Hello, rcmango!
A challenging problem . . .
Ship A travels south at a speed of 8 mph.
Ship B travels east at a speed of 16 mph.
The pic shows both ships at 7am.
With matlab, plot the distance between the ships as a function of time for the next 4 hours.
The horizontal axis will show the time of day starting at 7.
and the vertical will show the distance.
If visiblity is at 8 miles, estimate the time when people on both ships can see each other. Code:
P
*

8t

* A
* 
d * 
* 148t
* 
* 
*      *      * C
Q 16t B 2516t
Ship A starts at P and sails south at 8 mph.
In t hours, it has sailed 8t miles to point A.
Note that: .AC .= .14  8t
Ship B starts at Q and sails east at 16 mph.
In t hours, it has sailed 16t miles to point B.
Note that: .BC .= .25  16t
Using Pythagorus, the distance between them is:
. . . . . . . . . . . . ___________________
. . d .= .AB .= .√(25  16t)² + (14  8t)² . ← (graph this)
. . . . . . . . . . . . . . . . . . . . .___________________
If the distance is 8 miles: . √(25  16t)² + (14  8t)² .= .8
Square both sides: .625  800t + 256t² + 196  224t + 64t² .= .8
. . which simplifies to the quadratic: .320t²  1024t + 757 .= .0
. . . . . . . . . . . . . . . . . . . . . . . . _________________
. . . . . . . . . . . . . . . . . .1024 ± √1024²  4(320)(757)
Quadratic Formula: . t .= .
. . . . . . . . . . . . . . . . . . . . . . . . 2(320)
. . and we get two roots: .t .= .1.16, 2.04
We discard the larger answer.
. . In 2 hours, both ships will be beyond point C.
Hence, they will be 8 miles apart in 1.16 hours . . . at about 8:10 am.