The instruments with an aircraft show that it is headed east with an airspeed of 180 mph. After 2 hours, the aircraft is actually 375 miles east and 20 miles north of its starting point. Find the wind speed and its direction.

the plane flew 360 mi east in the air

but the air moved east and north
... this is the wind speed

15 mi E and 20 mi N forms a 3-4-5 triangle

the wind is ... 25 mi / 2 hr
... on a bearing N of E

the tangent of the bearing angle is
... 20 / 15

To find the wind speed and direction, we need to use the concept of groundspeed.

Let's assume that the aircraft is flying at a heading of east (90°) with an airspeed of 180 mph. The wind has two components, one along the east direction (headwind/tailwind) and the other perpendicular to the east direction (crosswind).

Let's denote the headwind/tailwind component as Wx and the crosswind component as Wy.

Given that the aircraft is heading east (90°) and after 2 hours it is 375 miles east, we can calculate the groundspeed using the Pythagorean theorem:

Groundspeed = √((375)^2 + (20)^2)
Groundspeed = √(140625 + 400)
Groundspeed = √140,825
Groundspeed ≈ 375.3 mph

We can formulate the relationship between groundspeed (GS), airspeed (AS), and wind components as follows:

GS = AS + Wx
0 = Wy

Since there is no crosswind component, Wy = 0.

Therefore,

GS = AS + Wx
375.3 mph = 180 mph + Wx
Wx = 375.3 mph - 180 mph
Wx = 195.3 mph

The wind speed, W, is the magnitude of Wx.

W = |Wx|
W = |195.3 mph|
W ≈ 195.3 mph

So, the wind speed is approximately 195.3 mph.

Now let's find the wind direction:

To find the wind direction, we can use trigonometry. We know that the crosswind component, Wy, is zero, and the headwind/tailwind component, Wx, is 195.3 mph.

The direction of the wind (θ) can be calculated using the inverse tangent function:

θ = tan^(-1)(Wy/Wx)

Since Wy = 0, we have:

θ = tan^(-1)(0/Wx)
θ = tan^(-1)(0)
θ = 0°

Therefore, the wind direction is 0°, which indicates the wind is coming from the east (opposite to the aircraft's heading).

In conclusion, the wind speed is approximately 195.3 mph, and the wind is coming from the east.

To find the wind speed and direction, we can use the concept of vectors. Let's consider the true airspeed of the aircraft as V_aircraft and the wind speed as V_wind.

Given information:
Aircraft airspeed (V_aircraft) = 180 mph
Time taken (t) = 2 hours
Distance traveled east (D_east) = 375 miles
Distance traveled north (D_north) = 20 miles

We can break down the total displacement of the aircraft into two components using vector addition:

- The eastward displacement component due to the aircraft's airspeed.
- The northward displacement component due to the wind's effect.

First, let's calculate the eastward displacement (D_east_aircraft) due to the aircraft's airspeed:

D_east_aircraft = V_aircraft * t
= 180 mph * 2 hours
= 360 miles east

Now, let's calculate the northward displacement (D_north_wind) due to the wind's effect:

D_north_wind = D_north - V_wind * t

We can find the time component of the wind's effect by dividing the northward displacement by the northward wind speed:

t = D_north / V_wind
= 20 miles / V_wind

Now, substituting the value of time t into the equation for northward displacement, we have:

D_north_wind = 20 miles - V_wind * (20 miles / V_wind)
= 20 miles - 20 miles
= 0 miles north

Since the aircraft has not traveled any distance north due to the wind's effect, we can deduce that the wind speed (V_wind) is 0 mph in the north direction.

Therefore, the wind speed is 0 mph, indicating that there is no wind, and its direction is not applicable (N/A) since wind speed is zero.