Difficult relative velocity question involving boats

Click For Summary
The discussion revolves around calculating the speed of a passenger on Boat 2 relative to the shore, considering multiple frames of reference. The participants emphasize the importance of understanding relative velocities and using vector addition to solve the problem. The equation V(13) = V(12) + V(23) is highlighted as a key formula for relating the velocities of the boats and the passenger. Drawing vectors is recommended to visualize the problem and aid in calculations. The conversation encourages a step-by-step approach to simplify the complexity of relative motion.
carrbaseball
Messages
17
Reaction score
0
Here is the question...Two boats are heading away from shore. Boat 1 heads due north at a speed of 3 m/s relative to the shore. Relative to Boat 1, Boat 2 is moving 40° north of east at a speed of 1.7 m/s. A passenger on Boat 2 walks due east across the deck at a speed of 1.1 m/s relative to Boat 2. What is the speed of the passenger relative to the shore?

This is confusing me with the numerous relativities.
I am using the equation V(13)= V(12)+V(23)
where V(13) is the velocity of object 1 relative to object 3.
Im not sure how to get this problem started.
?
 
Physics news on Phys.org
Draw your vectors.

\vec{v}_{P, S} = \vec{v}_{P, B2} + \vec{v}_{B2, S}
\vec{v}_{B2, S} = \vec{v}_{B2, B1} + \vec{v}_{B1, S}
\vec{v}_{P, S} = \vec{v}_{P, B2} + \vec{v}_{B2, B1} + \vec{v}_{B1, S}

P - passenger ; S - shore; B1, B2 - boats 1 and 2.
 
Ok, ihave drawn up my vectors and I am trying to plug in my values. I hate to sound stupid, but I'm lost in this process.
 
What's important here is to understand the concept of relative frames. There is always a stationary observer who observes his reference to have zero velocity, if you will, while the other two are either moving towards or away at some rate. You can't just use equations without knowing where they come from.

So, picture yourself at some intersection where you are pointing north. Some guy on a motorcycle goes away from you. Now there is the guy on the motorcycle who sees you moving south, and now a runner ahead of him crossing the street to some store. Okay, now take a break and figure out what is happening before you try to digest the last piece of information. You see a motorcycle moving away from you, and another man running really fast away from him because he is about to get creamed by traffic. So now, put the last piece in. I'll let you do this yourself and see if you get the idea.

Just take it slow, and it isn't all that tough. Draw a huge picture so that you can use vector lengths to aid your thought.
 
You just need to keep on "expanding" the relative formula until you can calculate all the required vectors. When you run into a vector you do not have in terms of given information set up a new formula for its calculation the same way. In the end you need to work your way backwards towards the first formula once you have all the vectors in terms of known vectors.

Show us how you progress with the problem.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
25
Views
3K
Replies
31
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K