A plane flies on a bearing of 120 degrees at a constant speed of 550 km/h. If the velocity of the wind is 50 km/h on a bearing 220 degrees, what is the velocity of the plane with respect to the ground?
Textbook Answer: 543.5 km/h at a bearing of 125.2 degrees
My answer: 543.6 km/h at a bearing of 175 degrees
My work:
Let vector w represent 50 km/h
Let vector v represent 550 km/h
Let vector r represent the resultant
I'm using component vectors. . .
w = (50cos220 , 50sin220)
w = (-38.3 , -32.1)
v = (550cos120 , 550sin120)
v = (-275 , 476.3)
r = (-38.3 , -32.1) + (-275 , 476.3)
r = (-313.3 , 444.2)
To find the resultant:
| r | = (-313.3)^2 + (444.2)^2
| r | = (sqrt)295 470.53
| r | = 543.6 km/h
To find the angle:
tanx = 444.2/-313.3
x = 55 degrees
120 + 55 = 175 (To find the bearing)
----------------What did I do wrong? How can I get the 125 degree instead of 175 degrees?
Just the last line is wrong. The angle 55 degrees references the resultant from the -x axis. Subtract that value from 180 to find the reference from the positive x axis
180 - 55=125
I see where the confusion is coming from. The angle you calculated using the inverse tangent function is the angle between the resultant vector (r) and the horizontal axis. However, the bearing angle is measured clockwise from the north direction.
To find the bearing angle, you need to subtract the angle you found from 180 degrees because the angle you calculated is greater than 90 degrees. This is because the plane is flying in the southeast direction.
180 degrees - 55 degrees = 125 degrees
So, the correct bearing angle is 125 degrees, not 175 degrees. Everything else in your solution is correct, including the magnitude of the resultant velocity being 543.6 km/h.