As noted in Exercise 1.26, a spelunker is surveying a cave. She follows a passage 180 m straight west, then 210 m in a direction 45o east of south, and then 280 m at 30o east of north. After a fourth displacement, she finds herself back where she started. Use the method of components to determine the magnitude and direction of the fourth displacement. Draw the vector-addition diagram and show that it is in qualitative agreement with your numerical solution.

Solution 69P Step 1 of 6: Solving for numerical solution, Let North and East be positive. First displacement: 180 m west. 0 0 A=<180cos 180 ,180 sin 180 >= o Second displacement: 230 m, 45 east of south. 0 0 B =<230cos -45 , 230 sin -45 >= <115 2, -115 2> Step 2 of 6: Third displacement: 280 m, 30 east of north. C =<280cos 30 , 280 sin 30 >= <140, 140 3> Adding all this three vectors, we will get the fourth vector(final displacement) D = A + B + C = <122.63, 79.85> Vector D is 122.63 m east and 79.85 m north of point A. Step 3 of 6: To find the direction(angle) Using basic trigonometry, tan = R y R x R y Solving for angle = tan ( ) R x Using R = x22.63 and R = 79.85y = tan (1 122.63) 79.85 =59.9 0 =60 0 o The direction of the last displacement is: 60 west of south.