Calculating Relative Velocity: Tips for Solving Tricky Problems

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SUMMARY

The forum discussion centers on calculating relative velocity in a scenario involving two ships, A and B, with specific velocities and directions. The participants discuss the correct approach to vector addition using the triangle law and emphasize the importance of understanding component-wise addition of vectors. Key insights include the necessity of having the same direction for vector addition and the use of calculus to determine maximum angles for optimal paths. The final conclusion is that for ship B to intercept ship A, its northern component of velocity must equal 5 m/s.

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  • Understanding of vector addition and components
  • Familiarity with the triangle law of vector addition
  • Basic knowledge of calculus for optimizing angles
  • Concept of relative velocity in physics
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  • Study vector addition techniques in physics
  • Learn about the triangle law of vector addition in detail
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Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators looking for effective methods to teach relative velocity concepts.

  • #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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