East distance = E = 0 + 30 sin 30 - 25 - 40 sin 45
North distance = N = 50 + 30 cos 30 + 0 - 40 cos 45
distance from origin (displacement magnitude) = sqrt (E^2+N^2)
tan (angle north of east) = N/E
taylor decides to search for hard evidence to support his "they never landed on the moon theory" so he gets into his cactus powered mobile and drives. he begins by driving 50.0km [N]. he then turns and drives 30.0k [N30E]. he comes his futile search by driving 25.0 km [w] and then 40.0 km [sw]. if taylors entire trip took 8.5 hours, find his displacement
2 answers
All angles are measured CW from +y-axis.
Disp. = 50i + 30[30o] - 25 + 40[225o]
X = 30*sin30 - 25 + 40*sin225 = 15 - 25 - 28.3 = -38.3.
Y = 50i + 30*cos30 + 40*cos225 = 50i + 26i - 28.3i = 52.3i
Disp.. = -38.3 + 52.3i = 64.8km[-36.2] = 64.8km[36.2o] W. of N. = 324o CW
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Disp. = 50i + 30[30o] - 25 + 40[225o]
X = 30*sin30 - 25 + 40*sin225 = 15 - 25 - 28.3 = -38.3.
Y = 50i + 30*cos30 + 40*cos225 = 50i + 26i - 28.3i = 52.3i
Disp.. = -38.3 + 52.3i = 64.8km[-36.2] = 64.8km[36.2o] W. of N. = 324o CW
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