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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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- Thread starter PleaseAnswerOnegai
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In summary, the problem is asking to determine the x- and y-components of a velocity vector with a magnitude of 100.0 m/s and an angle of 160° with the positive x-axis. The correct sketch for this would be on the second quadrant. The confusion may have arisen from the use of "positive x-axis" in the problem. The angle should be measured counterclockwise from the horizontal x-axis, making the x-coordinate negative. The correct equations to use are ##\sin(\pi-x)=\sin x## and ##\cos(\pi-x)=-\cos x##.

- #1

- 13

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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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- #2

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Welcome to PhysicsForums.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

View attachment 269386

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...

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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...

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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:

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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:

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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! :)

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Is the angle ##20°## or ##160°##? Is the x-coordinate positive or negative?

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It's true that ##\sin(\pi-x)=\sin x##, but ##\cos(\pi-x)=-\cos x##, not ##\cos x##.PleaseAnswerOnegai said:

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