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, [tex]d_1 = 1\times610=610[/tex] miles.
The distance covered at 310 degrees, [tex]d_2 = 3\times610=1830[/tex] miles.
Total distance covered, d, is the magnitude of the sum of vectors [tex]\vec{d_1}[/tex] and [tex]\vec{d_2}[/tex] as shown in the figure.
The angle between the vectors [tex]\vec{d_1}[/tex] and [tex]\vec{d_2}[/tex], [tex]\theta=310-270=40[/tex] degree.
Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},
[tex]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})}[/tex]
[tex]\Rightarrow d=2330.51[/tex] miles
Hence, at 1:00 a.m, the airplane is at a distance of 2330.51 miles from the airport.
