# Boats?

#### sheva2291

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
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
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: .$$\displaystyle t \;=\;\frac{9}{25}\text{ hours} \;=\;21.6\text{ minutes}$$

• sheva2291

#### sheva2291

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