Question

An airplane leaves an airport at 9:00 p.M. With a heading of 270 and a speed of 610 mph. At 10:00 p.M. The pilot changes the heading to 310. Determine how far the plane is from the airport at 1:00 a.M.

Answer 1

2330.51 miles

Explanation:

Given that the speed of the airplane = 610 mph.

The airplane leaves an airport at 9:00 P.M. with a heading of 270 degrees and at 10:00 P.M., the pilot changes the heading to 310 degrees.

So, for 1 hour the plane is heading at 270 degrees and for 3 hours, from 10:00 p.m to 1:00 a.m, the plane is heading at 310 degrees as shown in the figure.

As, distance = time x speed, so

The distance covered at 270 degrees,
`d_1 = 1*610=610` miles.

The distance covered at 310 degrees,
`d_2 = 3*610=1830` miles.

Total distance covered, d, is the magnitude of the sum of vectors
`\vec{d_1}` and
`\vec{d_2}` as shown in the figure.

The angle between the vectors
`\vec{d_1}` and
`\vec{d_2}`,
`\theta=310-270=40` degree.

Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},

`d= \sqrt{|\vec{d_1}|^2+|\vec{d_2}|^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\theta} \\\\\Rightarrow d=\sqrt{610^2+1830^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\(40^(\circ))}`

`\Rightarrow d=2330.51` miles

Hence, at 1:00 a.m, the airplane is at a distance of 2330.51 miles from the airport.