You are traveling on an airplane. The velocity of the plane with respect to the air is 140 m/s due east. The velocity of the air with respect to the ground is 34 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?
just add the vectors:
v = <140,0> + <-17,29.4> = <123,29.4>
so the speed is |v| = 126.47
the heading is 90-θ where tanθ=29.4/123
the eastward speed is |v|cosθ
1. 30o W. of N. = 90+30 = 120o CCW from +X-axis.
V = 140 + 34[120o].
V = 140+34*Cos120 + i34*sin120 = 123 + 29.4i = 126.5 m/s[13.5o]N. of E.
To solve this problem, we can use vector addition to find the resultant velocity of the plane with respect to the ground.
1) The speed of the plane with respect to the ground can be found by finding the magnitude of the resultant velocity vector. We can use the Pythagorean theorem to find this magnitude.
Magnitude of resultant velocity = √(velocity of the plane with respect to the air)^2 + (velocity of the air with respect to the ground)^2
= √(140 m/s)^2 + (34 m/s)^2
Solving this equation gives us:
Magnitude of resultant velocity = √19600 + 1156
= √20756
≈ 144 m/s
So, the speed of the plane with respect to the ground is approximately 144 m/s.
2) The heading of the plane with respect to the ground can be found by finding the angle between the resultant velocity vector and the north direction. We can use trigonometry to find this angle.
Angle = arctan((velocity of the air with respect to the ground) / (velocity of the plane with respect to the air))
= arctan(34 m/s / 140 m/s)
= arctan(0.2429)
Solving this equation gives us:
Angle ≈ 13.91°
Since the velocity of the air is 30° west of due north, we subtract this angle from 90° to get the heading.
Heading = 90° - 30°
= 60°
So, the heading of the plane with respect to the ground is 60°.
3) To find how far east the plane will travel in 1 hour, we can use the formula:
Distance = (speed of the plane with respect to the ground) * (time)
In this case:
Distance = (144 m/s) * (1 hour)
= 144 m/s * 3600 s
= 518400 m
So, the plane will travel approximately 518400 meters east in 1 hour.
To solve these problems, we can use vector addition to add the velocities of the plane with respect to the air and the velocity of the air with respect to the ground.
1) To find the speed of the plane with respect to the ground, we need to find the magnitude of the resultant velocity vector.
To do this, we can use the Pythagorean theorem.
Let's label the velocity of the plane with respect to the ground as vPG.
Using the Pythagorean theorem, we have:
vPG^2 = (vP)^2 + (vG)^2
Given:
vP = 140 m/s (velocity of the plane with respect to the air)
vG = 34 m/s (velocity of the air with respect to the ground)
Plugging in the values:
vPG^2 = (140 m/s)^2 + (34 m/s)^2
Calculating:
vPG^2 = 19600 m^2/s^2 + 1156 m^2/s^2
vPG^2 = 20756 m^2/s^2
Taking the square root of both sides to find vPG:
vPG ≈ 144.1 m/s
Therefore, the speed of the plane with respect to the ground is approximately 144.1 m/s.
2) To find the heading of the plane with respect to the ground, we need to find the angle it makes with due north.
We can use trigonometry to find this angle.
Let's label the angle between the resultant velocity vector and due north as θ.
Using trigonometry, we can find the tangent of θ:
tan(θ) = |vG| / |vP|
Given:
|vP| = 140 m/s
|vG| = 34 m/s
Plugging in the values:
tan(θ) = 34 m/s / 140 m/s
Calculating:
tan(θ) ≈ 0.243
Taking the inverse tangent (arctan) of both sides to find θ:
θ ≈ arctan(0.243)
Calculating:
θ ≈ 13.9°
Therefore, the heading of the plane with respect to the ground is approximately 13.9°.
3) To find how far east the plane will travel in 1 hour, we need to find the horizontal component of the velocity vector.
We can use trigonometry to find this component.
Let's label the horizontal component as dE (distance east).
Using trigonometry, we can find the cosine of the angle:
cos(θ) = |dE| / |vPG|
Given:
|vPG| = 144.1 m/s
θ ≈ 13.9°
Plugging in the values:
cos(13.9°) = |dE| / 144.1 m/s
Calculating:
|dE| ≈ 144.1 m/s * cos(13.9°)
Calculating:
|dE| ≈ 139.9 m/s
Therefore, the plane will travel approximately 139.9 meters east in 1 hour.