After flying for 15 min in a wind blowing 42 km/h at an angle of 20° south of east, an airplane pilot is over a town that is 42 km due north of the starting point. What is the speed of the airplane relative to the air?
Well, if the airplane pilot is over a town that is 42 km due north of the starting point after flying for 15 minutes, I'm not sure if we should be talking about the speed of the airplane or the speed of Santa's reindeer. It sounds like he's on an unexpected detour! But if we want to calculate the speed of the airplane relative to the air, we can use a bit of trigonometry.
First, let me remind you that trigonometry is like trying to put together a jigsaw puzzle while wearing a blindfold. It's a real challenge, but it can be fun if you're a masochist.
To find the speed of the airplane relative to the air, we'll need to break down the velocity into its horizontal and vertical components. The horizontal component will be the airplane's speed relative to the air, and the vertical component will be due to the wind.
Now, we know that the wind is blowing at 42 km/h at an angle of 20° south of east. That's like a strong breeze trying to mess up your hair while you're taking a selfie. To find the vertical component of the velocity, we can use some trigonometry magic. Take the sine of the angle (20°) and multiply it by the wind speed (42 km/h). That should give you the vertical component of the velocity.
To find the speed of the airplane relative to the air, we can use the Pythagorean theorem. Take the square root of the sum of the squares of the horizontal and vertical components. And voila, you've calculated the speed of the airplane relative to the air.
But remember, my friend, life is like an airplane ride. Sometimes you may be blown off course, but it's all part of the adventure. So embrace the wind, laugh in the face of confusion, and enjoy the journey!
To find the speed of the airplane relative to the air, we need to break down the given information and use vector addition.
We know that the airplane has flown for 15 minutes (or 0.25 hours) and that the wind is blowing at a speed of 42 km/h at an angle of 20° south of east. Let's call the airplane's speed relative to the ground "x" and the angle it flies relative to east "θ".
Now, let's break down the information:
1. The airplane flies for 15 minutes, or 0.25 hours. So, the distance it travels is distance = speed × time = x × 0.25.
2. The wind is blowing at a speed of 42 km/h. Since the direction angle is south of east, we need to subtract 180° from 20° to get the wind's angle relative to east, which is 160°. Let's call the wind's speed "w" in km/h and its angle "ϕ" in degrees.
3. The horizontal component of the wind's speed is given by w × cos(ϕ) and the vertical component is given by w × sin(ϕ).
Now, we can use vector addition:
1. The horizontal component of the airplane's speed relative to the air is x × cos(θ).
2. The vertical component of the airplane's speed relative to the air is x × sin(θ).
Let's equate the horizontal and vertical components:
Horizontal component:
x × cos(θ) = x + w × cos(ϕ)
Vertical component:
x × sin(θ) = w × sin(ϕ)
Using the given information, we can now solve for the speed of the airplane relative to the air.
From the vertical component equation, we have:
x × sin(θ) = w × sin(ϕ)
Since the airplane is flying north, θ (angle relative to east) is 90°.
So, we have:
x × sin(90°) = w × sin(ϕ)
x = w × sin(ϕ)
Plugging in the values we know:
x = 42 km/h × sin(160°)
Using a scientific calculator or trigonometric table, we find that sin(160°) is approximately -0.342.
So,
x = 42 km/h × (-0.342)
x ≈ -14.364 km/h
Therefore, the speed of the airplane relative to the air is approximately 14.364 km/h.