Question

Two airplanes leave an airport at the same time. The velocity of the first airplane is 690 m/h at a heading of 10.1◦. The velocity of the second is 560 m/h at a heading of 159◦. How far apart are they after 2.8 h?

Answer in units of m.

Answer 1

To find the separation distance between two airplanes after 2.8 hours with different velocities and headings, calculate the distances they travel, decompose them into x and y components, and then use the Pythagorean theorem to get the total distance apart.

Step-by-step explanation:

To determine how far apart two airplanes are after 2.8 hours when one airplane leaves an airport at a velocity of 690 m/h at a heading of 10.1° and the second airplane at a velocity of 560 m/h at a heading of 159°, we can use vector analysis. First, we convert their speeds into distances traveled by multiplying their speeds by the time, which is 2.8 hours. Then, we decompose these distances into x (east-west) and y (north-south) components using trigonometric functions.

Plane 1: Distance = 690 m/h * 2.8 h = 1932 m

x-component: 1932 m * cos(10.1°)

y-component: 1932 m * sin(10.1°)

Plane 2: Distance = 560 m/h * 2.8 h = 1568 m

x-component: 1568 m * cos(159°) (Since it is west of north, we take the cosine of (180° - 159°))

y-component: 1568 m * sin(159°)

Next, we calculate the differences in x and y components between the two planes and use the Pythagorean theorem to find the total separation distance.

Separation distance = √[(x1 - x2)^2 + (y1 - y2)^2]