One ship is sailing south at a rate of 5 knots, and another is sailing east at a rate of 10 knots. At 2 P.M. the second ship was at the place occupied by the first ship one hour before. at what time was the distance between the ships not changing.?

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One ship is sailing south at a rate of 5 knots, and another is sailing east at a rate of 10 knots. At 2 P.M. the second ship was at the place occupied by the first ship one hour before. at what time was the distance between the ships not changing.?

Mathematics
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Did you draw a picture yet. i find this really helps.
@ChmE |dw:1348980587824:dw|

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So let \(t\) be our variable, and it will represent hours.
what happen next? @wio
Let f(t) be the position of ship sailing south, and g(t) be the position ship sailing east.
f(0) = g(2)
then?
I'm interested in the sol'n as well. How did you come up with f(0) = g(2)
I thought it would be one hour
You're right, it's f(0) = g(1)
@bii17 sry for not responding earlier. I wasn't confident in my sol'n and didn't want to possibly steer you down a wrong path at the risk of confusing you.
ChmE, what is your solution?
@ChmE that's ok.. @wio can i see your solution.. i really dont get the problem..
My solution is to find the position functions, then use the Pythagorean theorem on them to get a 'change in distance' function. Then take the derivative. Then find when that is 0
can you show it step by step ?? @wio
Okay bii17. What is the anti-derivative of 5?
anti derivative??? @wio
I don't understand what the question is asking. Is it looking for at what time is the distance the same as the initial change in distance? Thats how I read it
Yes, what does \(\int5dx\) equal?
we havent discussed it in our class.. no idea
Hmmmm, okay, well let's say that the ship started as position 0, it is moving 5 knots, how many knot-hours has it traveled in t hours?
Let a knot-hour be the distance you travel in an hour at the speed of a knot. Our ship will have traveled 5t
Since distance = speed * time.
So f(t) = 5t. Now g(t) = 2t + c. Where c is the initial position. We don't know it yet, but we can use f(0) = g(1) to find out c.
5(0) = 2(1) + c => c = -2
So g(t) = 2t-2 Are you following?
@wio yes..
Now, since they are traveling perpendicular to each other... how can we find the distance between them with respect to time?
|dw:1348982375609:dw|
See where I'm headed ChmE?
The reason is because we can give them any initial position we want, so long as they are consistent. Right?
I made it so the initial position would be the point that they both eventually reach. Then I made it so that at time 0, the first boat was at that position.
No problem, it's a bit complicated but I can try to explain it better if you'd like.
Basically I made the origin the point where they both reach at some point.
I made t=0 when the first boat is at that position.
so the second boat doesn't get into that position until 1 hour later
so since it is moving at 2 knots, it is 2knot-hours behind, giving us that -2 in 2t-2
No!
It is the time at which the distance between the ships isn't changing.
So first we find the position of the ships with respect to the origin. Then.... well look at that triangle I drew up top!
what should we do next?

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