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 May 15th, 2010, 09:48 PM #1 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 shortest distance 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? Attempt : the velocity A relative to B is 25. And in one hour, A travels 15 km to reach the dock. That's all I can decipher from the question. What else do I have to look into?
 May 15th, 2010, 10:06 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,538 Thanks: 920 Math Focus: Elementary mathematics and beyond Re: shortest distance Let the dock be the origin. Let the position of the ship heading east at 15 km/h be x. Let the position of the ship heading south at 20 km/h be y. x = 15 - 15t and y = 20t, where t is the time in hours. Can you take it from there?
May 15th, 2010, 10:54 PM   #3
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Re: shortest distance

Quote:
 Originally Posted by greg1313 Let the dock be the origin. Let the position of the ship heading east at 15 km/h be x. Let the position of the ship heading south at 20 km/h be y. x = 15 - 15t and y = 20t, where t is the time in hours. Can you take it from there?
Thanks Greg, not quite . The usual way I do questions like this is I combine the relative displacement and relative velocity diagrams, then find the shortest distance. Can that be applied to this question?

I thought the dock should be the point where both ships are heading towards?

May 15th, 2010, 11:04 PM   #4
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Re: shortest distance

"A boat leaves a dock at 3:00 P.M. and travels due south . . ." implies the ship is headed away from the dock.

"Another boat has been heading due east at 15 km/h and reaches the same dock . . ." implies the ship is headed towards the dock.

I'll finish up using calculus:

Let $d$ be the distance between the two ships.

$d^2\,=\,(15\,-\,15t)^2\,+\,(20t)^2$

$d^2\,=\,625t^2\,-\,450t\,+\,225$

Take the derivative, set it equal to 0 to minimize:

$1250t\,-\,450\,=\,0$

$t\,=\,\frac{450}{1250}\,=\,0.36$ so the ships are closest together at $21.6$ minutes after 3:00.

Quote:
 Originally Posted by mikeportney The usual way I do questions like this is I combine the relative displacement and relative velocity diagrams, then find the shortest distance. Can that be applied to this question?
Don't know but I imagine so (oops I kinda missed that part).

May 16th, 2010, 01:00 AM   #5
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Re: shortest distance

Quote:
Originally Posted by greg1313
"A boat leaves a dock at 3:00 P.M. and travels due south . . ." implies the ship is headed away from the dock.

"Another boat has been heading due east at 15 km/h and reaches the same dock . . ." implies the ship is headed towards the dock.

I'll finish up using calculus:

Let $d$ be the distance between the two ships.

$d^2\,=\,(15\,-\,15t)^2\,+\,(20t)^2$

$d^2\,=\,625t^2\,-\,450t\,+\,225$

Take the derivative, set it equal to 0 to minimize:

$1250t\,-\,450\,=\,0$

$t\,=\,\frac{450}{1250}\,=\,0.36$ so the ships are closest together at $21.6$ minutes after 3:00.

Quote:
 Originally Posted by mikeportney The usual way I do questions like this is I combine the relative displacement and relative velocity diagrams, then find the shortest distance. Can that be applied to this question?
Don't know but I imagine so (oops I kinda missed that part).
Thanks a lot Greg, though I am still interested to know a geometrical approach.

 May 16th, 2010, 02:56 PM #6 Global Moderator   Joined: Dec 2006 Posts: 17,698 Thanks: 1353 That geometrical approach works. What do you get if you try it and do the diagram you suggested?

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