Question

A passenger plane flew from Texas to Frankfurt and back. The round trip is 9900 km.

Answer 1

We want to find the average speed of the plane in still air, we will get that the average speed of the plane in still air is 355.3 km/hr

So we know that the total trip is 9900km, so from Texas to Frankfurt the distance is half of that:

9900km/2 = 4950km

Now, The speed of the plane in still air is x km/hr

The speed of the air is 50 km/hr, and it flows from Frankfurt to Texas.

So in the first part of the trip, the wind blows in opposite direction to the motion of the plane, thus the speed of the plane will be:

(x - 50) km/hr

While in the return trip the wind blows in favor of the plane, in this case, the speed is:

(x + 50)km/hr

Then if the first part of the trip lasts for T hours (the second for T - 4 hours) we can write the system of equations:

(x - 50)*T = 4950

(x + 50)*(T - 4) = 4950

Where I removed the units so it is easier to read.

If we isolate T in the first equation we get:

T = 4950/(x - 50)

Now we can replace this in the other equation to get:

(x + 50)*(4950/(x - 50)  - 4) = 4950

If we multiply both sides by (x - 50) we get:

(x + 50)*4950 - 4*(x + 50)*(x - 50) = 4950*(x - 50)

Now we can solve this for x:

(x + 50)*4950 - 4*(x^2 - 2500) - 4950*(x - 50) = 0

100*4950 - 4*(x^2 - 2500)  = 0

495000 - 4*x^2 + 10000 = 0

505000 = 4*x^2

√(505000/4) = x = 355.3

The average speed of the plane in still air is 355.3 km/hr

If you want to learn more about systems of equations, you can read:

brainly.com/question/13476446