Question
The cruising speed of a Boeing 767 in still air is 530 mph. Suppose that a 767 is cruising directly east when it encounters an 80 mph wind blowing 40 degrees south of west. Sketch the vectors for the velocities of the airplane and the wind. Express both vectors in ordered pair notation. Find the sum of the vectors. Find the magnitude and direction of the resultant vector.
Answers
first vector = (530cos0, 530sin0
= (530,0)
2nd vector = (80cos220 , 80sin 220
= (-61.28355, -51.423
resultant = ((468.71644 , -51.423)
maginitude = √((468.71644^2 + (-51.423)^2 )
= 471.529
direction:
tan ฿ = -51.423/468.71644 = -.839..
฿ = E 40° S
= (530,0)
2nd vector = (80cos220 , 80sin 220
= (-61.28355, -51.423
resultant = ((468.71644 , -51.423)
maginitude = √((468.71644^2 + (-51.423)^2 )
= 471.529
direction:
tan ฿ = -51.423/468.71644 = -.839..
฿ = E 40° S
edit to reiny's answer, they had an error in the final step.
first vector = (530cos0, 530sin0
= (530,0)
2nd vector = (80cos220 , 80sin 220
= (-61.28355, -51.423
resultant = ((468.71644 , -51.423)
maginitude = √((468.71644^2 + (-51.423)^2 )
= 471.529
direction:
tan ฿ = -51.423/468.71644 = -0.11
฿ = 6° S of E
first vector = (530cos0, 530sin0
= (530,0)
2nd vector = (80cos220 , 80sin 220
= (-61.28355, -51.423
resultant = ((468.71644 , -51.423)
maginitude = √((468.71644^2 + (-51.423)^2 )
= 471.529
direction:
tan ฿ = -51.423/468.71644 = -0.11
฿ = 6° S of E
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