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.
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. . .
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)
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
It seems like you made a couple of errors in your calculations.
First, when finding the components of the wind vector, you used the wrong angle. The angle should be N80°W, not N80°E. So the correct calculation should be:
w = (40cos100°, 40sin100°)
w = (-9.4, 39.3)
Next, the ground velocity vector is the sum of the plane's velocity vector and the wind vector. So the correct calculation should be:
r = (175cos8°, 175sin8°) + (-9.4, 39.3)
r = (172.9, 24.1) + (-9.4, 39.3)
r = (163.5, 63.4)
Finally, you can find the magnitude of the ground velocity vector using the Pythagorean theorem. So the correct calculation should be:
|r|^2 = 163.5^2 + 63.4^2
|r| = √(163.5^2 + 63.4^2)
|r| = 175.9 km/h
To find the direction, you can use trigonometry. The angle θ is given by:
tanθ = 63.4 / 163.5
θ = arctan(63.4 / 163.5) = 20.5 degrees
Since the plane is traveling on a bearing of N8°E, you subtract this angle from 90° to get the angle with respect to the North direction:
90° - 20.5° = 69.5°
Therefore, the plane's ground velocity is 175.9 km/h at a bearing of N69.5°W.
It seems like there is a mistake in your calculations. Let me guide you through the correct steps in determining the plane's ground velocity.
To find the ground velocity of the plane, we need to add the velocity of the plane relative to the ground and the velocity of the wind. We will use vector addition to calculate this.
1. Convert the velocities of the plane and wind into their vector form:
- The velocity of the plane can be written as u = (175cos8°, 175sin8°).
- The velocity of the wind can be written as w = (40cos80°, 40sin80°).
Make sure to convert the angles to radians before taking the cosine and sine.
2. Add the two vectors to find the resultant velocity vector r:
r = u + w
To add the vectors, add their corresponding components:
r = ((175cos8°) + (40cos80°), (175sin8°) + (40sin80°)).
3. Calculate the magnitude of the resultant velocity vector r:
| r | = √((r_x)^2 + (r_y)^2).
Square each component of r, sum the squares, and then take the square root.
4. Calculate the direction angle of the resultant velocity vector:
θ = tan^(-1)(r_y / r_x),
where θ is the angle measured counterclockwise from the positive x-axis.
5. Write the resultant velocity vector in terms of its magnitude and direction angle:
r = | r |, θ.
Now, plug in the values for u, w, and r into these steps and redo the calculations to find the correct magnitude and direction angle of the resultant velocity vector.