You have recently been promoted and are now a senior pilot with BHS Exprress, providing courteous and timely service to countless cities across the Northeast. In order to arrive in New York City at the scheduled time, you know you must fly at a speed of 380mph on a course of 160 degrees. You check the weather conditions and find that the wind is blowing at 42 mph from 35 degrees East of North. Find the following information: Eastern Component of the Air velocity, southern component of the air velocity, air speed, heading. I have already solved for the velocities so I just need air speed and heading
can you convert those speeds and headings to vectors? That's the first step in all these problems. If not, it's time for review. If so, what do you get?
1. Vw = 42mi/h[35o]E. of N.
X = 42*sin35 = 24.1mi/h = E. component.
Y = 42*Cos35 = 34.4mi/h = S. component.
2. 380mi/h[160o] Bearing.
X = 380*sin160 = 130mi/h.
Y = 380*Cos160 = -357mi/h.
Add the components of the wind and plane:
X = 24.1 + 130 = 154.1 mi/h.
Y = 34.4 + (-357) = -322.7 mi/h.
X + Yi = 154.1-322.7i = 358 mi/h[154.5o].
Tan(A) = (-322.7)/154.1 = -2.09409,
A = -64.5o = 64.5o S. of E. = 154.5o CW from +Y-axis. = Direction.
Okay so for the heading I got 334.474 degrees south of east but I'm confused on the air speed. We only did this for two days in class so I'm very confused on the vector unit, which is why I'm having so much trouble. It's break now and my teacher won't answer any questions.
Sarah, it is impossible to have 334o South of East. 334o CW from the +Y-axis is possible. 334o CCW from the +X-axis is also possible.
I would like to see your calculations.
To find the air speed and heading, we will need to use vector addition. Let's start by breaking down the velocities into their components:
1. Air velocity: The air velocity is the sum of the wind velocity and the velocity of the airplane. We already have the wind velocity as 42 mph at 35 degrees East of North. The velocity of the airplane is the desired speed and course, which is 380 mph at 160 degrees.
To find the components of the wind vector, we will use trigonometry.
Wind Velocity (magnitude and direction):
Magnitude: √(42^2 + 380^2 - 2*42*380*cos(35))
Direction: tan^(-1)((380*sin(160) - 42*sin(35))/(380*cos(160) - 42*cos(35)))
2. Air speed: The magnitude of the air speed is the magnitude of the air velocity.
Air Speed: √(Wind Velocity magnitude^2)
3. Heading: The direction of the air speed is the direction of the air velocity.
Heading: Wind Velocity direction
By solving the equations above, you'll be able to find the air speed and heading for your flight.