2D motion finding displacement direction and velocity

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Homework Help Overview

The problem involves analyzing the motion of a person walking in three different directions, requiring the calculation of displacement and velocity, including their magnitudes and directions. The context is 2D motion, focusing on vector components and trigonometric relationships.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the breakdown of displacement into x and y components, with attempts to apply trigonometric functions to find these components. Questions arise regarding the interpretation of vector notation and the application of the Pythagorean theorem for magnitude calculation.

Discussion Status

Some participants have made progress in calculating the magnitude of displacement using the Pythagorean theorem. However, there remains uncertainty about how to determine the direction of displacement and velocity, with multiple suggestions being explored regarding the use of inverse tangent functions.

Contextual Notes

Participants express confusion over the notation involving hatted components and the proper application of equations. There is also mention of emotional responses to difficulties encountered in the problem-solving process.

u0467750
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Homework Statement


some one walking three directions 0.35 east then 0.75 south then 2.95 33 degrees north of west. The whole trip took 2 hours. Magnitude and direction of displacement as well as the magnitude and direction of velocity.


Homework Equations


r (displacement vector) x (I think in the equation it's the x components) y ( y components) a and I am confused with the x^ and y^ the hats are supposed to be on top. The equation is r=xx^+yy^ where x=d1-d3cos(theta) and where y=-d2+d3sin(theta) where the d's are distance one = 03.5 distance two =0.75 and distance 3 = 2.95


The Attempt at a Solution


so I solved the equations of x and y and for x I got -2.124 then for y I got 0.856 then I pluged those into the other equation and got 1.606 which i thought would be the magnitude but it was not correct and I don't really know what to do with the x^ and y^ or how to find the direction of this displacement any help would be great
 
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Actually showing your work step by step for how you got your anwers would be helpful.
 
ok so for the x equation I put x=0.35-2.95(cos(33)) then I put that in my calculator and got the answer I gave then for the y equation I did y=-0.75+2.95(sin(33)) and got y=0.856 then in the other equation where I'm finding r I just added those two numbers together because I don't know what to do with the x and y hats . . . I am retarded
 
u0467750 said:
ok so for the x equation I put x=0.35-2.95(cos(33)) then I put that in my calculator and got the answer I gave then for the y equation I did y=-0.75+2.95(sin(33)) and got y=0.856 then in the other equation where I'm finding r I just added those two numbers together because I don't know what to do with the x and y hats . . . I am retarded

You're not retarded. Don't say that.

The hatted characters are just there to indicate the direction of the component of the vector. In the 2D case, the magnitude is found by using the pythagorean theorem with the components you found as the legs of the triangle. It should be smooth sailing from there.
 
ok so I used the pythagorean theorem and got the right answer for the magnitude of the displacement, but how do I find the direction of the displacement
 
and I found the velocity but also do not know how to find the direction is it the inverse tangent of the angle 33 with the x and y components I found?
 
It's the inverse tangent of y divided x. Be sure you have the correct signs on the components.
 

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