- #1
PleaseAnswerOnegai
- 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
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
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: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.
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: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! :)
It's true that ##\sin(\pi-x)=\sin x##, but ##\cos(\pi-x)=-\cos x##, not ##\cos x##.PleaseAnswerOnegai said:
A velocity vector is a mathematical representation of an object's speed and direction of motion. It includes both magnitude (speed) and direction, and is typically represented graphically as an arrow.
To find the components of a velocity vector, you will need to know the magnitude (speed) of the vector and the angle at which it is pointing. You can then use trigonometric functions to calculate the horizontal and vertical components of the vector.
Finding the components of a velocity vector allows you to break down the overall motion of an object into its individual horizontal and vertical components. This can be useful in analyzing and predicting the motion of objects in different directions.
Finding the components of a velocity vector is important in many fields, including physics, engineering, and navigation. It is used in projectile motion calculations, aircraft and spacecraft design, and determining the trajectory of objects in motion.
Yes, there are several tools and formulas that can assist in finding the components of a velocity vector. These include trigonometric functions, vector addition and subtraction, and the Pythagorean theorem. There are also many online calculators and software programs available for calculating and graphing velocity vectors.