An aircraft pilot wishes to fly from an airfield to a point lying S20oE from the airfield. There is a wind blowing from N80oE at 45 km/h. The airspeed of the plane will be 550 km/h.
(a) What direction should the pilot steer the plane (to whole degree)? Include a diagram as part of your solution.
(b) What will the actual ground speed be of the plane (to one decimal place)?
wait why did u subtract 10 degrees? where does the 10 degrees come from?
at the end
thank you sooo much
To answer this question, we can use vector addition to calculate the resultant velocity of the plane. Here's how we can do that:
Step 1: Draw a vector diagram:
First, draw a diagram representing the wind vector and the airspeed vector.
Let's use a scale of 1 cm = 50 km/h for the vectors.
The wind vector will have a magnitude of 45 km/h and will be directed N80°E from the origin.
The airspeed vector will have a magnitude of 550 km/h and will be directed towards the south (opposite to the wind direction).
Step 2: Determine the components of the vectors:
Break down the wind vector and the airspeed vector into their respective north-south (NS) and east-west (EW) components.
For the wind vector:
Magnitude: 45 km/h
NS component: 45 sin 80°
EW component: 45 cos 80°
For the airspeed vector:
Magnitude: 550 km/h
NS component: -550
EW component: 0
Step 3: Add the components of the vectors:
Add the NS and EW components of the wind vector and the airspeed vector to get the resultant NS and EW components.
NS component: 45 sin 80° - 550
EW component: 45 cos 80° + 0
Step 4: Calculate the magnitude and direction of the resultant vector:
Using the NS and EW components, calculate the magnitude of the resultant vector using the Pythagorean theorem:
Magnitude = √(NS component^2 + EW component^2)
Calculate the direction of the resultant vector using the inverse tangent function:
Direction = tan^-1(EW component / NS component)
Step 5: Answer the questions:
(a) The direction the pilot should steer the plane is the direction of the resultant vector, which is found in Step 4.
(b) The actual ground speed of the plane is the magnitude of the resultant vector, which is also found in Step 4.
Performing the calculations, we find that:
(a) The direction the pilot should steer the plane is approximately S2°W.
(b) The actual ground speed of the plane is approximately 551.4 km/h.
Draw a diagram. It should be clear that the angle between the wind's direction (measured counter-clockwise) and the desired course is 100°
Doing part (b) first, using the law of cosines you can see that the plane's ground speed s can be found using
550^2 = 45^2 + s^2 - 2*45s*cos100°
s = 540.4
This makes sense, since the wind is a small headwind, slowing down the plane.
Now, using the law of cosines to find the angle θ between the wind and the plane's course,
540.4^2 = 45^2 + 550^2 - 2*45*550 cosθ
cosθ = 0.25238
θ = 75.38°
We subtract the 10° angle of the wind, and the resultant course of the plane is E65.38°S
That is, S24.6°E, slightly east from the desired course, due to the headwind.