A plane has an airspeed of 200 miles per hour and a heading of 28.0°. The ground speed of the plane is 207 miles per hour, and its true course is in the direction of 38.0°. Find the speed and direction of the air currents, assuming they are constants. (Round your answers to one decimal place.)
_____mi/hr at ____° from due north
I got calculated 36.1 to one decimal point. And the wind direction I believe should be 200. Do not trust my answer. Thanks!
To find the speed and direction of the air currents, we need to use vector addition.
Let's consider the plane's velocity relative to the ground as the resultant vector, and the airspeed of the plane as one of the vectors contributing to it. The other vector would be the velocity of the air currents.
We can break down the airspeed and the ground speed into their respective components.
Using trigonometry, we can find the horizontal component of the airspeed by multiplying the airspeed by the cosine of the heading angle:
Horizontal component of airspeed = 200 mph * cos(28.0°)
= 200 mph * 0.8839
= 176.78 mph (rounded to one decimal place)
Similarly, we can find the vertical component of the airspeed by multiplying the airspeed by the sine of the heading angle:
Vertical component of airspeed = 200 mph * sin(28.0°)
= 200 mph * 0.4695
= 93.90 mph (rounded to one decimal place)
Next, let's find the horizontal component of the air currents. The horizontal component affects the ground speed but not the true course. We can find it by subtracting the horizontal component of the airspeed from the ground speed:
Horizontal component of air currents = Ground speed - Horizontal component of airspeed
= 207 mph - 176.78 mph
= 30.22 mph (rounded to one decimal place)
Finally, we can find the direction of the air currents using the tangent of the angle:
Direction of air currents = arctan(vertical component of airspeed/horizontal component of air currents)
= arctan(93.90 mph/30.22 mph)
= 73.9° (rounded to one decimal place)
Therefore, the speed of the air currents is approximately 30.2 mph, and the direction is approximately 73.9° from due north.
To find the speed and direction of the air currents, we can use the concept of vector addition.
Let's represent the airspeed of the plane as a vector A, the ground speed of the plane as a vector B, and the air currents as a vector C.
Since the airspeed of the plane is the result of the vector addition of the ground speed and the air currents, we can write the equation:
A = B + C
Given values:
A (airspeed of the plane) = 200 miles per hour
B (ground speed of the plane) = 207 miles per hour
We need to solve for C (speed and direction of the air currents).
To find the speed of the air currents, we can calculate the magnitude of vector C:
|C| = |A - B|
Let's substitute the known values into the equation:
|C| = |200 - 207|
Simplifying this, we get:
|C| = |-7| = 7
So, the speed of the air currents is 7 miles per hour.
Now, let's find the direction of the air currents. We can use trigonometry to determine the angle between the air currents and due north.
We know that the true course of the plane is in the direction of 38.0° from due north. Let's call this angle θ.
Also, the heading of the plane (angle between the airspeed vector and due north) is 28.0°.
To find the direction of the air currents, we need to find the angle between the air currents and the heading of the plane. This can be done using the angle difference formula:
ϕ = θ - 28.0°
Substituting the given values:
ϕ = 38.0° - 28.0° = 10.0°
Therefore, the direction of the air currents is 10.0° from due north.
In summary, the speed of the air currents is 7.0 miles per hour, and the direction is 10.0° from due north.
I assume you are studying vectors.
Draw a line at 28° of length 200 (your resultant)
Draw a line at 38° of length 207, join its tail to the first line.
That smaller line will be the vector representing the wind
I see a triangle with sides 200 and 207, with a contained angle of 10°
Let the length of the smaller side be x
by cosine law:
x^2 = 200^2 + 207^2 - 2(200)(207)cos10°
etc.
x will be you speed of the wind, and from your diagram it should be straightforward to find the wind direction