The piolet of a single-engine airplane flies at a constant speed of 200 km/h at a bearing of N 25 degree E. There is a 40 km/h crosswind blowing southeast (S 45 degress E). What are the plane's actual speed and direction? Round the speed to the nearest tenth and the directiuon to the nearest degree.
Do this one just like your hiker problem. For each part, figure the x- and y-distances, add them up, then at the end find the angle whose tangent is the final y/x value.
If X = 2 inches, Y = 4 inches, and Z = 4 inches, what is the surface area of the box?
190.1 km/h
54 degrees, or N 36 degrees E
To find the plane's actual speed and direction, we can use vector addition.
Step 1: Convert the given wind direction and speed into vector form.
The wind is blowing in the southeast direction (S 45 degrees E) with a speed of 40 km/h.
To convert this into vector form, we break it down into its north and east components.
North Component = Wind Speed * cos(Southeast Angle)
East Component = Wind Speed * sin(Southeast Angle)
North Component = 40 km/h * cos(45 degrees)
North Component ≈ 40 km/h * 0.7071
North Component ≈ 28.284 km/h
East Component = 40 km/h * sin(45 degrees)
East Component ≈ 40 km/h * 0.7071
East Component ≈ 28.284 km/h
So, the wind vector is 28.284 km/h to the north and 28.284 km/h to the east.
Step 2: Convert the given aircraft speed and bearing into vector form.
The plane is flying at a constant speed of 200 km/h at a bearing of N 25 degrees E.
To convert this into vector form, we break it down into its north and east components.
North Component = Aircraft Speed * cos(Bearing Angle)
East Component = Aircraft Speed * sin(Bearing Angle)
North Component = 200 km/h * cos(25 degrees)
North Component ≈ 200 km/h * 0.9063
North Component ≈ 181.3 km/h
East Component = 200 km/h * sin(25 degrees)
East Component ≈ 200 km/h * 0.4226
East Component ≈ 84.5 km/h
So, the aircraft vector is 181.3 km/h to the north and 84.5 km/h to the east.
Step 3: Add the aircraft vector and wind vector together.
To calculate the resulting vector, you simply add the north and east components separately.
Resultant North Component = Aircraft North Component + Wind North Component
Resultant East Component = Aircraft East Component + Wind East Component
Resultant North Component = 181.3 km/h + 28.284 km/h
Resultant North Component ≈ 209.58 km/h
Resultant East Component = 84.5 km/h + 28.284 km/h
Resultant East Component ≈ 112.784 km/h
So, the resultant vector is 209.58 km/h to the north and 112.784 km/h to the east.
Step 4: Find the magnitude and direction of the resultant vector.
To find the magnitude, we use the Pythagorean theorem:
Resultant Magnitude = sqrt(Resultant North Component^2 + Resultant East Component^2)
Resultant Magnitude = sqrt(209.58 km/h^2 + 112.784 km/h^2)
Resultant Magnitude ≈ sqrt(43950 km^2/h^2 + 12712.04656 km^2/h^2)
Resultant Magnitude ≈ sqrt(56662.04656 km^2/h^2)
Resultant Magnitude ≈ 238.1 km/h (rounded to the nearest tenth)
To find the direction, we use the inverse tangent function:
Resultant Direction = arctan(Resultant East Component / Resultant North Component) - 180 degrees
Resultant Direction = arctan(112.784 km/h / 209.58 km/h) - 180 degrees
Resultant Direction ≈ arctan(0.5384) - 180 degrees
Resultant Direction ≈ 28 degrees (rounded to the nearest degree)
Therefore, the plane's actual speed is approximately 238.1 km/h (rounded to the nearest tenth), and its direction is approximately 28 degrees (rounded to the nearest degree).