Calculating the sum of displacement vectors

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The discussion focuses on calculating the sum of displacement vectors in a given scenario involving multiple movements. The original poster describes movements of 100m right, 300m down, and 150m left diagonally, seeking clarification on the correct vector representation. Community members emphasize the importance of a step-by-step analysis and correcting the initial attempt, noting that the poster incorrectly identified the number of vectors. They advise on the correct signs and components for each vector, guiding the poster to refine their approach. The conversation highlights the necessity of understanding vector addition and the proper application of trigonometric functions in solving the problem.
student120
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Homework Statement
No, I'm trying to learn.
Relevant Equations
Vector 1: V1=(0,100)
Vector 2: V2=(100,0)
Vector 3: V3=(0,-300)
Vector 4: V4=(-150,0)
Vector 5: V5=(-200cos(60°),-200sin(60°))
but after that I couldn't go any further
I greet everyone

I am faced with such a question

First 100m to the right, then 300m down, then 150 left diagonally, even I don't think this is exactly right, I think there will be 150x sin30 or cos30, but I'm not sure, I need to do 200xcos60 afterwards, but I couldn't settle the question. can anyone help me solve it step by step?

Every answer from the community will be honored, thanks in advance


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student120 said:
can anyone help me solve it step by step?
No, what we CAN do, based on forum rules, is show you where you are going wrong once you show YOUR step by step analysis. We don't spoon feed answers, we help you figure out how to get your own answers.
 
Consider a step-by-step analysis where you write down 4 displacement vectors which is what your diagram shows. The attempt at it that you posted under "Relevant Equations" is not correct and shows 5 vectors whereas you have 4. Please try again.
 
student120 said:
Vector 1: V1=(0,100)
Where is that in the diagram?
student120 said:
Vector 2: V2=(100,0)
Vector 3: V3=(0,-300)
ok
student120 said:
Vector 4: V4=(-150,0)
Vector 5: V5=(-200cos(60°),-200sin(60°))
No and no.
Use the same approach in 4 that you appear to have used in 5, but be careful with the signs.
 
And for “Vector 5” (which is actually the 4th vector in the chain), you show both x and y components as negative. According to your diagram, that vector points to the left, which is indeed the negative x direction, and upward, which is the ___?___ y direction.
 
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