anonymous
  • anonymous
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
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
So let \(t\) be our variable, and it will represent hours.
anonymous
  • anonymous
what happen next? @wio
anonymous
  • anonymous
Let f(t) be the position of ship sailing south, and g(t) be the position ship sailing east.
anonymous
  • anonymous
f(0) = g(2)
anonymous
  • anonymous
then?
anonymous
  • anonymous
I'm interested in the sol'n as well. How did you come up with f(0) = g(2)
anonymous
  • anonymous
I thought it would be one hour
anonymous
  • anonymous
You're right, it's f(0) = g(1)
anonymous
  • anonymous
@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
  • anonymous
ChmE, what is your solution?
anonymous
  • anonymous
@ChmE that's ok.. @wio can i see your solution.. i really dont get the problem..
anonymous
  • anonymous
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
anonymous
  • anonymous
can you show it step by step ?? @wio
anonymous
  • anonymous
Okay bii17. What is the anti-derivative of 5?
anonymous
  • anonymous
anti derivative??? @wio
anonymous
  • anonymous
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
anonymous
  • anonymous
Yes, what does \(\int5dx\) equal?
anonymous
  • anonymous
we havent discussed it in our class.. no idea
anonymous
  • anonymous
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?
anonymous
  • anonymous
Let a knot-hour be the distance you travel in an hour at the speed of a knot. Our ship will have traveled 5t
anonymous
  • anonymous
Since distance = speed * time.
anonymous
  • anonymous
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.
anonymous
  • anonymous
5(0) = 2(1) + c => c = -2
anonymous
  • anonymous
So g(t) = 2t-2 Are you following?
anonymous
  • anonymous
@wio yes..
anonymous
  • anonymous
Now, since they are traveling perpendicular to each other... how can we find the distance between them with respect to time?
anonymous
  • anonymous
|dw:1348982375609:dw|
anonymous
  • anonymous
See where I'm headed ChmE?
anonymous
  • anonymous
The reason is because we can give them any initial position we want, so long as they are consistent. Right?
anonymous
  • anonymous
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.
anonymous
  • anonymous
No problem, it's a bit complicated but I can try to explain it better if you'd like.
anonymous
  • anonymous
Basically I made the origin the point where they both reach at some point.
anonymous
  • anonymous
I made t=0 when the first boat is at that position.
anonymous
  • anonymous
so the second boat doesn't get into that position until 1 hour later
anonymous
  • anonymous
so since it is moving at 2 knots, it is 2knot-hours behind, giving us that -2 in 2t-2
anonymous
  • anonymous
No!
anonymous
  • anonymous
It is the time at which the distance between the ships isn't changing.
anonymous
  • anonymous
So first we find the position of the ships with respect to the origin. Then.... well look at that triangle I drew up top!
anonymous
  • anonymous
what should we do next?

Looking for something else?

Not the answer you are looking for? Search for more explanations.