An airplane's speed in still air is 240 km/h. (a) If the pilot wishes the resultant motion of the plane to be due north when a 60 km/h wind is blowing the plane due east, in what direction should the plane be headed? (b) What will then be the resultant speed of the airplane over the ground?
Draw a triangle. If the plane's speed due north is v, then
60^2 + v^2 = 240
direction θ such that sinθ = 60/240
To find the answer to this question, we need to break it down into two parts:
(a) determining the direction the plane should be headed, and
(b) calculating the resultant speed of the airplane over the ground.
(a) To determine the direction the plane should be headed, we need to consider the effect of the wind on the airplane's motion. Since the wind is blowing due east, it will exert a force on the airplane, causing it to have a resulting motion that is not purely north.
To find the direction the plane should be headed, we can use vector addition. We can subtract the velocity of the wind from the velocity of the airplane to find the resulting velocity.
Let's denote the northward direction as positive y-axis and the eastward direction as positive x-axis. The velocity of the airplane in still air is 240 km/h, and the wind is blowing at 60 km/h due east.
Using vector subtraction, we can find the resulting northward and eastward components of the airplane's velocity.
The northward component will be the same as the velocity of the airplane in still air since the wind is blowing directly east. So, the northward component is 240 km/h.
The eastward component will be the velocity of the airplane in still air minus the velocity of the wind. So, the eastward component is 240 km/h - 60 km/h = 180 km/h.
Using the northward and eastward components, we can find the direction using trigonometry. The direction can be determined using the tangent function:
tan(theta) = eastward component / northward component
Substituting the values, we get:
tan(theta) = 180 km/h / 240 km/h
Simplifying, we find:
tan(theta) = 0.75
To find theta, we can take the inverse tangent or arctan of 0.75:
theta = arctan(0.75)
Evaluating this using a calculator, we find:
theta ≈ 36.87 degrees
Therefore, the plane should be headed approximately 36.87 degrees north of due east.
(b) To calculate the resultant speed of the airplane over the ground, we can use the Pythagorean theorem. The resultant speed is the magnitude of the resulting velocity.
Using the northward and eastward components, we can calculate the resultant velocity:
resultant velocity = sqrt((northward component)^2 + (eastward component)^2)
Substituting the values, we get:
resultant velocity = sqrt((240 km/h)^2 + (180 km/h)^2)
Calculating this using a calculator, we find:
resultant velocity ≈ 304 km/h
Therefore, the resultant speed of the airplane over the ground is approximately 304 km/h.
In summary, (a) the plane should be headed approximately 36.87 degrees north of due east, and (b) the resultant speed of the airplane over the ground is approximately 304 km/h.