After class, a student walks a distance of 22.3 m in a direction of 40 degrees n

Question

After class, a student walks a distance of 22.3 m in a direction of 40 degrees north of east in 29.4 sec. He then walks a distance of 33.6 m in a direction 15 degrees west of north in 40.2 sec. The student then walks a distance of 28 m in a direction of 38 degrees west of south in 30.2 sec. He then turns and walks a distance of 25.1 m in a direction of 22 degrees south of east in 30.3 seconds. If the x component of final displacement is 14.420 m and the y component is 21.697 m, what is the AVERAGE velocity (direction and magnitude) of the student during the walk ?

Explanation / Answer

let the starting point be ( 0 , 0 ) ;

the final x component =  22.3 cos 40 - 33.6 sin 15 - 28 sin 38 + 25.1 cos 22 = 14.420 m ;

the final y component = 22.3 sin 40 + 40.2 cos 15 - 28 cos 38 -25.1 sin 22 = 21.697 ; 

final displacement = 14.420 i + 21.697 j  

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