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A motor boat leaves a ship and travels due north at 80 km/h.

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A motor boat leaves a ship and travels due north at 80 km/h. [#permalink] New post 20 Oct 2005, 08:29
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A motor boat leaves a ship and travels due north at 80 km/h. The ship proceeds 30 degrees south/east at 32 km/h. If the motorboat has enough fuel for 4 hours what is the maximum distance north that it can travel so that it can safely return to the ship?
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 [#permalink] New post 20 Oct 2005, 11:00
hmm. Reminds me of physics in school. This one is tough (at least it is for me). If I remember correctly it involves some trig. If I had some answers I would pick one and try it. I will leave this one to the more advanced people.
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 [#permalink] New post 20 Oct 2005, 21:27
This is my approach ....Not so sure if it's correct :wink:
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 [#permalink] New post 21 Oct 2005, 00:14
Laxie, I think it should be more like this but I can not find the path to get a result :?
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 [#permalink] New post 21 Oct 2005, 00:35
I'm not so sure, Antmavel. That's also what I'm confused coz I'm not familar with how a navigator locates the direction with those jargons :oops:
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 [#permalink] New post 21 Oct 2005, 00:37
I bet you're right! :wink: ...but anyway, you can come down to the solution using cosine law in a triangle ^_^
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 [#permalink] New post 21 Oct 2005, 00:45
I correct it.
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 [#permalink] New post 21 Oct 2005, 05:28
What I'd do is this. Assuming angle posted by Antmavel is correct, then all you need to do is to draw a perpendicular line from the ship (S) to the y axis, call it T. Let's call the original starting point O and the position of motorboat M. You'll see that the length of these segments always keep the same ratio. For simplistic sake let's don't use 80 and 32, instead lets use 5 versas 2 (80/32=5/2). It's easy to see that ST=1 if angel SOT is 30 degree and also MS= 5+sqrt(3). You can then derive that MT=sqrt(29+10sqrt(3)). This of course is hard to calculate, you only know that it's about sqrt(46) which is a little less than 7. Then you know the share of time the motorboat spend going north would be 5/(5+7). If the motor boat has 4 hours total, then it can at least go north for 1.67 hours before it must go find the ship.
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 [#permalink] New post 08 Nov 2005, 00:33
I got 112 miles north.

Took me a about 12 minutes to get thoughts straight, and 2 mins of calculations afterwards.

x = distance boat travelled north.
y = distance from northest point back to ship's final place after 4 hours

ship travelled 128 miles, that is 64 vertical and 64*root3 horizontal.


so, we get a rt angled triangle.
(x + 64)^2 + 64^2*3 = y^2
and x + y is distance boat can travel, which is 320 (80 * 4)

solving, we get x = 112.
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  [#permalink] 08 Nov 2005, 00:33
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A motor boat leaves a ship and travels due north at 80 km/h.

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