Lesson A kayaker and an orca - simple, elegant and unexpectedly short solution

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A kayaker and an orca - simple, elegant and unexpectedly short solution


Problem 1

A kayaker was paddling due  East at  3 mph and an orca was swimming due  North at  10 mph.
If they started  1 mile apart,  how long did it take for the orca to swim underneath the kayaker?

Solution

Let 't' be the time under the problem's question.


To get the meeting point, the kayaker paddled 3t miles to the east.
while the orca swam 10t miles to the North.


According to the problem, these distances, 3t miles and 10t miles, 
are the legs of a right-angled triangle, whose hypotenuse is 1 mile.


So, we write the Pythagorean equation

    %283t%29%5E2 + %2810t%29%5E2 = 1,

    9t%5E2 + 100t%5E2 = 1,

    109t%5E2 = 1,

    t%5E2 = 1%2F109,

    t = sqrt%281%2F109%29 = 0.095782629 of an hour, or 0.095782629*60 = 5.746957711 minutes.


ANSWER.  The time is about 5.747 minutes.

This problem is nice and interesting.
You can solve it considering the events in reverse time, assuming that the kayaker and the orca
move in perpendicular directions back from the meeting point to their starting positions.

Then after t hours the distance between them will be 1 mile, and it is easy to find time 't'.

Thus we have simple, elegant and unexpectedly short solution.


My additional lessons on  Travel and Distance  problems  (section  3)  in this site are
    - Traveling on moving walkway
    - Two ships in parallel courses in ocean at restricted visibility
    - Three kids get home from school using a tandem bike and walk
    - One car passing another car

    - OVERVIEW of additional lessons on Travel and Distance - section 3

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.



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