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Relative Motion of a ship

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data

    A ship is traveling at 8.0 m/s at an angle of 50 degrees north of east relative to the water. The water moves at 3 m/s due south relative to the shore.
    a- Determine the velocity of the ship relative to the shore.
    b- A person running along the deck of the ship moves at 2 m/s at 45 degrees south of west. What is his velocity relative to the shore?

    2. Relevant equations

    V= (Va cos angle a + Vb cos angle b) i-hat + (Va sin angle a + Vb sin angle b) j-hat
    3. The attempt at a solution

    part a-
    (8cos50 + 3 cos -90) i-hat + (8sin 50 + 3 sin -90) j-hat =

    5.14 m/s i-hat + -3.12 m/s j-hat

    I am confused about part b. Do I need to factor in the motion of the ship that the person is on? Or do I combine his velocity with that of the water since it is measured relative to the shore?
     
  2. jcsd
  3. Sep 18, 2008 #2
    Well, for part a you wrote

    V (ship w/res to water) + V (water w/res to shore) = V ship w/res to shore

    Assuming the info for part b is a velocity w/res to the ship, where in the equation do you think you should add the new vector?
     
  4. Sep 18, 2008 #3
    Since the man is running in a different direction on the ship than the ship is moving, I thought I would have to create a new vector for his direction.
     
  5. Sep 19, 2008 #4
    That is correct. You need to write V (person w/res to ship).

    Now, based on how you did part a, how would you use this additional vector to find the person's velocity w/res to the shore?
     
  6. Sep 19, 2008 #5
    I created a vector for the guy, which was (2 m/s cos 225) i + (2 m/s sin 225) j and added it to the ship to shore vector, giving me 3.73 m/s i + 1.72 m/s j.

    Does that seem correct?
     
  7. Sep 19, 2008 #6
    Yes, it does.

    What you've done is

    V (person w/res ship) +V (ship w/res to water) + V (water w/res to shore) = V (person w/res to shore)
     
  8. Sep 19, 2008 #7
    I see. Thanks a lot for your help.
     
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