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bii17

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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  1. ChmE
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    Did you draw a picture yet. i find this really helps.

    • one year ago
  2. bii17
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    @ChmE |dw:1348980587824:dw|

    • one year ago
  3. bii17
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    @ChmE ??

    • one year ago
  4. wio
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    So let \(t\) be our variable, and it will represent hours.

    • one year ago
  5. bii17
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    what happen next? @wio

    • one year ago
  6. wio
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    Let f(t) be the position of ship sailing south, and g(t) be the position ship sailing east.

    • one year ago
  7. wio
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    f(0) = g(2)

    • one year ago
  8. bii17
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    then?

    • one year ago
  9. ChmE
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    I'm interested in the sol'n as well. How did you come up with f(0) = g(2)

    • one year ago
  10. ChmE
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    I thought it would be one hour

    • one year ago
  11. wio
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    You're right, it's f(0) = g(1)

    • one year ago
  12. ChmE
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    @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
  13. wio
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    ChmE, what is your solution?

    • one year ago
  14. bii17
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    @ChmE that's ok.. @wio can i see your solution.. i really dont get the problem..

    • one year ago
  15. wio
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    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
  16. bii17
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    can you show it step by step ?? @wio

    • one year ago
  17. wio
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    Okay bii17. What is the anti-derivative of 5?

    • one year ago
  18. bii17
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    anti derivative??? @wio

    • one year ago
  19. ChmE
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    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
  20. wio
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    Yes, what does \(\int5dx\) equal?

    • one year ago
  21. bii17
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    we havent discussed it in our class.. no idea

    • one year ago
  22. wio
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    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?

    • one year ago
  23. wio
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    Let a knot-hour be the distance you travel in an hour at the speed of a knot. Our ship will have traveled 5t

    • one year ago
  24. wio
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    Since distance = speed * time.

    • one year ago
  25. wio
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    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
  26. wio
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    5(0) = 2(1) + c => c = -2

    • one year ago
  27. wio
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    So g(t) = 2t-2 Are you following?

    • one year ago
  28. bii17
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    @wio yes..

    • one year ago
  29. wio
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    Now, since they are traveling perpendicular to each other... how can we find the distance between them with respect to time?

    • one year ago
  30. wio
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    |dw:1348982375609:dw|

    • one year ago
  31. wio
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    See where I'm headed ChmE?

    • one year ago
  32. wio
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    The reason is because we can give them any initial position we want, so long as they are consistent. Right?

    • one year ago
  33. wio
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    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
  34. wio
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    No problem, it's a bit complicated but I can try to explain it better if you'd like.

    • one year ago
  35. wio
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    Basically I made the origin the point where they both reach at some point.

    • one year ago
  36. wio
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    I made t=0 when the first boat is at that position.

    • one year ago
  37. wio
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    so the second boat doesn't get into that position until 1 hour later

    • one year ago
  38. wio
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    so since it is moving at 2 knots, it is 2knot-hours behind, giving us that -2 in 2t-2

    • one year ago
  39. wio
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    No!

    • one year ago
  40. wio
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    It is the time at which the distance between the ships isn't changing.

    • one year ago
  41. wio
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    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
  42. roe_gael
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    what should we do next?

    • one year ago
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