Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

NEED HELP DUE AT 11:59
Two airplanes leave an airport at the same

time. The velocity of the first airplane is

740 m/h at a heading of 39.3

◦

. The velocity

of the second is 620 m/h at a heading of 87◦

.

How far apart are they after 3.5 h?

Answer in units of m.

Sagot :

onyebuchinnaji

The distance between the two airplanes after 3.6 h is 4,356.85 m.

What is displacement?

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.

Resultant velocity of the two planes

The resultant velocity of the two planes is calculated as follows.

Net horizontal speed of the planes;

  • first plane = 740 m/h x cos(39.3) = 572.64 m/h
  • second plane = 620 m/h x cos(87) = 32.45 m/h

Total horizontal speed = 572.64 m/h + 32.45 m/h = 605.1 m/h

Net vertical speed of the planes

  • first plane = 740 m/h x sin(39.3) = 468.7m/h
  • second plane = 620 m/h x sin(87) = 619.15 m/h

Total horizontal speed = 468.7 m/h + 619.15 m/h = 1,087.85 m/h

Resultant velocity of the planes

v = √(1,087.85² + 605.1²)

v = 1,244.81 m/h

Distance between the two planes

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.

Learn more about resultant velocity here: https://brainly.com/question/24767211

#SPJ1

semsee45

Answer:

1962.6 m (nearest tenth)

Explanation:

[tex]\boxed{\sf Distance=Velocity \times Time}[/tex]

First airplane

  • Velocity = 740 m/h
  • Time = 3.5 h
  • Distance = 740 × 3.5 = 2590 m
  • Angle = 39.3°

Second airplane

  • Velocity = 620 m/h
  • Time = 3.5 h
  • Distance = 620 × 3.5 = 2170 m
  • Angle = 87°

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):

  • a = 2170
  • b = 2590
  • c = x
  • C = 87° - 39.3° = 47.7°

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).

View image semsee45
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.