Question

Two airplanes leave an airport at the same. The velocity of the first airplane is 660m/h at a heading of 52.1 degrees. The velocity of the second is 640m/h at a heading of 123 degrees. How far apart are they after 2.5h?

Answer 1

To find the distance between the two airplanes after 2.5 hours, we can use the formula for displacement, find the horizontal and vertical components of each airplane's velocity using trigonometry, and then calculate their respective displacements and use the distance formula.

Step-by-step explanation:

To find the distance between the two airplanes after 2.5 hours, we can use the formula for displacement:

Displacement = velocity × time

First, we need to find the horizontal and vertical components of each airplane's velocity. To do this, we can use trigonometry:

Plane 1 horizontal velocity = 660m/h × cos(52.1°)

Plane 1 vertical velocity = 660m/h × sin(52.1°)

Plane 2 horizontal velocity = 640m/h × cos(123°)

Plane 2 vertical velocity = 640m/h × sin(123°)

Once we have the horizontal and vertical components for each airplane, we can calculate their respective displacements by multiplying the components by the time (2.5 hours). Finally, we can use the distance formula to find the distance between the two airplanes:

Distance = √[(Plane 2 horizontal displacement - Plane 1 horizontal displacement)^2 + (Plane 2 vertical displacement - Plane 1 vertical displacement)^2]