Question

You are on an airplane traveling 30° south of due west at 140 m/s with respect to the air. The air is moving with a speed 31 m/s with respect to the ground due north. 1) What is the speed of the plane with respect to the ground? 126.764 m/s 2) What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due easth).

Answer 1

Answer:

1) Vp = 127.36 m/s

2) According to the reference given: 252.17°

Explanation:

For the first part, we know that

$V_{p/a} = V_{p} - V_{a}$

$V_{p} = V_{p/a} + V_{a} =(-121.24,-70)+(0,31)=(-121.24,-39)m/s$

So, the speed of the plane respect to ground is:

Vp = 127.36 m/s

For the second part, we calculate the angle:

$\alpha = 17.83°$  on the third quadrant. According to the reference given, we need to make 270° - α, so:

The final angle is 252.17°

Answer 2

The given velocity of the plane and air can be resolved as vectors from

which the resultant velocity of the plane can be calculated.

Correct response:

• The speed of the airplane is approximately 127.36 m/s

• The heading of the plane is approximately 252.17°

Method used to find the magnitude and direction of the plane

Given parameters are;

Direction airplane is traveling = 30° South of due west

Speed of airplane = 140 m/s

Speed of the air = 31 m/s

Direction of the air = Due north

1) The speed of the plane with respect to the ground is given as follows:

The velocities in component form are;

$\vec{v}_{plane}$ = -140×cos(30°)·i - 140 × sin(30°)·j

Which gives;

$\vec {v}_{plane}$ = -70·√3·i - 70·j

$\vec{v}_{air}$ = 31·j

The resultant velocity of the plane, $\vec {v}$ = -70·√3·i - 39·j

The magnitude of the speed, $| \vec {v} |$ = $\sqrt{(-70 \cdot \sqrt{3} )^2 + (-39)^2}$ ≈ 127.36

• The speed of the plane with respect to the ground, $| \vec {v} |$ ≈ 127.36 m/s

$\theta = \arctan \left(\dfrac{-39}{-70 \cdot \sqrt{3} } \right) \approx 17.83^{\circ}$

The direction of the airplane south of west, θ ≈ 17.83°

Therefore;

• The direction of the airplane relative to the north, in the clockwise direction is ≈ 270° - 17.83° = 252.17°

Learn more about vectors here:

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