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Relative motion of ships

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Two ships, A and B, leave port at the same time. Ship A travels northwest at 22 knots and ship B travels at 29 knots in a direction 40° west of south. (1 knot = 1 nautical mile per hour; see Appendix D.)
    (a) What is the magnitude the velocity of ship A relative to B?
    _____knots

    (b) What is the direction of the velocity of ship A relative to B?
    _____° east of north

    (c) After what time will the ships be 125 nautical miles apart?
    ____h

    (d) What will be the bearing of B (the direction of B's position) relative to A at that time?
    _____° west of south



    2. Relevant equations
    Velocity of A=Velocity of B+Velocity of A with respect to be or

    V_a=V_b+V_ab



    3. The attempt at a solution
    I've tried to use the initial speeds:

    22=V_ab+29
    V_ab=-7
    so the magnitude would be 7 which is an incorrect answer

    I've also tried using the vector equations:

    V_a=(-15.55i+15.55j) or (22cos135i+22sin135j)( I used 45 degrees+90 to get the sign with respect to a y axis that has positive north, and x axis with positive east)
    V_b=(-22.22i+18.64j) or (29sin310i+29cos310j)(I used 40+270 to get the correct sign with respect to above names axis)

    Splitting the two equations I get:

    -15.55=V_abx-22.22
    and
    15.55=V_aby+18.64

    therefore:

    V_abx=-6.67
    and
    V_aby=-3.09

    so V_ab is (-6.67i-3.09j)and the magnitude of this vector is:
    7.35 however I've used all but one try left so I wanted to make sure this was the correct answer. Is this the proper way to solve this problem?
     
  2. jcsd
  3. Oct 24, 2008 #2
    Your first answer was incorrect because you have to subtract the vectors, not just the sizes of the vectors. Also, in your second answer you have V_b pointing in the wrong direction.
     
  4. Oct 28, 2008 #3
    Try the following procedure to get the answers to this problem:
    • Start by assuming that angles are measured positive counterclockwise from the conventional x axis, with north at 90 degrees and west at 180 degrees. You can adjust your angular answers to conform to the requirements of the problem.
    • Convert the polar coordinates of the two velocities to Cartesian. This gives you the velocity vectors vA and vB for boats A and B.
    • The velocity vector of boat A relative to boat B is vX = vA-vB.
    • Convert vX to polar coordinates. The polar coordinates are the magnitude and direction of the velocity of ship A relative to ship B ( The answers to parts a and b )
    • The product of the velocity magnitude and time is the distance between the boats. Set this product equal to 125 and solve the equation for time. This gives the answer for part c.
    • Add 180 degrees to the answer to part b to get the answer to part c.
     
  5. Sep 27, 2010 #4
    I have a similar problem to this, but the velocity of boat A is 19 knots, and the velocity of boat B is 28 knots. I found the direction of ship A relative to ship B to be 82.548 degrees east of north, but for some reason it says I have the wrong answer. Any advice?
     
  6. Sep 29, 2010 #5
    (b) 82.5 degrees is the direction counterclockwise from the x axis. That is about 7.5 degrees east of north.
     
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