Boats?

May 2010
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0
A boat leaves a dock at 3:00 P.M. and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 P.M. How many minutes after 3:00 P.M. were the two boats closest together?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
A boat leaves a dock at 3:00 P.M. and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 P.M. How many minutes after 3:00 P.M. were the two boats closest together?
Start by setting up an appropriate coordinate system. It will be simplest to set the origin at the dock, the positive x-axis west and the positive y axis south with x and y in km from the dock. Take t in hours with t= 0 at 3:00 P.lM.

The first boat travels due south so its x coordinate is always 0. Its speed is 20 km/h so its distance from the dock, the y coordinate, after t hours is 20t.

The second boat travels due east so its y coordinate is always 0. its speed is -15 (because the positive x-axis points west and the boat is going east) so x= -15t+ b. You can find b by using the fact that the boat "reaches the same dock at 4:00 P.M."


The distance between the two boats is \(\displaystyle \sqrt{x^2+ y^2}\). You can find the minimum distance by differentiating with respect to t and setting the derivative equal to 0 or, for this simple problem, by completing the square.
 

Soroban

MHF Hall of Honor
May 2006
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6,341
Lexington, MA (USA)
Hello, sheva2291!

A boat leaves a dock at 3:00 PM and travels due south at 20 km/hr.
Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 PM.
How many minutes after 3:00 PM were the two boats closest together?
Code:
      : - - -  15 - - - - :
      P  15t  B   15-15t  D
      o-------o-----------o
                *         |
                  *       |
                 x  *     | 20t
                      *   |
                        * |
                          o A

Ship \(\displaystyle A\) starts at dock \(\displaystyle D\) and sails south at 20 km/hr,
In \(\displaystyle t\) hours, it has sailed \(\displaystyle 20t\) km to point \(\displaystyle A\).

Ship \(\displaystyle B\) starts at \(\displaystyle P\) (15 km west of \(\displaystyle D\)) and sails east at 15 km/hr.
In \(\displaystyle t\) hours, it has sailed \(\displaystyle 15t\) km to point \(\displaystyle B\).
. . Hence: .\(\displaystyle BD \:=\:15-15t\)

Let \(\displaystyle x \:=\:AB.\)

In right triangle \(\displaystyle BDA\!:\;\;x \;=\;\sqrt{(20t)^2 + (15-15t)^2} \)

. . Hence, we have: .\(\displaystyle x \;=\;\left(625t^2 - 450t + 225\right)^{\frac{1}{2}}\)

And that is the function we must minimize.


I believe the answer is: .\(\displaystyle t \;=\;\frac{9}{25}\text{ hours} \;=\;21.6\text{ minutes}\)

 
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May 2010
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ahhh thank you very much, you explained it very clearly. i changed around the numbers and got it right =)