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Question 1182063: A plane flying with a heading of S65°E and a speed of 530 km/h encounters a headwind blowing directly from the east at 70 km/h. Determine the ground speed and direction of the plane.
Thank you
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
The notation S65°E means that we start looking directly south, and then turn 65 degrees toward the east.
When looking at a unit circle, we see that 270 degrees corresponds to directly south. This is because 0 degrees corresponds to directly east and we rotate counterclockwise.
After aiming at the 270 degree marker, we move rotate an additional 65 degrees. So the angle theta is at 270+65 = 335
The direction S65°E corresponds to the angle 335 degrees
This is the direction of the plane's vector where we don't account for the wind.
The speed of the plane without wind is 530 km/hr
Let's find the component form of the vector
x = r*cos(theta)
x = 530*cos(335)
x = 480.343127
y = r*sin(theta)
y = 530*sin(335)
y = -223.987679
The values are approximate.
If we ignore wind, the plane's airspeed vector is approximately < x, y > = < 480.343127, -223.987679 >
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The wind's vector is <-70, 0> since the wind is blowing from the east, aka the direction is east to west. So that means the angle is 180 degrees and the speed is 70
We can compute the component form like so
x = r*cos(theta)
x = 70*cos(180)
x = -70
y = r*sin(theta)
y = 70*sin(180)
y = 0
So that explains where < x,y > = < -70,0 > comes from.
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Let's add the two vectors we found
< 480.343127, -223.987679 > + < -70,0 >
< 480.343127+(-70), -223.987679+0 >
< 410.343127, -223.987679 >
This represents the ground speed vector of the plane since we're now accounting for the wind.
The speed is the magnitude of the vector we just found.
For any vector < a,b >, the magnitude is sqrt(a^2+b^2)
speed = sqrt(a^2+b^2)
speed = sqrt((410.343127)^2+(-223.987679)^2)
speed = 467.495414
speed = 467 km/hr
I'm rounding to the nearest whole number since the other speed value is given as a whole number. Round however your teacher instructs.
As expected, the ground speed is slower than the airspeed because the wind is slowing down the plane.
The term "headwind" means the air is coming from the head of the plane and pushing against it to slow it down.
In contrast, tailwinds come from the tail to push the plane forward and speed it up (making the ground speed faster than the airspeed).
I should clarify that "airspeed" refers to the plane's speed without the wind affecting it. Be sure not to mix up the terms "airspeed" and "wind speed" (they're two different concepts). As the name implies, "ground speed" is the speed of the plane relative to the ground.
Now let's compute the angle theta
theta = arctan(b/a)
theta = arctan(-223.987679/410.343127)
theta = -28.628122
Add on 360 degrees to get to a positive coterminal angle, that's between 0 and 360
So we get -28.628122+360 = 331.371878
When rounding to the nearest whole degree, we get 331 degrees.
Converting to the compass notation given earlier (similar to the format S65°E), we can note that
331-270 = 61
So this means we start looking directly south (270 degrees) and then turn 61 degrees toward the east to arrive at the compass bearing S61°E
We can see that the westward wind is nudging the plane to aim slightly more to the west, but overall its direction is still in the southeast quadrant.
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Answers:
Ground speed = 467 km/hr
Direction = S61°E
This is after wind is taken into account. The values are approximate and rounded to the nearest whole number. Round otherwise if your teacher instructs.
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