1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relative Velocity; Help needed please

  1. Feb 10, 2015 #1
    2. (a) is the question I'm stuck with
    upload_2015-2-10_18-8-14.png


    2. Relevant equations
    I know that 3m/s is the Velocity of the man relative to the water and the Velocity of the man is 8m/s and is represented by the vector pointing in the direction theta downstream but I just can't get my head around this question, probably something simple that's alluding me! I have part (b) completed and it is meant to be the more difficult of the two.


    3. The attempt at a solution
    I've attempted to add the vectors via triangle law but I can't seem to get it right my vector triangle always ends up having the hypotenuse as one of the shorter sides, any help would be appreciated!
     
  2. jcsd
  3. Feb 10, 2015 #2

    Nathanael

    User Avatar
    Homework Helper

    Please explain

    Which vectors are you attempting to add? Can you please show your work?
     
  4. Feb 10, 2015 #3
    I just realised what I was doing wrong, but I'm still not finished the question, here's what I have so far

    image.jpg
     
  5. Feb 10, 2015 #4
    Here you go! :)
     

    Attached Files:

  6. Feb 10, 2015 #5

    Nathanael

    User Avatar
    Homework Helper

    You can only add the magnitude of vectors if they are in the same direction!! Add them component by component.
     
  7. Feb 10, 2015 #6
    Thank you!
     
  8. Feb 10, 2015 #7
    I'm a bit curious, how did you do b)? :)
     
  9. Feb 10, 2015 #8

    Nathanael

    User Avatar
    Homework Helper

    It is good that you want to help, but you are not supposed to give full solutions. The point is for the OP to find the solution so that they can better understand.
     
  10. Feb 10, 2015 #9
    I'm sorry Sir!
     
  11. Feb 10, 2015 #10
    I prefer to be shown a solution and work it backwards to see how one arrives at that solution, I learn better that way!
     
  12. Feb 10, 2015 #11

    Nathanael

    User Avatar
    Homework Helper

    When problems get harder, this method will break down. As a problem gets more complex, it gets increasingly difficult to understand other people's work. People have their own perspectives and their own ideas for solving things (there is very often many different ways to solve a problem) it's not always easy to understand what they were doing (even if you've solved the problem yourself!)
    I'm glad you got something out of it, but if I may make a suggestion; I suggest trying to be as independent as possible when solving problems.
     
  13. Feb 10, 2015 #12
    turns out I didn't do part b correct, any pointers?
     
  14. Feb 10, 2015 #13

    Nathanael

    User Avatar
    Homework Helper

    The shortest path is when θ is largest.
    (Ideally we would want θ=90°, but this can't happen because even if the person swam directly against the current, they would still be moving downstream.)
    Do you know how to find the maximum θ using calculus?
     
  15. Feb 10, 2015 #14
    I mean the actual second part of the question about the ships moving in subsequent motion this is what I have so far

    ImageUploadedByPhysics Forums1423598011.976621.jpg
     
  16. Feb 10, 2015 #15

    Nathanael

    User Avatar
    Homework Helper

    Oh! Sorry, I'm a little blind o0)

    I assume by "intercept" they mean something like "crash" :-p

    Sorry I can't understand your thought process based off the image (I can't even read it very well). It would be helpful if you gave a brief explanation of your approach.

    [[EDIT: Deleted my method, I didn't think it through; it is simpler than the method I proposed]]

    What is your method?
     
    Last edited: Feb 10, 2015
  17. Feb 10, 2015 #16
    It gives the velocity the ship is travelling and the direction in which it is travelling, so I broke this into i and j components, if given the velocity then the velocity vector will always be the hypotenuse of the triangle so using that and the angle by sin30 = y/10 then 10sin30 = y, which is 5 so 5j is one vector component of the ships movement and the other is 5root3 i and then I took the velocity of A relative to B which is the Velocity of A - Velocity of B so that was (5root3 - v)i + 5j and usually the find the magnitude of the relative velocity it would the square root of the i component squared + the j component squared but then I get an expression in V which is messy and I have no idea where to go from there.

    also i is the horiztonal and j is the verticle component
     
  18. Feb 10, 2015 #17

    Nathanael

    User Avatar
    Homework Helper

    If the two boats crash, then they will have the same position at the same time.
    Right now, the boats are at the same position along the North-South axis.
    In the future, (when they crash) they will also have the same position along the North-South axis.
    So what do you know about their North-South velocities?
     
  19. Feb 10, 2015 #18
    ahhh, the N/S velocity of A is 5j and the N/S velocity of B is 0j as it travels along the west-east axis
     
  20. Feb 10, 2015 #19

    Nathanael

    User Avatar
    Homework Helper

    But if B travels along the E/W axis, then will they ever crash?

    Remember, the problem does not give a direction for boat B, you can make it whatever direction you want!
     
  21. Feb 10, 2015 #20
    as B travels along the E/W from east to west, A approaches from the south west (travelling E30degreesN and the problem is we have to find the minimum value of the i component of B's velocity for the two to intercept, and in the beginning B is 10km due east of A
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Relative Velocity; Help needed please
Loading...