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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.?
 one year ago
 one year 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.?
 one year ago
 one year ago

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ChmEBest ResponseYou've already chosen the best response.0
Did you draw a picture yet. i find this really helps.
 one year ago

bii17Best ResponseYou've already chosen the best response.0
@ChmE dw:1348980587824:dw
 one year ago

wioBest ResponseYou've already chosen the best response.0
So let \(t\) be our variable, and it will represent hours.
 one year ago

wioBest ResponseYou've already chosen the best response.0
Let f(t) be the position of ship sailing south, and g(t) be the position ship sailing east.
 one year ago

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

ChmEBest ResponseYou've already chosen the best response.0
I thought it would be one hour
 one year ago

wioBest ResponseYou've already chosen the best response.0
You're right, it's f(0) = g(1)
 one year ago

ChmEBest 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.
 one year ago

wioBest ResponseYou've already chosen the best response.0
ChmE, what is your solution?
 one year ago

bii17Best ResponseYou've already chosen the best response.0
@ChmE that's ok.. @wio can i see your solution.. i really dont get the problem..
 one year ago

wioBest ResponseYou've already chosen the best response.0
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
 one year ago

bii17Best ResponseYou've already chosen the best response.0
can you show it step by step ?? @wio
 one year ago

wioBest ResponseYou've already chosen the best response.0
Okay bii17. What is the antiderivative of 5?
 one year ago

ChmEBest ResponseYou've already chosen the best response.0
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
 one year ago

wioBest ResponseYou've already chosen the best response.0
Yes, what does \(\int5dx\) equal?
 one year ago

bii17Best ResponseYou've already chosen the best response.0
we havent discussed it in our class.. no idea
 one year ago

wioBest ResponseYou've already chosen the best response.0
Hmmmm, 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?
 one year ago

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

wioBest ResponseYou've already chosen the best response.0
Since distance = speed * time.
 one year ago

wioBest ResponseYou've already chosen the best response.0
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.
 one year ago

wioBest ResponseYou've already chosen the best response.0
So g(t) = 2t2 Are you following?
 one year ago

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

wioBest ResponseYou've already chosen the best response.0
See where I'm headed ChmE?
 one year ago

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

wioBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

wioBest ResponseYou've already chosen the best response.0
Basically I made the origin the point where they both reach at some point.
 one year ago

wioBest ResponseYou've already chosen the best response.0
I made t=0 when the first boat is at that position.
 one year ago

wioBest ResponseYou've already chosen the best response.0
so the second boat doesn't get into that position until 1 hour later
 one year ago

wioBest ResponseYou've already chosen the best response.0
so since it is moving at 2 knots, it is 2knothours behind, giving us that 2 in 2t2
 one year ago

wioBest ResponseYou've already chosen the best response.0
It is the time at which the distance between the ships isn't changing.
 one year ago

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

roe_gaelBest ResponseYou've already chosen the best response.0
what should we do next?
 one year ago
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