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

Answer:

2330.51 miles

Step-by-step 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, $d_1 = 1\times610=610$ miles.

The distance covered at 310 degrees, $d_2 = 3\times610=1830$ 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.