In the figure, a radar station detects an airplane approaching directly from the east. At first observation, the airplane is at distance d1 = 350 m from the station and at angle θ1 = 31° above the horizon. The airplane is tracked through an angular change Δθ = 123° in the vertical east–west plane; its distance is then d2 = 770 m. Find the (a) magnitude and (b) direction of the airplane's displacement during this period. Give the direction as an angle relative to due west, with a positive angle being above the horizon and a negative angle being below the horizon.
To find the magnitude and direction of the airplane's displacement during this period, we can break it down into components and use trigonometry.
(a) Magnitude of Displacement:
We can find the magnitude of displacement using the two distances, d1 and d2. The change in distance Δd can be found by subtracting d1 from d2.
Δd = d2 - d1
Δd = 770 m - 350 m
Δd = 420 m
So, the magnitude of the airplane's displacement is 420 m.
(b) Direction of Displacement:
To find the direction, we need to consider the change in angle, Δθ. The direction will be relative to due west.
Since the airplane is initially approaching directly from the east, it will have an initial direction perpendicular to due west. The angle θ1 is given as 31° above the horizon, so the initial direction relative to due west is 90° + 31° = 121°.
The airplane then moves through an angular change of Δθ = 123° in the vertical east-west plane. We need to add this angular change to the initial direction to get the final direction.
Final direction = Initial direction + angular change
Final direction = 121° + 123°
Final direction = 244°
Therefore, the direction of the airplane's displacement during this period is 244° relative to due west.
To find the magnitude and direction of the airplane's displacement, we can break it down into horizontal and vertical components.
First, let's find the horizontal displacement:
The horizontal displacement is given by the difference in distances in the east-west direction, d2 - d1.
So, Δx = d2 - d1 = 770 m - 350 m = 420 m.
Next, let's find the vertical displacement:
The vertical displacement is determined by the change in angle, Δθ.
Using trigonometry, we can find the vertical component of the displacement, Δy = Δd * sin(Δθ).
Here, Δd is the change in the distance from the station, which is d2 - d1 = 770 m - 350 m = 420 m.
So, Δy = (d2 - d1) * sin(Δθ) = 420 m * sin(123°).
Now, we can calculate the magnitude of the displacement using the Pythagorean theorem:
Magnitude of displacement, Δs = √(Δx² + Δy²).
Plug in the values we calculated earlier:
Δs = √(420 m)² + (420 m * sin(123°))².
Finally, for the direction of the displacement, we can use the tangent function:
Direction, θ = tan⁻¹(Δy/Δx).
Substitute the values:
θ = tan⁻¹((420 m * sin(123°)) / 420 m).
To summarize:
(a) The magnitude of the airplane's displacement is given by √(420 m)² + (420 m * sin(123°))².
(b) The direction is given by tan⁻¹((420 m * sin(123°)) / 420 m).
what's the trouble? Viewing from the side, the plane has moved from
(350 cos31°,350 sin31°) to (770 sin154°,770cos154°)
That is, from
(300,180) to (-692,338)
Now you can crank it out, right?