Finding the components of this velocity vector

AI Thread Summary
The discussion revolves around determining the x- and y-components of a velocity vector with a magnitude of 100.0 m/s at an angle of 160° from the positive x-axis. Participants clarify that angles are measured counterclockwise, placing the vector in the second quadrant, contrary to initial assumptions of it being in the first quadrant. There is confusion about the angle's implications for the x and y coordinates, particularly regarding their signs. The conversation emphasizes the importance of correctly interpreting the angle's position to accurately calculate the vector's components.
PleaseAnswerOnegai
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Homework Statement
A velocity vector has magnitude 100.0 m/s and make an angle of 160° with the positive x-axis. Determine the x- and y-components of the vector.
Relevant Equations
Trig functions
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PleaseAnswerOnegai said:
Homework Statement:: A velocity vector has magnitude 100.0 m/s and make an angle of 160° with the positive x-axis. Determine the x- and y-components of the vector.
Relevant Equations:: Trig functions

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Welcome to PhysicsForums. :smile:

That sketch looks wrong. What line is 180 degrees from the x-axis? So 160 degrees would be pretty close to that line, not vertical like that...
 
Just started Physics and I wasn't sure if my answer and solution was right. I attempted to project it on the cartesian plane based on my understanding. It said "an angle of 160 with the positive x-axis." What I understood from this is that angles of the triangle that makes contact to the x-axis equal to 160 so I assumed the angles would be 70 and 90.
 
berkeman said:
Welcome to PhysicsForums. :smile:

That sketch looks wrong. What line is 180 degrees from the x-axis? So 160 degrees would be pretty close to that line, not vertical like that...
Hi! What would the correct sketch look like? While I was sketching it I knew something was fishy and I just had to ask hahaha
 
PleaseAnswerOnegai said:
Just started Physics and I wasn't sure if my answer and solution was right. I attempted to project it on the cartesian plane based on my understanding. It said "an angle of 160 with the positive x-axis." What I understood from this is that angles of the triangle that makes contact to the x-axis equal to 160 so I assumed the angles would be 70 and 90.
No, usually the angle is measured counterclockwise from the horizontal x-axis. So an angle of 90 degrees is straight up (along the y-axis), and an angle of 180 degrees is pointing left along the -x-axis. So where would 160 degrees counterclockwise from the x-axis be? And what would the x- and y- components of that vector be then?
 
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berkeman said:
So it would then be on the 2nd Quadrant, correct? What confused me was the usage of "positive x-axis" so I wrongfully assumed that it would be strictly in the 1st Quadrant (where x values are positive). I will solve it and send my answer! :)
 
PleaseAnswerOnegai said:
So it would then be on the 2nd Quadrant, correct? What confused me was the usage of "positive x-axis" so I wrongfully assumed that it would be strictly in the 1st Quadrant (where x values are positive). I will solve it and send my answer! :)
physics_question.JPG
 
Is the angle ##20°## or ##160°##? Is the x-coordinate positive or negative?
 
  • #10
PleaseAnswerOnegai said:
It's true that ##\sin(\pi-x)=\sin x##, but ##\cos(\pi-x)=-\cos x##, not ##\cos x##.
 
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