Question

An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. What is the rate of the plane in still air and what is the rate of the wind

Answer 1

Speed of the plane in still air:
`779\; {\rm km \cdot h^(-1)}`.

Windspeed:
`100\; {\rm km \cdot h^(-1)}`.

Explanation:

Assume that
`x\; {\rm km \cdot h^(-1)}` is the speed of the plane in still air, and that
`y\; {\rm km \cdot h^(-1)}` is the speed of the wind.

• When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be
  `(x - y)\; {\rm km \cdot h^(-1)}`.
• When this plane is travelling in the same direction as the wind, the ground speed of this plane would be
  `(x + y)\; {\rm km \cdot h^(-1)}`.

The question states that when going against the wind (
`v = (x - y)\; {\rm km \cdot h^(-1)}`,) the plane travels
`6111\; {\rm km}` in
`9\; {\rm h}`. Hence,
`9\, (x - y) = 6111`.

Similarly, since the plane travels
`7911\; {\rm km}` in
`9\; {\rm h}` when travelling in the same direction as the wind (
`v = (x + y)\; {\rm km \cdot h^(-1)}`,)
`9\, (x + y) = 7911`.

Add the two equations to eliminate
`y`. Subtract the second equation from the first to eliminate
`x`. Solve this system of equations for
`x` and
`y`:
`x = 779` and
`y = 100`.

Hence, the speed of this plane in still air would be
`779\; {\rm km \cdot h^(-1)}`, whereas the speed of the wind would be
`100\; {\rm km \cdot h^(-1)}`.