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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
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 = 52.90o below the -ve X axis
θ = 52.9o + 1800
θ = 232.90o counter clockwise from the +ve X axis.
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