Your answer is a new experiment, since it's something that hasn't been tried before, or has, but resulted in with lots of errors.
Answer:

Explanation:
F = Force
C = Drag coefficient equal for both aircrafts
ρ = Density of air
A = Surface area equal for both aircrafts
v = Velocity



Dividing the above two equations we get

The ratio of the drag forces is 
they are heated to extremely high temperatures. they are cooled to extremely low temperatures. they are made to be very thick.
Answer:
6.78 km
Explanation:
Length of path due east = 4.3km
Length of path south = 2.48km
Unknown:
Distance covered = ?
Solution:
The distance covered is the total length of path from start to finish. It takes cognizance of the turns and every direction moved.
Unlike displacement which only considers the net direction from start to finish, distance sums up the total path.
So;
Distance = 4.3km + 2.48km = 6.78km
Answer:
v = 26.7 mph
Explanation:
During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:
Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi
Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:
Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi
So, the total displacement during all the trip wil be:
Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi
So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:
v = Δx / Δt = 160 mi / 6 h = 26.7 mph