:         |
:        A*
48        | \
:         |   \
:  48-24t |     \R
:         |       \
:         |         \
-        Q+ - - - - - *B
18t

Ship A starts at $\displaystyle P$ and sails south at 24 km/hr.
. . In $\displaystyle t$ hours, it has sailed $\displaystyle 24t$ km to point $\displaystyle A.$

Ship B starts at $\displaystyle Q$ and sails east at 18 km/hr.
. . In $\displaystyle t$ hours, it has sailed $\displaystyle 18t$ km to point $\displaystyle B.$

Let $\displaystyle R$ = distance $\displaystyle AB.$
Note that $\displaystyle AQ\:=\:48-24t$

Pythagorus says: .$\displaystyle R^2\:=\:(48-24t)^2 + (18t)^2 \:=\:900t^2 - 2304t + 2304$ [1]

Differentiate with respect to time: .$\displaystyle 2R\left(\frac{dR}{dt}\right)\:=\:1800t - 2304$

. . and we have: .$\displaystyle \frac{dR}{dt}\:=\:\frac{900t - 1152}{R}$ [2]

When $\displaystyle t = 1$, from [1] we have:
. . $\displaystyle R^2\:=\:900\cdot1^2 - 2304\cdot1 + 2304 \:=\:900\quad\Rightarrow\quad R \,=\,30$

Substitute into [2]: .$\displaystyle \frac{dR}{dt}\:=\:\frac{900(1) - 1152}{30}\:=\:-\frac{252}{30}$

They are approaching each other at $\displaystyle \boxed{8.4\text{ km/hr}}$

They stop approaching each other when $\displaystyle \frac{dR}{dt} = 0$

. . We have: .$\displaystyle \frac{900t - 1152}{R} \:=\:0\quad\Rightarrow\quad900t \,=\,1152\quad\Rightarrow\quad\boxed{t \,=\,1.28\text{ hours}}$

From [1], we have: .$\displaystyle R^2\;=\;900(1.28^2) - 2304(1.27) + 2304\;=\;829.44$

. . Therefore, the ships are: .$\displaystyle R \:=\:\boxed{28.8\text{ km apart}}$

That is the minimum distance between the two ships.