An airplane pilot sets a compass course due west and maintains an airspeed of 225km/h . After flying for a time of 0.520h , she finds herself over a town a distance 122km west and a distance 15km south of her starting point.
Part A
Find the magnitude of the wind velocity.
v = km/h
Part B
Find the direction of the wind velocity.
Express your answer as an angle measured south of west
θ =∘ south of west
Part C
If the wind velocity is 42km/h due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 225km/h .
Express your answer as an angle measured north of west
θ = ∘ north of west
Part A:
To find the magnitude of the wind velocity, we can use the Pythagorean theorem.
Let's assume the wind velocity is denoted by v (km/h), and the horizontal component of the wind velocity (due west) is denoted by v_west, and the vertical component of the wind velocity (due south) is denoted by v_south.
v_west = 122km / 0.520h = 235.38 km/h
v_south = 15km / 0.520h = 28.85 km/h
Using the Pythagorean theorem:
v^2 = v_west^2 + v_south^2
v^2 = (235.38 km/h)^2 + (28.85 km/h)^2
v^2 ≈ 55253.64 km^2/h^2
v ≈ √(55253.64) km/h
v ≈ 235.00 km/h
Therefore, the magnitude of the wind velocity is approximately 235.00 km/h.
Part B:
To find the direction of the wind velocity (θ measured south of west), we can use trigonometry.
tan(θ) = v_south / v_west
θ = arctan(v_south / v_west)
θ = arctan(28.85 km/h / 235.38 km/h)
Using a calculator, we get:
θ ≈ 7.02° south of west
Therefore, the direction of the wind velocity is approximately 7.02° south of west.
Part C:
To find the direction the pilot should set her course to travel due west with a wind velocity of 42 km/h due south, we need to find the resultant wind velocity.
Let's denote the resultant wind velocity (due west) as v_resultant, and the vertical component of the wind velocity (due south) as v_south.
v_resultant = airspeed - v_south
v_resultant = 225 km/h - 42 km/h
v_resultant = 183 km/h
Now, let's find the angle (θ) measured north of west:
tan(θ) = v_south / v_resultant
θ = arctan(v_south / v_resultant)
θ = arctan(42 km/h / 183 km/h)
Using a calculator, we get:
θ ≈ 13.01° north of west
Therefore, to travel due west with a wind velocity of 42 km/h due south, the pilot should set her course approximately 13.01° north of west.
Part A:
To find the magnitude of the wind velocity, we need to consider the displacement of the plane caused by both its own airspeed and the wind velocity.
Let's start by finding the displacement caused by the plane's airspeed alone. The pilot has flown west for 0.520 hours at an airspeed of 225 km/h. The distance covered is equal to the product of the airspeed and time:
Distance (d) = Airspeed (v) × Time (t)
d = 225 km/h × 0.520 h
d ≈ 117 km (rounded to the nearest kilometer)
So, the plane would have covered a distance of approximately 117 km directly west without any influence from the wind.
Now we can determine the displacement caused by the wind. The displacement in the north-south direction is 15 km south, and in the east-west direction, it is 122 km west.
Since it has a westward component, we subtract the displacement caused by the plane's airspeed from the total westward displacement:
Wind displacement = Total westward displacement - Plane's westward displacement
Wind displacement = 122 km - 117 km
Wind displacement = 5 km west
Therefore, the magnitude of the wind velocity is 5 km/h.
Part B:
To find the direction of the wind velocity, we need to calculate the angle measured south of west.
We can use trigonometry to find this angle (θ). The tangent of the angle is equal to the ratio of the northward displacement to the westward displacement:
Tangent (θ) = (Southward displacement / Westward displacement)
Tangent (θ) = 15 km / 122 km (remember, we consider southward displacement as negative)
Now we can find θ by taking the inverse tangent (or arctan) of this ratio:
θ = arctan (15 km / 122 km)
θ ≈ 6.88°
Therefore, the direction of the wind velocity, measured south of west, is approximately 6.88°.
Part C:
Given that the wind velocity is 42 km/h due south, and the pilot wants to fly due west, we need to calculate the angle measured north of west at which the plane should set its course.
Let's find the angle (θ) using trigonometry again. The tangent of the angle is equal to the ratio of the northward displacement to the westward displacement (in this case, we assume the southward displacement to be zero):
Tangent (θ) = 0 / 225 km/h (since there is no southward displacement)
Now we can find θ by taking the inverse tangent (or arctan) of this ratio:
θ = arctan (0 / 225 km/h)
θ = 0°
Therefore, the pilot should set her course due west at an angle of 0°, or essentially straight west.
X = -122 km
Y = -15 km
Vw = d/t = 15/0.52 = 28.8 km/h
B. Tan Ar = Y/X = -15/-122 = 0.12295
Ar = 7.0o = Reference angle.
A = 7 + 180 = 187o = 7o S of W.
C. X = 225 km/h.
Y = 42 km/h.
Tan A = 42/225 = 0.18667
A = 10.57o S of W.
Pilot should head 10.57o N of W.