Determining Wind Velocity From Light Plane Flight

[mikenash]

A light plane attains an airspeed of 470 km/h. The pilot sets out for a destination 800 km due north but discovers that the plane must be headed 17.0° east of due north to fly there directly. The plane arrives in 2.00 h. What were the (a) magnitude (in km/h) and (b) direction of the wind velocity? Give the direction as an angle relative to due west, where north of west is a positive angle, and south of west is a negative angle.

Pyj-800j=wj
Pxi=Wxi

attempt

tan inverse
(wx/WY)

then i got lost and do not know how to start the problem

 

[Andrew Mason]

A light plane attains an airspeed of 470 km/h. The pilot sets out for a destination 800 km due north but discovers that the plane must be headed 17.0° east of due north to fly there directly. The plane arrives in 2.00 h. What were the (a) magnitude (in km/h) and (b) direction of the wind velocity? Give the direction as an angle relative to due west, where north of west is a positive angle, and south of west is a negative angle.

Pyj-800j=wj
Pxi=Wxi

attempt

tan inverse
(wx/WY)

then i got lost and do not know how to start the problem

Draw a vector diagram. One vector $\vec v_{pa}$ represents the velocity of the plane relative to the air. The other vector $\vec v_{ag}$ represents the velocity of the air relative to the ground. What does the resultant vector ($\vec v_{pa}+\vec v_{ag} = \vec v_{pg}$ represent? (Hint: I have given you a hint). Do we know the length of this vector? Do we know the length of $\vec v_{pa}$? Do we know its angle relative to $\vec v_{pg}$? Resolve the North and West components of $\vec v_{pa}$. What do these components plus the North and West components of $\vec v_{ag}$ have to add up to?

AM

 

[baba89]

I would approach this problem by first breaking it down into smaller parts and using known mathematical formulas and principles to solve for the unknown variables.

Firstly, let's define the variables:
Vp = plane's airspeed (470 km/h)
Vw = wind velocity (unknown)
θ = angle at which the plane must be headed (17.0° east of due north)
d = distance traveled (800 km)
t = time taken (2.00 h)

To determine the wind velocity, we can use the formula:
Vp = Vw + Vw cosθ
Substituting in the known values, we get:
470 km/h = Vw + Vw cos17.0°
Solving for Vw, we get:
Vw = 470 km/h / (1 + cos17.0°)
Vw = 470 km/h / 1.032
Vw = 455.81 km/h

So, the magnitude of the wind velocity is approximately 455.81 km/h.

To determine the direction of the wind velocity, we can use the formula:
tanθ = Vw sinθ / Vw cosθ
Substituting in the known values, we get:
tanθ = Vw sin17.0° / Vw cos17.0°
Solving for θ, we get:
θ = tan^-1 (Vw sin17.0° / Vw cos17.0°)
θ = tan^-1 (455.81 km/h * sin17.0° / 455.81 km/h * cos17.0°)
θ = tan^-1 (0.298)
θ = 16.8°

Since the plane had to be headed 17.0° east of due north, the wind velocity must be in the opposite direction, which is 180° - 16.8° = 163.2° west of due north.

Therefore, the direction of the wind velocity is approximately 163.2° west of due north.

In summary, the wind velocity is approximately 455.81 km/h directed 163.2° west of due north.