Question

Can you please answer this question by giving more details to understand

A person going for a walk follows the path shown in the figure, where y1 = 270 m and θ = 60.0°. The total trip consists of four straight-line paths. At the end of the walk, what is the person's resultant displacement measured from the starting point?

down below the image of continuing question you have to see the image and trying to solve the question from the image

https://www.webassign.net/pse/p3-57alt.gif

Answer #1

There are four displacement vectors.

S1 = 100 i m

S2 = - 270 j m

S3 = -150 cos30 i - 150 sin30 j m = - 129.9 i - 75 j m

S4 = - 200 cos60 i + 200 sin60 j = -100 i + 173.2 j m

Resulatant displacement S = S1 + S2 + S3 + S4

S = (100 i m) + (- 270 j m) + (- 129.9 i - 75 j m) + (-100 i + 173.2 j m)

**S = -129.9 i -171.8 j m**

S = Sqrt[(-129.9)^2 + (-171.8)^2]

**S = 215.38 m**

Direction :

θ = Tan^-1[171.8/129.9]

θ = **52.90 ^{o} below the -ve X axis**

θ = 52.9^{o} + 180^{0}

**θ = 232.90 ^{o} counter clockwise from the
+ve X axis.**

**if any problem comment and rate the answer
plz..**

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