• Support PF! Buy your school textbooks, materials and every day products Here!

Relative motion

  • Thread starter perdita_x
  • Start date
  • #1
5
0

Homework Statement


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.


Homework Equations


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

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.
 

Attachments

Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
I tried using the relative acceleration between the two boats in the formula, but that didnt give the right answer.
There's nothing wrong with that approach. What did you get for the relative acceleration? (Be sure to treat the accelerations as vectors.)
 
  • #3
5
0
There's nothing wrong with that approach. What did you get for the relative acceleration? (Be sure to treat the accelerations as vectors.)
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.
 
  • #4
Doc Al
Mentor
44,882
1,129
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.
OK, but that doesn't make use of the given angles. Do the accelerations have the same magnitude?
 
  • #5
5
0
OK, but that doesn't make use of the given angles. Do the accelerations have the same magnitude?
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?
 
  • #6
Doc Al
Mentor
44,882
1,129
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?
Yes, that's the idea. (But be careful with signs.)
 
  • #7
5
0
Thanks! And yes, must watch out for signs :D
 

Related Threads for: Relative motion

  • Last Post
Replies
3
Views
832
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
591
Top