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anonymous
 3 years ago
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.?
anonymous
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you draw a picture yet. i find this really helps.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ChmE dw:1348980587824:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So let \(t\) be our variable, and it will represent hours.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what happen next? @wio

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let f(t) be the position of ship sailing south, and g(t) be the position ship sailing east.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm interested in the sol'n as well. How did you come up with f(0) = g(2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I thought it would be one hour

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You're right, it's f(0) = g(1)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ChmE, what is your solution?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ChmE that's ok.. @wio can i see your solution.. i really dont get the problem..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0My 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you show it step by step ?? @wio

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay bii17. What is the antiderivative of 5?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anti derivative??? @wio

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, what does \(\int5dx\) equal?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we havent discussed it in our class.. no idea

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmmm, okay, well let's say that the ship started as position 0, it is moving 5 knots, how many knothours has it traveled in t hours?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let a knothour be the distance you travel in an hour at the speed of a knot. Our ship will have traveled 5t

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since distance = speed * time.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.05(0) = 2(1) + c => c = 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So g(t) = 2t2 Are you following?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, since they are traveling perpendicular to each other... how can we find the distance between them with respect to time?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348982375609:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0See where I'm headed ChmE?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The reason is because we can give them any initial position we want, so long as they are consistent. Right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No problem, it's a bit complicated but I can try to explain it better if you'd like.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Basically I made the origin the point where they both reach at some point.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I made t=0 when the first boat is at that position.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the second boat doesn't get into that position until 1 hour later

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so since it is moving at 2 knots, it is 2knothours behind, giving us that 2 in 2t2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is the time at which the distance between the ships isn't changing.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So first we find the position of the ships with respect to the origin. Then.... well look at that triangle I drew up top!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what should we do next?
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