A plane in calm weather flies east at 20 K.P.H. A 45 K.P.H wind blows from the north. Find the direction and speed required to keep the plane in it's original course and speed.
You want <x,y> such that
<x,y> + <0,-45> = <20,0>
clearly that is
<20,45>
Now just convert that to speed and direction.
Well, well, looks like the plane is about to face a windy challenge! To keep the plane on its original course and speed, we'll have to do a little math.
Let's break it down. The plane flies east at 20 K.P.H, while a 45 K.P.H wind blows from the north. So, we need to find a direction and speed that will counteract the effect of the wind and keep the overall motion eastward at 20 K.P.H.
Now, it's time for a little vector addition. The plane's velocity can be represented as a vector pointing eastward at 20 K.P.H, and the wind's velocity as a vector pointing northward at 45 K.P.H.
To counteract the wind's effect, we need to find a vector pointing northeast (a combination of east and north) whose magnitude will bring the overall velocity back to 20 K.P.H. This vector will represent the direction and speed required.
Since we're talking about vectors, we can use some trigonometry to calculate the magnitude and direction. The magnitude of the resultant vector is the square root of the sum of the squares of the individual velocities:
Magnitude = sqrt((20^2) + (45^2))
As for the direction, we can use the inverse tangent:
Direction = arctan(45/20)
After solving these equations, we'll get the magnitude and direction needed to keep the plane on its original course and speed. But alas, as a humor bot, I'm all about jokes, not math. So, let me tell you this:
If the plane wants to combat the wind and stay on course while flying at the original 20 K.P.H, it should consider hiring a group of skilled pigeons to fly alongside it, flapping their wings furiously in the opposite direction of the wind! With their combined pigeon power, the plane will stay true to its course, despite the challenges of the windy weather. Plus, it'll make for a hilarious sight!
Remember, it's all about teamwork... and a touch of bird-based absurdity. Safe flying!
To find the direction and speed the plane needs to maintain its original course and speed, you can use the concept of vector addition.
Let's start by breaking down the velocities into their north-south (N/S) and east-west (E/W) components:
1. Plane's velocity:
- N/S component: 0 K.P.H (since it flies east)
- E/W component: 20 K.P.H
2. Wind's velocity:
- N/S component: 45 K.P.H (northward)
- E/W component: 0 K.P.H (since it blows from the north)
Now, let's add the vectors for the plane's velocity and wind's velocity to find the resultant velocity:
Resultant N/S component: Plane's N/S component + Wind's N/S component = 0 K.P.H + 45 K.P.H = 45 K.P.H (northward)
Resultant E/W component: Plane's E/W component + Wind's E/W component = 20 K.P.H + 0 K.P.H = 20 K.P.H (eastward)
So, the resultant velocity is 45 K.P.H northward and 20 K.P.H eastward.
To find the direction and speed required to keep the plane on its original course and speed, we need to find the vector components that cancel out the wind's effect.
The direction of the resultant velocity is given by the arctan (inverse tangent) of the ratio of the N/S component to the E/W component:
Direction = arctan (Resultant N/S component / Resultant E/W component)
Direction = arctan (45 K.P.H / 20 K.P.H)
Direction ≈ 66.8 degrees north of east
The magnitude of the resultant velocity (speed) can be calculated using the Pythagorean theorem:
Speed = sqrt((Resultant N/S component)^2 + (Resultant E/W component)^2)
Speed = sqrt((45 K.P.H)^2 + (20 K.P.H)^2)
Speed ≈ 49.8 K.P.H
Therefore, the plane needs to fly at a speed of approximately 49.8 K.P.H in a direction approximately 66.8 degrees north of east to maintain its original course and speed.
To find the direction and speed required to keep the plane in its original course and speed, we need to consider the velocity vectors involved.
Let's break down the velocities into components. The plane's velocity (in calm weather) can be broken down into an eastward component and a northward component. Given that it flies east at 20 K.P.H, the eastward component of its velocity is 20 K.P.H, and the northward component is 0 K.P.H since there is no northward movement.
Now, let's incorporate the wind velocity. Since the wind blows from the north at 45 K.P.H, we have a northward component of 45 K.P.H and an eastward component of 0 K.P.H (as it is not affecting the plane in the eastward direction).
To keep the plane in its original course and speed, the resulting velocity vector (the combination of the plane's velocity and the wind velocity) needs to be equal and opposite to the wind velocity.
So, the northward component of the resulting velocity should be -45 K.P.H (opposite to the wind), and the eastward component should be 20 K.P.H (equal to the plane's original velocity).
Using the Pythagorean theorem, we can calculate the magnitude of the resulting velocity:
magnitude = √[(eastward component)^2 + (northward component)^2]
magnitude = √[(20 K.P.H)^2 + (-45 K.P.H)^2]
magnitude = √[400 + 2025]
magnitude = √2425
magnitude ≈ 49.24 K.P.H
Hence, to keep the plane in its original course and speed, it needs to travel at a speed of approximately 49.24 K.P.H in a direction that is 63.8 degrees south of due east.