Asked by Bobby
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)?
(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)?
Answers
Answered by
Steve
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.
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.
Answered by
Bobby
thank you sooo much
Answered by
Bobby
wait why did u subtract 10 degrees? where does the 10 degrees come from?
at the end
at the end