I tried doing this problem, but I still don't understand it at all...
Pilot Reem Ahmad's flight plan has her leaving Dulles International Airport and flying a Boeing 727 due east at 450 mph. There is a 65 mph wind with the bearing 60 degrees.
a. Determine the compass heading that Reem should follow.
b. Determine the airplane's ground speed. (Assume the speed with no wind is 450 mph)
The E,W component from the wind. Iassume the wind is going toward 060, so the E component of that is 65Cos60
So now, you have plane east velocity + wind east velocity= 450
plane east velocity= 450-65Cos60
calculate that.
The NS component of wind is 65Sin60, and the plane has to eliminate that.
(450-65cos60)SinHeading=-65Sin60
solve for Heading
To solve this problem, we'll break it down into two parts:
a. Determining the compass heading:
To determine the compass heading that Reem should follow, we need to take into account the wind's direction. The angle between the airplane's direction (due east) and the direction of the wind will give us the compass heading.
In this case, the angle between the airplane's direction and the wind's direction is 60 degrees. Since the wind is coming from the northeast (bearing 60 degrees), we need to subtract this angle from 90 degrees (east) to get the compass heading.
Compass Heading = 90 degrees - 60 degrees = 30 degrees
Therefore, Reem should follow a compass heading of 30 degrees.
b. Determining the airplane's ground speed:
The airplane's ground speed is the actual speed at which it is moving with respect to the ground. It is a combination of the airplane's speed and the wind speed.
Since the wind is blowing in the direction opposite to the airplane's eastward flight, it will reduce the effective speed of the airplane.
To find the ground speed, we can use the concept of vector addition. The ground speed can be determined by finding the magnitude of the resultant vector formed by adding the airplane's speed vector and the wind speed vector.
Using the Pythagorean theorem, we can calculate the ground speed:
Ground Speed = √(Airplane Speed^2 + Wind Speed^2)
Ground Speed = √(450^2 + 65^2) = √(202500 + 4225) = √206725 ≈ 454.66 mph
Therefore, the airplane's ground speed is approximately 454.66 mph.
To solve this problem, we need to consider the concept of vector addition. We can break down the motion of the airplane into two components: the motion due to the airplane's speed (450 mph) and the motion due to the wind (65 mph).
a. To determine the compass heading that Reem should follow, we need to find the resultant of these two vectors. The angle of this resultant vector will give us the compass heading.
First, let's draw a diagram to help visualize the situation. Draw a straight line to represent the airplane's velocity (450 mph) pointing due east. Then, draw another line to represent the wind's velocity (65 mph) with a bearing of 60 degrees.
To add these vectors, we can use the cosine rule. The formula is:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, a is the magnitude of the airplane's velocity (450 mph), b is the magnitude of the wind's velocity (65 mph), and C is the angle between them (60 degrees).
Calculating the resultant vector:
c^2 = (450 mph)^2 + (65 mph)^2 - 2 * 450 mph * 65 mph * cos(60 degrees)
Simplifying:
c^2 = 202,500 + 4,225 - 58,500 * 0.5
c^2 = 207,725 - 29,250
c^2 = 178,475
c ≈ 422.75 mph (rounded to two decimal places)
Now that we have the magnitude of the resultant vector, we can find its angle. We can use the sine rule, since we have both the opposite side (450 mph) and the hypotenuse (422.75 mph). The formula is:
sin(A) = a / c
Solving for A, the angle we're looking for:
A = sin^(-1)(450 mph / 422.75 mph)
A ≈ 52.45 degrees (rounded to two decimal places)
Therefore, Reem should follow a compass heading of approximately 52.45 degrees.
b. To determine the airplane's ground speed, we need to find the magnitude of the resultant vector. In this case, the resultant vector represents the airplane's motion relative to the ground.
We have already calculated the magnitude of the resultant vector in part a, which is approximately 422.75 mph. Therefore, the airplane's ground speed is approximately 422.75 mph (rounded to two decimal places).