A light plane is travelling at 175 km/h on a bearing of N8°E in a 40 km/h wind from N80°E. Determine the plane's ground velocity using algebraic vectors.

Question

The answer for this question is:

167 km/h , N5°W

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My Work:

(a , b) = ( | r | * cosθ , | r | * sinθ )

u = (175cos82, 175sin82)
u = (24.4 , 173.3)

w = (40cos10 , 40sin10)
w = (39.4 , 6.9)

r = (24.4 + 39.4 , 6.9 + 173.3)
r = (63.8 , 180.2)

| r |^2 = 180.2^2 + 63.8^2
(sqrt)| r |^2 = (sqrt)(180.2^2 + 63.8^2)
| r | = 191 km/h

tanθ = 180.2 / 63.8
θ = 70.5

---------- What am I doing wrong? This question is so frustrating. . .

Answer 1

your wind is coming from N80E so your w vector should be (-39.4,-6.9)

then │r│ = (24.4-39.4,173.3-6.9)

it works out

(BTW, I just did the same question by using the cosine law, then the sine law and got the answer much faster, but then again, it said using algebraic vectors)

Answer 2

V(with respect to ground) = V'(with respect to air) + v(air with respect to ground)
Write that in vector form. Call +y the N direction and +x the E direction.
Vx = V'x + vx = 175 sin 8 - 40 sin 80
Vy = V'y + vy = 175 cos 8 - 40 cos 80
The "-" signs are needed for the air speed relative to ground, since the wind is coming FROM the N 80 E direction.
Vx = -15.0
Vy = 166.4
|V| = sqrt[166.4^2 + 15^2] = 167.1 km/h
Direction is just W of N, by an amount tan^-1 15/166.4 = 5.1 degrees