The distance between the two airplanes after 3.6 h is 4,356.85 m.
Displacement is the change in the position of an object. The distance between the two planes after the given time can be determined by finding the displacement of the two planes.
The resultant velocity of the two planes is calculated as follows.
Net horizontal speed of the planes;
Total horizontal speed = 572.64 m/h + 32.45 m/h = 605.1 m/h
Net vertical speed of the planes
Total horizontal speed = 468.7 m/h + 619.15 m/h = 1,087.85 m/h
v = √(1,087.85² + 605.1²)
v = 1,244.81 m/h
d = 1,244.81 m/h x 3.5 h
d = 4,356.85 m
Thus, the distance between the two airplanes after 3.6 h is 4,356.85 m.
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Answer:
1962.6 m (nearest tenth)
Explanation:
[tex]\boxed{\sf Distance=Velocity \times Time}[/tex]
First airplane
Second airplane
Draw a diagram with the calculated distances and given angles (see attachment).
To find how far apart the planes are, connect the ends of the line segments to create a triangle and use the cosine rule to find x.
Cosine rule
[tex]\boxed{c^2=a^2+b^2-2ab \cos C}[/tex]
[tex]\textsf{Where $a, b$ and $c$ are the sides and $C$ is the angle opposite side $c$}.[/tex]
From inspection of the diagram (attached):
Substitute the values into the formula and solve for x:
[tex]\begin{aligned}\implies x^2 & =2170^2+2590^2-2(2170)(2590) \cos 47.7^{\circ}\\x^2 & =4708900+6708100-11240600\cos 47.7^{\circ}\\x^2 & =11417000-11240600\cos 47.7^{\circ}\\x^2 & =3851935.541...\\x & =\sqrt{3851935.541...}\\x & = 1962.634846...\end{aligned}[/tex]
Therefore, the airplanes are 1962.6 m apart after 3.5 hours (nearest tenth).
