Calculating Displacement with Vectors

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To calculate the resultant displacement of a car traveling from point A to point C, a vector diagram can be constructed, with the car first moving 6 km east and then 4 km northwest. The correct approach involves breaking down the vectors into their components using trigonometric functions for both the eastward and northwestward movements. The resultant vector can be found by summing the x and y components of the two vectors. The initial measurement of 4.3 km at a bearing of 50 degrees may not be accurate, emphasizing the importance of using calculations over measurements for precision. Understanding and applying the vector component equations is crucial for solving such problems effectively.
Angus Dolan
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Homework Statement


Hello, I have a homework question that I'm having difficulty with, it is:
A car travels due east (bearing 90) from point A for 6km to point B and then North-west (bearing 315) for 4.0km to point C.
By constructing a vector diagram, or otherwise, find the resultant displacement (magnitude and direction) of the car at point C from A.


Homework Equations


I don't think pythagoras theorem works here, as there is no right-angled triangle present. Although, I may be incorrect.



The Attempt at a Solution


I've drawn a vector diagram, head to tail, and measured the displacement, it comes to 4.3km at a bearing of 50. Is that correct?
 
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You can try other methods of finding vector C, you can find the components of vector A, and vector b

Remember finding the magnitude or resultant vector is
\vec{R} = \vec{A} + \vec{B}

which is really
\vec{R} = (A_{x} ) (B_{x} ) + (A_{y} ) (B_{y} )

A_{y} = (\vec{A}) (sin \theta)
A_{x} = (\vec{A}) (cos \theta)


B_{y} = (\vec{A}) (sin \theta)
B_{x} = (\vec{A}) (cos \theta)

The Ax, Ay, Bx, By equations are derived from this idea
xycomponents.gif
 
I would plug and chug for you, but I'm kind of confused by the bearings.
 
Don't rely on measuring the displacement vector. Sometimes the ruler isn't as precise, also it will help you in the long run to have these equations in your tool box.
 
Let me know if you solved it, or if it helped. So I can get some personal feedback, because I don't want to be misinforming people lol.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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