The discussion focuses on 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 from the x-axis, placing the vector in the second quadrant. The correct interpretation of the angle leads to the conclusion that the x-component is negative and the y-component is positive, aligning with the properties of trigonometric functions.
PREREQUISITESStudents beginning their studies in physics, particularly those learning about vectors and trigonometry, as well as educators looking for clarification on teaching these concepts.
Welcome to PhysicsForums.PleaseAnswerOnegai said:
Hi! What would the correct sketch look like? While I was sketching it I knew something was fishy and I just had to ask hahahaberkeman said:Welcome to PhysicsForums.
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...
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?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! :)berkeman said:
PleaseAnswerOnegai said:
It's true that ##\sin(\pi-x)=\sin x##, but ##\cos(\pi-x)=-\cos x##, not ##\cos x##.PleaseAnswerOnegai said: