Answer to Question #233597 in Physics for lovely

Question

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

 

Answer 1

Find the total displacement due East and North:

$$D_{E1}=231\text{ km},\\ D_{E2}=209\cos42.1°=155\text{ km},\\ D_{E3}=262\cos54.8°=151\text{ km},\\ D_{E4}=0\text{ km}.\\\space\\ D_{N1}=0\text{ km},\\ D_{N2}=-209\sin42.1°=-140\text{ km},\\ D_{N3}=262\sin54.8°=214\text{ km},\\ D_{N4}=100\text{ km}.$$

Determine the magnitude and direction of the resultant displacement:

$$D_N=D_{N1}+D_{N2}+D_{N3}+D_{N4}=174\text{ km},\\ D_E=D_{E1}+D_{E2}+D_{E3}+D_{E4}=537\text{ km}.\\ D=\sqrt{D_E^2+D_N^2}=564\text{ km},\\\space\\ \theta=\arctan\frac{174}{537}=18.0°\text{ N of E}.$$