An airplane flies due north at 200 km/h relative to the air. There is a wind blowing at 70 km/h to the northeast relative to the ground. What are the plane's speed and direction relative to the ground?
v=200 km/h, v₀= 70 km/h,
Angle that make v and v₀ is 45° =>
An angle α in cosine law is
α (360°-2•45°)/2 =135°
Cosine law:
V=sqrt(v²+v₀²-2vv₀cos α)=
=sqrt[40000+4900-2•200•70•(-0.707)]=254 km/h
Projections:
x: v•cos90 +v₀cos45° =254cosβ
y:vsin90+ v₀sin45°=254sinβ
0.707•70 =254 cosβ
200+0.707•70 = 254 sinβ
49.49 =254 cosβ
249.49 =254 sinβ
sinβ/cosβ=tanβ =249.49/49.49 = 5.04
β =tan⁻¹5.04 = 78.8° (northwest)
To find the plane's speed and direction relative to the ground, we can use vector addition.
1. First, let's break down the velocity of the plane and the wind into their respective horizontal and vertical components. We'll take the positive x-axis to the east and the positive y-axis to the north.
The plane's velocity relative to the air (Vpa) is purely northward and has no eastward component since it is flying due north. Therefore, its horizontal component Vpxa = 0 km/h and its vertical component Vpya = 200 km/h.
The wind's velocity relative to the ground (Vwg) is going to the northeast, which can be split into a northward and an eastward component. Using the given information, we can calculate these components.
The angle between the northerly wind component and the ground is the complement of the angle between the wind and the northeast direction, which is 45 degrees.
So, the wind's northward component (Vwn) = Vwg cos(45°) = 70 km/h * cos(45°) = 49.5 km/h, and the eastward component (Vwe) = Vwg sin(45°) = 70 km/h * sin(45°) = 49.5 km/h.
2. Now, let's add the horizontal and vertical components of the plane's velocity relative to the air to the corresponding components of the wind's velocity relative to the ground.
The horizontal component of the plane's velocity relative to the ground (Vpxg) is the sum of the plane's horizontal component and the wind's eastward component. Therefore, Vpxg = Vpxa + Vwe = 0 km/h + 49.5 km/h = 49.5 km/h.
The vertical component of the plane's velocity relative to the ground (Vpyg) is the sum of the plane's vertical component and the wind's northward component. Therefore, Vpyg = Vpya + Vwn = 200 km/h + 49.5 km/h = 249.5 km/h.
3. Finally, we can use these components of velocity to calculate the magnitude (speed) and direction of the plane's velocity relative to the ground.
The speed of the plane relative to the ground (Vpg) can be found using the Pythagorean theorem:
Vpg = √(Vpxg² + Vpyg²) = √(49.5 km/h)² + (249.5 km/h)²) ≈ 254.2 km/h.
The direction of the plane's velocity relative to the ground (θ) can be found using the inverse tangent function:
θ = tan⁻¹(Vpyg / Vpxg) = tan⁻¹(249.5 km/h / 49.5 km/h) ≈ 78.6°.
Therefore, the plane's speed relative to the ground is approximately 254.2 km/h, and its direction is approximately 78.6° north of east.
To find the plane's speed and direction relative to the ground, we can use vector addition.
Step 1: Split the velocity of the airplane into its northward and eastward components. Since the airplane is flying due north, the northward component of its velocity is 200 km/h, and the eastward component is 0 km/h.
Step 2: Split the velocity of the wind into its northward and eastward components. Since the wind is blowing to the northeast, we can use trigonometry to find the components. The magnitude of the wind's velocity is 70 km/h, so the northward component is 70*cos(45°) km/h and the eastward component is 70*sin(45°) km/h.
Step 3: Add the northward components and the eastward components separately.
Northward component: 200 km/h + 70*cos(45°) km/h = 200 km/h + 70*(√2/2) km/h ≈ 247.43 km/h
Eastward component: 0 km/h + 70*sin(45°) km/h = 70 km/h*(√2/2) km/h ≈ 49.49 km/h
Step 4: Use the Pythagorean theorem to find the magnitude of the resultant velocity (speed).
Resultant velocity magnitude = √[(northward component)^2 + (eastward component)^2]
≈ √[(247.43 km/h)^2 + (49.49 km/h)^2]
≈ √[61224.9149 km²/h² + 2449.9601 km²/h²]
≈ √[63674.8750 km²/h²]
≈ 252.32 km/h
Step 5: Find the direction of the resultant velocity using trigonometry.
Direction = arctan(eastward component / northward component)
= arctan(49.49 km/h / 247.43 km/h)
≈ arctan(0.20)
≈ 11.31°
Therefore, the plane's speed relative to the ground is approximately 252.32 km/h, and its direction relative to the ground is approximately 11.31° north of east.