Respuesta :
Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity
[tex]v_{12} = v_1 - v_2[/tex]
here
speed of first plane is 700 mi/h at 31.3 degree
[tex]v_1 = 700 cos31.3\hat i + 700 sin31.3\hat j[/tex]
[tex]v_1 = 598.12\hat i + 363.7\hat j[/tex]
speed of second plane is 570 mi/h at 134 degree
[tex]v_2 = 570 cos134 \hat i + 570 sin134 \hat j[/tex]
[tex]v_2 = -396\hat i + 410\hat j[/tex]
now the relative velocity is given as
[tex]v_{12} = (598.12 + 396)\hat i + (363.7 - 410)\hat j[/tex]
[tex]v_{12} =994.12\hat i -46.3 \hat j[/tex]
now the distance between them is given as
[tex]d = v* t[/tex]
[tex]d = (994.12 \hat i - 46.3 \hat j)* 3[/tex]
[tex]d = 2982.36\hat i - 138.9\hat j [/tex]
so the magnitude of the distance is given as
[tex]d = \sqrt{2982.36^2 + 138.9^2}[/tex]
[tex]d = 2985.6[/tex] miles
so the distance between them is 2985.6 miles
Answer:
Planes are 2985.48 miles far after 3 hours.
Explanation:
Let right represents positive x- axis and up represent positive y - axis. Horizontal component is i and vertical component is j.
We have velocity of the first airplane is 700 m/h at an angle 31.3° to the horizontal.
So Velocity of first plane = 700 cos 31.3 i + 700 sin 31.3 j
= (598.12 i + 363.66 j) m/h
We also have velocity of the second airplane is 570 m/h at an angle 134° to the horizontal.
Velocity of second plane = 570 cos 134 i + 570 sin 134 j
= (-395.96 i + 410.02 j) m/h
Displacement of first plane after 3 hours = Velocity * Time
= (598.12 i + 363.66 j) *3
= (1794.36 i + 1090.98 j) mi
Displacement of second plane after 3 hours = (-395.96 i + 410.02 j)*3
= (-1187.88 i + 1230.06 j) mi
Displacement vector between two planes = (1794.36 i + 1090.98 j) - (-1187.88 i + 1230.06 j)
= (2982.24 i - 139.08 j) mi
Magnitude of displacement = [tex]\sqrt{2982.24^2+(- 139.08)^2} =2985.48 mi[/tex]
So planes are 2985.48 miles far after 3 hours.
