Calculating Relative Velocity: Tips for Solving Tricky Problems

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The discussion centers on solving a relative velocity problem involving a man swimming in water and two ships. The original poster struggles with vector addition and understanding how to calculate the relative velocities correctly. Participants emphasize the importance of adding vectors component-wise and suggest that the shortest path occurs when the angle θ is maximized. The conversation also touches on the conditions for two ships to intercept each other, highlighting that they must have the same North-South position at the same time. Ultimately, the group works towards clarifying the conditions for interception and the necessary calculations for the ships' velocities.
  • #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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