Question #233597

 cruise ship leaves port and sails due east for a distance of 231 km. To avoid a storm, it turns and sails 42.1o south of east for 209 km, 54.8o north of east for 262 km, and then sails 100 km north. Determine the magnitude and direction of the resultant displacement.

 


1
Expert's answer
2021-09-07T09:51:09-0400

Find the total displacement due East and North:


DE1=231 km,DE2=209cos42.1°=155 km,DE3=262cos54.8°=151 km,DE4=0 km. DN1=0 km,DN2=209sin42.1°=140 km,DN3=262sin54.8°=214 km,DN4=100 km.D_{E1}=231\text{ km},\\ D_{E2}=209\cos42.1°=155\text{ km},\\ D_{E3}=262\cos54.8°=151\text{ km},\\ D_{E4}=0\text{ km}.\\\space\\ D_{N1}=0\text{ km},\\ D_{N2}=-209\sin42.1°=-140\text{ km},\\ D_{N3}=262\sin54.8°=214\text{ km},\\ D_{N4}=100\text{ km}.

Determine the magnitude and direction of the resultant displacement:



DN=DN1+DN2+DN3+DN4=174 km,DE=DE1+DE2+DE3+DE4=537 km.D=DE2+DN2=564 km, θ=arctan174537=18.0° N of E.D_N=D_{N1}+D_{N2}+D_{N3}+D_{N4}=174\text{ km},\\ D_E=D_{E1}+D_{E2}+D_{E3}+D_{E4}=537\text{ km}.\\ D=\sqrt{D_E^2+D_N^2}=564\text{ km},\\\space\\ \theta=\arctan\frac{174}{537}=18.0°\text{ N of E}.


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