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May 15th, 2010, 09:48 PM   #1
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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?
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May 15th, 2010, 10:06 PM   #2
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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?
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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?
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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 be the distance between the two ships.





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



so the ships are closest together at 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).
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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 be the distance between the two ships.





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



so the ships are closest together at 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.
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May 16th, 2010, 02:56 PM   #6
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That geometrical approach works. What do you get if you try it and do the diagram you suggested?
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let t be the time, in hours, after 4:00 pm if the position of the boat at t = 0 corresponds to the origin, then the position of the boat heading south at time t is
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