Calculating Relative Velocity: Tips for Solving Tricky Problems

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Homework Help Overview

The discussion revolves around a problem involving relative velocity, specifically focusing on a man swimming in a river and the motion of two ships. The original poster expresses difficulty in understanding how to correctly add velocity vectors in the context of the problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the addition of vectors using the triangle law and question the correctness of their vector representations. There are inquiries about which vectors are being added and requests for clarification on the original poster's work. Some participants suggest considering component-wise addition of vectors.

Discussion Status

The discussion is active, with participants providing guidance on vector addition and encouraging the original poster to clarify their approach. There is an acknowledgment of the need for independence in problem-solving, and some participants are exploring different interpretations of the problem setup.

Contextual Notes

There are indications of confusion regarding the direction of motion for one of the ships, as well as the assumptions made about the velocities involved. The original poster has noted that they are unsure about the implications of the problem's constraints and the relationships between the velocities of the two ships.

  • #31
Nathanael said:
I doubt it. If this is what they mean then there are infinite different directions which ship B could travel. But to crash, there are exactly 2 angles which it can travel.

Yes, i see, well how do I go about calculating the direction is v=6
 
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  • #32
Aaron Curran said:
Yes, i see, well how do I go about calculating the direction is v=6
For which directions is the component of the velocity in the northern direction equal to 5 m/s?
 
  • #33
Nathanael said:
For which directions is the component of the velocity in the northern direction equal to 5 m/s?

is E33.56N one of them?
 
  • #34
Aaron Curran said:
is E33.56N one of them?
Yep. And the other?
 
  • #35
Nathanael said:
Yep. And the other?
I want to say 90-33.56 but I have a feeling that is wrong, other than that I really don't know
 
  • #36
Aaron Curran said:
E33.56N
By E33.56N you mean "33.56 degrees East of North," right?

Is it possible for it to travel west of north and crash? How many degrees west of north would it need to go?
 
  • #37
Nathanael said:
By E33.56N you mean "33.56 degrees East of North," right?

Is it possible for it to travel west of north and crash? How many degrees west of north would it need to go?

Yeah that's what I meant, I should have expressed it as N33.56E, I get where you're coming from but I still don't know how to go about calculating it
 
  • #38
Aaron Curran said:
Yeah that's what I meant, I should have expressed it as N33.56E, I get where you're coming from but I still don't know how to go about calculating it
Well, I don't know how else to explain it except to just tell you. It would be the same angle as you calculated, 33.56°, except west of north. Draw (or imagine) a picture to understand it better. The angle is measured against the y axis. No matter if you measure it to the left of the y-axis or to the right of the y axis, it still gives you the same y component (5 m/s)
 
  • #39
Nathanael said:
Well, I don't know how else to explain it except to just tell you. It would be the same angle as you calculated, 33.56°, except west of north. Draw (or imagine) a picture to understand it better. The angle is measured against the y axis. No matter if you measure it to the left of the y-axis or to the right of the y axis, it still gives you the same y component (5 m/s)

I thought about that but then I thought it was too simple, thanks so much for all your help! It seems so simple now that I know, makes me feel stupid for not seeing it earlier lol, thanks again!
 

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