Question

A model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s

What is the airplanes resultant speed (magnitude of vector)

What is the airplanes heading (direction of velocity vector

Answer 1

Using the Pythagorean theorem, the resultant speed of the airplane is approximately 19.21 m/s. To determine the airplane's heading, the arctangent function is used, resulting in a direction of approximately 38.66 degrees east of north.

Step-by-step explanation:

To find the resultant speed (magnitude of velocity vector) of the airplane, we must consider the horizontal and vertical components of velocity as a right triangle where the northward velocity (vertical) and eastward wind (horizontal) are perpendicular to each other. The Pythagorean theorem is applied here for this purpose.

The northward velocity of the airplane is 15 m/s, while the eastward wind velocity is 12 m/s. Using the Pythagorean theorem:

Resultant speed = √(15² + 12²) = √(225 + 144) = √(369) ≈ 19.21 m/s

To find the direction of the velocity vector (airplane's heading), we use the tangent function since we have the opposite (eastward wind) and adjacent (northward flight) sides:

θ = tan^{-1}(opposite/adjacent) = tan^{-1}(12/15) ≈ 38.66 degrees
The angle is measured east from the north, thus the airplane's heading is approximately 38.66 degrees east of north.

Answer 2

1.) 19.21 m/s

2.) 57 degree

Explanation: Given that a model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s

The airplanes resultant speed can be calculated by using pythagorean theorem

R = sqrt ( 15^2 + 12^2 )

R = sqrt( 225 + 144 )

R = sqrt( 369 )

R = 19.21 m/s

The magnitude of the resultant vector is 19.21 m/s

The direction of velocity vector will be:

Tan Ø = 15 /12

Tan Ø = 1.25

Ø = tan^-1(1.25)

Ø = 57.04

Ø = 57 degree.

Therefore, the airplanes is heading 57 degree in the horizontal direction of velocity vector