The route followed by a hiker consists of three displacement vectors , , and . V

Question

The route followed by a hiker consists of three displacement vectors , , and . Vector is along a measured trail and is 2870 m in a direction 34.0 ° north of east. Vector is not along a measured trail, but the hiker uses a compass and knows that the direction is 16.0 ° east of south. Similarly, the direction of vector is 20.0 ° north of west. The hiker ends up back where she started, so the resultant displacement is zero, or + + = 0. Find the magnitudes of (a) vector and (b) vector . i dont know where to start?

Explanation / Answer

SOLUTION: Consider east as +X , north as +Y               west as -X, south as -Y form the given data, we can write the vectors as, Consider east as +X , north as +Y               west as -X, south as -Y form the given data, we can write the vectors as, vector A =( 2870 m) [cos(34o) i + sin(34o) j]
vector B = B[sin(16o) i - cos(16o) j]
vector C = C[-cos(20o) i + sin(20o) j] but, given that, A + B + C = 0
thus,  
( 2870 m) [cos(34o) i + sin(34o) j] +B[sin(16o) i - cos(16o) j] +  C[-cos(20o) i + sin(20o) j] = 0 on equating i, j co-efficents speretely equal to 0, we have (2870 m) cos(34o) + Bsin(16o) - Ccos(20o) = 0   so, (2379.3 m) + (0.275) B - (0.939)C = 0         ......(1)     and (2870 m) sin(34o) - Bcos(16o) + Csin(20o) = 0, so, (1604.8 m) -(0.961) B + (0.342)C = 0           ......(2) In order to know B, C values, ( eqn: 1) (0.342) + (eqn: 2) (0.939) ( 813.7 m) + 0.094 B -0.321 C +( 1506.9 m)-0.902 B + 0.321 C =0 (2320.6 m) - 0.808 B = 0    so,   B = 2872 m plug this in equation (1) , we have (2379.3 m) + (0.275)(2872 m) - (0.939)C = 0        so, C =3374.9 m    thus, amgnitude of the vectors,      A= 2870 m      B= 2872 m      C= 3374.9 m       (2870 m) sin(34o) - Bcos(16o) + Csin(20o) = 0, so, (1604.8 m) -(0.961) B + (0.342)C = 0           ......(2) In order to know B, C values, ( eqn: 1) (0.342) + (eqn: 2) (0.939) ( 813.7 m) + 0.094 B -0.321 C +( 1506.9 m)-0.902 B + 0.321 C =0 (2320.6 m) - 0.808 B = 0    so,   B = 2872 m plug this in equation (1) , we have (2379.3 m) + (0.275)(2872 m) - (0.939)C = 0        so, C =3374.9 m    thus, amgnitude of the vectors,      A= 2870 m      B= 2872 m      C= 3374.9 m       In order to know B, C values, ( eqn: 1) (0.342) + (eqn: 2) (0.939) ( 813.7 m) + 0.094 B -0.321 C +( 1506.9 m)-0.902 B + 0.321 C =0 (2320.6 m) - 0.808 B = 0    so,   B = 2872 m plug this in equation (1) , we have (2379.3 m) + (0.275)(2872 m) - (0.939)C = 0        so, C =3374.9 m    thus, amgnitude of the vectors,      A= 2870 m      B= 2872 m      C= 3374.9 m      

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