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?
The reason why you are getting 175 degrees is because you are solving for the wrong angle. You are looking for the angle between the resultant and x - axis (-313.3) that is relative to the ground.
cosx = adj. / hypt.
x = cos- (-323.3/543.6)
x = 125 degrees
if you take the archtan of 444.2/-313= -54.80
than +180 then it will be 125.19
Your calculation of the magnitude of the resultant velocity vector is correct at 543.6 km/h. However, your calculation of the bearing angle seems to be incorrect.
To find the angle, you need to use the arctangent function (tan^(-1)) instead of the tangent function. In your calculation, you used the tangent function to find the angle, which resulted in 55 degrees.
To correct this, use the arctan function to find the angle:
angle = arctan(444.2/-313.3)
angle = -54.8 degrees
Bearing angle is usually measured clockwise from the north, so to convert the angle to a positive value, add 180 degrees:
angle = -54.8 degrees + 180 degrees
angle = 125.2 degrees
So, the correct bearing angle for the resultant velocity vector is 125.2 degrees.
To summarize:
Velocity magnitude: 543.6 km/h
Bearing angle: 125.2 degrees