Two airplanes leave an airport at the same time. the velocity of the first airplane is 740 m/h at a heading of 25.3◦. the velocity of the second is 570 m/h at a heading of 82◦.
how far apart are they after 1.5 h? answer in units of m.

They are 959.70 m apart after 1.5 h

Explanation:

Lets explain how to solve the problem

The given is:

→ The velocity of the 1st airplane is 740 m/h at a heading of 25.3°

→ The velocity of 2nd airplane is 570 m/h at a heading 82°

→ We need to find how far apart they are after 1.5 h

At first lets find the distance of each one after 1.5 h

→ d = v × t

→ = 740 × 1.5 = 1110 m

→ = 570 × 1.5 = 855 m

Assume that these two distance are two side of a triangle.

The angle between the two sides is the difference between their

heading.

The heading of the 1st airplane is 25.3° and the heading of the second

airplane is 82°

The angle between their distances = 82 - 25.3 = 56.7°

The angle between the two sides of the triangle is 56.7°

Lets use cosine rule to find the 3rd side of the triangle

→

→ = 1110 , = 855 , α = 56.7

Substitute these values in the rule

→ m

d represents the distance between the two airplanes after 1.5 h

They are 959.70 m apart after 1.5 h

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm