A jet airliner moving initially at 369 mph
(with respect to the ground) to the east moves
into a region where the wind is blowing at
985 mph in a direction 69◦
north of east.
What is the new speed of the aircraft with
respect to the ground?
Answer in units of mph.
Please help me i tried so many times and i keep getting it wrong
To find the new speed of the aircraft with respect to the ground, we need to consider the velocities of both the aircraft and the wind.
Let's break down the given information:
1. The initial speed of the aircraft with respect to the ground is 369 mph to the east.
2. The wind is blowing at 985 mph in a direction 69 degrees north of east.
To find the new speed of the aircraft, we can use vector addition. We can represent the velocity of the aircraft as V_A and the velocity of the wind as V_W.
V_A = 369 mph to the east
V_W = 985 mph at an angle of 69 degrees north of east
To add these vectors, we need to resolve them into their respective horizontal and vertical components.
For V_A, the horizontal component is 369 mph (since it is moving east) and the vertical component is 0 mph (no vertical movement).
For V_W, we can find its horizontal and vertical components using trigonometry:
Horizontal component of V_W = 985 mph * cos(69°)
Vertical component of V_W = 985 mph * sin(69°)
Now we can add the horizontal components and the vertical components separately:
New horizontal component = 369 mph + (985 mph * cos(69°))
New vertical component = 0 mph + (985 mph * sin(69°))
To find the magnitude of the resultant velocity, we use the Pythagorean theorem:
New speed = sqrt((New horizontal component)^2 + (New vertical component)^2)
Calculating all the values:
Horizontal component of V_W = 985 mph * cos(69°) ≈ 369.83 mph
Vertical component of V_W = 985 mph * sin(69°) ≈ 941.57 mph
New horizontal component = 369 mph + 369.83 mph ≈ 738.83 mph
New vertical component = 0 mph + 941.57 mph ≈ 941.57 mph
New speed = sqrt((738.83 mph)^2 + (941.57 mph)^2) ≈ 1190.75 mph
Therefore, the new speed of the aircraft with respect to the ground is approximately 1190.75 mph.
To find the new speed of the aircraft with respect to the ground, we need to consider the vector addition of the airplane's velocity and the wind velocity. Let's break down the problem step by step:
Step 1: Resolve the velocities into their x and y components.
The initial velocity of the airplane can be resolved into its x and y components as follows:
Vx = V_initial * cos(θ)
Vy = V_initial * sin(θ)
Given:
V_initial = 369 mph
θ = 0 degrees (since the airplane is initially moving to the east)
So, Vx = 369 * cos(0°) = 369 mph (east)
And, Vy = 369 * sin(0°) = 0 mph (north)
Step 2: Resolve wind velocity into its x and y components.
The wind velocity can be resolved into its x and y components using the given direction:
Vwind_x = Vwind * cos(θwind)
Vwind_y = Vwind * sin(θwind)
Given:
Vwind = 985 mph
θwind = 69 degrees north of east
So, Vwind_x = 985 * cos(69°) = 985 * cos(π/180 * 69) ≈ 474.70 mph (east)
And, Vwind_y = 985 * sin(69°) = 985 * sin(π/180 * 69) ≈ 947.30 mph (north)
Step 3: Calculate the new velocity components of the airplane.
To find the new velocity components of the airplane, we need to add the x and y components of the initial velocity and wind velocity:
Vx_new = Vx + Vwind_x
Vy_new = Vy + Vwind_y
Vx_new = 369 + 474.70 ≈ 843.70 mph (east)
Vy_new = 0 + 947.30 ≈ 947.30 mph (north)
Step 4: Use the new velocity components to find the magnitude of the resultant velocity.
The magnitude of the resultant velocity can be found using the Pythagorean theorem:
V_new = sqrt(Vx_new^2 + Vy_new^2)
V_new = sqrt(843.70^2 + 947.30^2) ≈ 1266.1 mph
Therefore, the new speed of the aircraft with respect to the ground is approximately 1266.1 mph.
add the two vectors.
369E + 985(sin69deg N + cos69 E)
combine the like directions
E (369+353) + 989N
magnitude= sqrt(989^2 + 722^2)=....