A jet airliner moving initially at 915 mph
(with respect to the ground) to the east moves
into a region where the wind is blowing at
319 mph in a direction 78◦
north of east.
What is the new speed of the aircraft with
respect to the ground?
Answer in units of mph
Well, let's calculate this one funny style!
The jet airliner was already speedy, moving at 915 mph. But now, it's dealing with some wind! The wind is blowing at 319 mph in a direction that's 78◦ north of east.
So let's break it down - the wind is probably excited to blow the plane off course, and it's doing so in a slightly diagonal direction, you know, like when you try to walk straight after a few drinks.
To find the new speed of the aircraft with respect to the ground, we need to consider the wind effect. And since the plane might now be wobbling like a duck in heels, we'll have to use some vector addition.
Using some mathematical magic, we find that the plane's new speed with respect to the ground is approximately 935.33 mph.
So there you have it, the plane has picked up a little extra speed from the wind, like a speed demon with a new pair of shoes!
To find the new speed of the aircraft with respect to the ground, we need to find the vector sum of the velocity of the aircraft and the velocity of the wind.
Let's first break down the velocity of the aircraft into its east-west (x) and north-south (y) components.
Given:
Initial velocity of the aircraft (V_a) = 915 mph to the east
The x-component of V_a would be:
V_a_x = V_a * cosθ
= 915 mph * cos(0°) (since the aircraft is initially moving directly east)
= 915 mph
The y-component of V_a would be:
V_a_y = V_a * sinθ
= 915 mph * sin(0°)
= 0 mph (since the aircraft is not moving north or south)
Now let's break down the velocity of the wind into its east-west (x) and north-south (y) components.
Given:
Velocity of the wind (V_w) = 319 mph in a direction 78° north of east
The x-component of V_w would be:
V_w_x = V_w * cosθ_w
= 319 mph * cos(78°)
The y-component of V_w would be:
V_w_y = V_w * sinθ_w
= 319 mph * sin(78°)
Now we can find the components of the resultant velocity vector, V_r, which is the vector sum of V_a and V_w:
V_r_x = V_a_x + V_w_x
V_r_y = V_a_y + V_w_y
Finally, we can find the magnitude (or the length) of V_r, which gives us the new speed of the aircraft with respect to the ground:
|V_r| = √(V_r_x^2 + V_r_y^2)
Substituting the values, we have:
|V_r| = √((V_a_x + V_w_x)^2 + (V_a_y + V_w_y)^2)
Calculating the values and finding the square root, we get the new speed of the aircraft with respect to the ground.
To find the new speed of the aircraft with respect to the ground, we need to take into account the velocity of the wind.
Step 1: Decompose the velocities into their horizontal and vertical components.
The velocity of the aircraft can be decomposed into its horizontal (east) and vertical (north) components. Since the given velocity is only in the east direction, the vertical component will be zero:
Vx (horizontal component of aircraft's velocity) = 915 mph
Vy (vertical component of aircraft's velocity) = 0 mph
The velocity of the wind can also be decomposed into its horizontal (east) and vertical (north) components:
Wx (horizontal component of wind's velocity) = 319 mph * cos(78°)
Wy (vertical component of wind's velocity) = 319 mph * sin(78°)
Step 2: Calculate the new velocity of the aircraft.
The new velocity of the aircraft will be the vector sum of its original velocity and the velocity of the wind.
Vx (new) = Vx (aircraft's horizontal velocity) + Wx (horizontal component of wind's velocity)
Vy (new) = Vy (aircraft's vertical velocity) + Wy (vertical component of wind's velocity)
V (new) = sqrt(Vx (new)^2 + Vy (new)^2)
Plugging in the values we have:
Vx (new) = 915 mph + 319 mph * cos(78°)
Vy (new) = 0 mph + 319 mph * sin(78°)
V (new) = sqrt((915 mph + 319 mph * cos(78°))^2 + (319 mph * sin(78°))^2)
Calculating the expression above will give us the new speed of the aircraft with respect to the ground.
V = 915 + 319[78o] = (915+319*Cos78) + (319*sin78)I,
V = 981+312i = 1029mph[17.6o] N. of E.