Thread: distance problem

heres a pic of whats going on: http://img405.imageshack.us/img405/415/untitledyp4.jpg

Ship A travels south at a speed of 8 miles/hr,

Ship B travels east at a speed of 16 miles/hr.

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 the people on both ships can see each other.'

..this may be a d=r*t problem.
dunno, how to set it up correctly.

thanks for any help.

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  *   |
                        *     |14-8t
                      *       |
                    *         |
      * - - - - - * - - - - - * C
      Q    16t    B   25-16t

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.