You are traveling on an airplane. The velocity of the plane with respect to the air is 180 m/s due east. The velocity of the air with respect to the ground is 40 m/s at an angle of 30° west of due north.
1)What is the speed of the plane with respect to the ground?
2)What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due east)?
3)How far east will the plane travel in 1 hour?
Last answer is wrong, but thanks for the first two.
1. Vpa = 180m/s. @ 0 deg.
Vag = 40m/s @ 120 deg,CCW.
Vpg = Vpa + Vag,
Vpg = (180 + 40cos120) + i40sin120,
Vpg = 160 + i34.64,
Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
2. tanA = Y / X = 34.64 / 160 = 0.2165,
A = 12.2 deg,CCW. = 12.2deg. North of
East.
3. 1 hr = 3600s.
d = Vt = 163.7m/s * 3600s = 589,320m.
For the distance you'll have to make a right triangle and solve for the missing side (x). Since the plane isn't traveling directly in the east direction it wont be exactly the m/s multiplied by one hour.
To solve these questions, we can use vector addition to combine the velocities of the plane with respect to the air and the velocity of the air with respect to the ground.
Let's label the given velocities:
Velocity of the plane with respect to the air = 180 m/s due east
Velocity of the air with respect to the ground = 40 m/s at an angle of 30° west of due north.
1) To find the speed of the plane with respect to the ground, we need to find the magnitude of the resultant velocity vector combining the plane's velocity and the wind's velocity.
To do this, we can consider the horizontal and vertical components of the wind's velocity.
Horizontal component = 40 m/s * sin(30°) = 20 m/s
Vertical component = 40 m/s * cos(30°) = 34.64 m/s
Now, we can add the horizontal component of the plane's velocity to the horizontal component of the wind's velocity:
180 m/s + 20 m/s = 200 m/s
Similarly, we can add the vertical component of the plane's velocity to the vertical component of the wind's velocity:
34.64 m/s + 0 m/s = 34.64 m/s
Now, we can find the magnitude of the resultant velocity vector using the Pythagorean theorem:
Speed = √(200 m/s)^2 + (34.64 m/s)^2 ≈ 204.63 m/s
Therefore, the speed of the plane with respect to the ground is approximately 204.63 m/s.
2) To find the heading of the plane with respect to the ground, we need to find the direction of the resultant velocity vector. We can use trigonometry to determine this.
The angle between the horizontal component of the resultant velocity (due east) and due north can be found as:
θ = tan^(-1)(vertical component / horizontal component)
θ = tan^(-1)(34.64 m/s / 200 m/s)
θ ≈ 9.88°
Since we defined 0° as due north, the heading of the plane with respect to the ground would be:
Heading = 90° - 9.88° ≈ 80.12°
Therefore, the heading of the plane with respect to the ground is approximately 80.12°.
3) To find how far east the plane will travel in 1 hour, we can use the speed of the plane with respect to the ground found in the first question.
Distance traveled = Speed * time
Distance traveled = 204.63 m/s * (1 hour * 3600 seconds/hour)
Distance traveled = 204.63 m/s * 3600 seconds
Now, we can calculate the distance:
Distance traveled = 737,868 meters
Therefore, the plane will travel approximately 737,868 meters east in 1 hour.