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 motion

  1. May 14, 2009 #1
    1. The problem statement, all variables and given/known data
    The raceboats A and B leave at the same time the beach at point O with zero initial velocity. The path of boat A forms an angle of 60º with the beach and the path of boat B forms an angle of 45º.
    Both boats move with constant acceleration. Compute the time elapsed until the boats are a
    distance d apart from each other.

    Given are the acceleration of boat A & B, and the distance d.


    2. Relevant equations
    s = (1/2)at^2
    aB/A=aB-aA

    3. The attempt at a solution
    This question has really bothered me because I know its not very hard, but somehow I cant figure out how to start. I tried using the relative acceleration between the two boats in the formula, but that didnt give the right answer. If someone could just help me get started that would be great.
     

    Attached Files:

    • q2.jpg
      q2.jpg
      File size:
      3.9 KB
      Views:
      35
  2. jcsd
  3. May 14, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    There's nothing wrong with that approach. What did you get for the relative acceleration? (Be sure to treat the accelerations as vectors.)
     
  4. May 14, 2009 #3
    We're always told that we should solve the problems symbolically first. So when I first tried, I just used it as in the formula with aB/A=aB-aA to find B relative to A.

    Im not exactly sure what you mean when you say I need to treat the accelerations as vectors. If you mean that I should split up each relative acceleration into x and y components, then say that aB/A=sqrt(a(B/A)x2 + a(B/A)y2), that doesnt give the correct answer either.
     
  5. May 14, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    OK, but that doesn't make use of the given angles. Do the accelerations have the same magnitude?
     
  6. May 14, 2009 #5
    Hm. Okay, they dont have the same magnitude. And I think I understand why I need to consider them as vectors.

    So something along the lines of a(B/A)x = aBcos45 - aAcos60, the same in y direction, then find the magnitude of aB/A and use that?
     
  7. May 14, 2009 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, that's the idea. (But be careful with signs.)
     
  8. May 14, 2009 #7
    Thanks! And yes, must watch out for signs :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relative motion
  1. Relative motion (Replies: 4)

  2. Relative Motion (Replies: 2)

Loading...