Results 1 to 5 of 5

Math Help - ship problem

  1. #1
    Member
    Joined
    Apr 2006
    Posts
    91

    ship problem

    A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
    Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.

    Iv found the co-ordinates of Q relative to P at t=0
    ---->X=(10,10)km

    iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
    -----.V= (0,40) - V(30,0)
    ----- V= (-30,40)km/h

    but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    At t=0, they are both 10 kms from their intersection.

    Think Pythagoras. Let D=square of distance between ships.

    D(t)=(10-30t)^{2}+(10-40t)^{2}

    D'(t)=2(10-30t)(-30)+2(10=40t)(-40)

    D'(t)=5000t-1400

    5000t-1400=0

    t=\frac{7}{25}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2006
    Posts
    91
    Quote Originally Posted by galactus
    At t=0, they are both 10 kms from their intersection.

    Think Pythagoras. Let D=square of distance between ships.

    D(t)=(10-30t)^{2}+(10-40t)^{2}

    D'(t)=2(10-30t)(-30)+2(10=40t)(-40)

    D'(t)=5000t-1400

    5000t-1400=0

    t=\frac{7}{25}
    thankz...but i was wondering if i did the first part right that i showed?(shown below) because im not too sure if i was meant to include negative signs with any of the answers? can u please just check that for me thankz

    Iv found the co-ordinates of Q relative to P at t=0
    ---->X=(10,10)km

    iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
    -----.V= (0,40) - V(30,0)
    ----- V= (-30,40)km/h
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by dopi
    A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
    Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.
    Hello, dopi,

    with your problem you use automatically a coordinate system: The pos. x-axis points to East and the pos. y-axis points to North.

    At the time t = 0 the ship P is at P(-10, 0) and the ship Q is at Q(0, 10). The movement of both ships is described by a straight line. The speed is described by a vector:
    \overrightarrow{v_P}=(30, 0)
    \overrightarrow{v_Q}=(0, -40)

    Thus P is moving along the line:
    \overrightarrow{x_P}=(-10, 0)+t*(30, 0)
    and Q is moving along the line:
    \overrightarrow{x_Q}=(0, 10)+t*(0, -40)

    The distance between the ships is: \vec{d}=\overrightarrow{x_P}-\overrightarrow{x_Q}




    Quote Originally Posted by dopi
    Iv found the co-ordinates of Q relative to P at t=0
    ---->X=(10,10)km

    iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
    -----.V= (0,40) - V(30,0)
    ----- V= (-30,40)km/h
    I don't understand why you want to calculate the relative coordinates. It isn't necessary with your problem.

    Quote Originally Posted by dopi
    but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz
    This was already done by galactus.

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails ship problem-dopi-ship.gif  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,802
    Thanks
    692
    Hello, dopi!

    You seem to be using a coordinate system and vectors
    . . and I don't know what you plan to do with them.

    galactus gave you the solution . . . so why struggle with your approach?

    I'll make some diagrams to accompany his excellent solution.


    Quote Originally Posted by galactus

    At t=0, they are both 10 kms from their intersection.
    Code:
                            * Q
                            |
                            |
                            | 10
                            |
                            |
                            |
        P * - - - - - - - - o
                 10

    t hours later, P has moved 30t km east to point A
    . . and Q has moved 40t km south to point B.
    Code:
                            *
                            |
                            | 40t
                            |
                            o B
                            |
                            | 10-40t
          * - - - o - - - - *
             30t  A  10-30t
    And we want to minimize the distance \overline{AB}.


    Quote Originally Posted by galactus

    Think Pythagoras. Let D = square of distance between ships.

    10-30t)^{2} + (10-40t)^{2}" alt="D(t) \:= \10-30t)^{2} + (10-40t)^{2}" />

    D'(t) \:= \:2(10-30t)(-30) + 2(10=40t)(-40)

    D'(t) \:= \:5000t - 1400

    5000t -1400\:=\:0

    t \:= \:\frac{7}{25}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Position of ship.
    Posted in the Trigonometry Forum
    Replies: 0
    Last Post: November 16th 2011, 12:33 PM
  2. How much lead on that ship?
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 17th 2011, 03:44 AM
  3. Find the distance of this ship
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 21st 2010, 07:52 PM
  4. Ship Near A Rock
    Posted in the Geometry Forum
    Replies: 2
    Last Post: November 20th 2008, 05:09 AM
  5. A ship...
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: July 29th 2006, 01:58 AM

Search Tags


/mathhelpforum @mathhelpforum